# Klein-Gordon Feynman propagator

Michael E. Peskin & Daniel V. Schroeder, An Introduction to Quantum Field Theory, (Perseus Books, 1995) – Chapter 2.

Although we’ve already gone through a derivation of the Feynman propagator for the Klein-Gordon field (see here and subsequent posts, listed as pingbacks at the bottom of that post), it’s worth revisiting it here to review the derivation in P&S, which is quite a bit shorter now that we have a few tools ready to deal with it.

In our original derivation of the Green’s function for the Klein-Gordon equation we defined the function

$\displaystyle D_{R}\left(x-y\right)=\frac{1}{\left(2\pi\right)^{3}}\int d^{3}p\int dp^{0}\left(\frac{-1}{2\pi i}\right)\frac{e^{-ip\left(x-y\right)}}{p^{2}-m^{2}} \ \ \ \ \ (1)$

The ${p^{0}}$ integral is done in the complex ${p^{0}}$ plane as a contour integral where the contour runs along the real axis from ${-\infty}$ to ${\infty}$, skirting the poles at ${p^{0}=\pm\left(\mathbf{p}^{2}+m^{2}\right)=\pm E_{\mathbf{p}}}$ with little semicircular arcs that go above the poles. For ${x^{0}>y^{0}}$, the contour is closed with a large semicircle in the lower half plane so that the contour encloses both poles, with the result that ${D_{R}\left(x-y\right)}$ is non-zero (and the contour is clockwise, which cancels the ${-1}$ in the integrand). For ${x^{0}, the contour is closed with a large semicircle in the upper half plane so that the contour excludes both poles, with the result that ${D_{R}\left(x-y\right)}$ is zero.

We can choose 3 other ways of skirting the two poles: we could use semicircles that go under both poles, or over the pole at ${-E_{\mathbf{p}}}$ and under the pole at ${+E_{\mathbf{p}}}$, or vice versa. The last choice (under ${-E_{\mathbf{p}}}$ and over ${+E_{\mathbf{p}}}$) gives the Feynman propagator. In this case, if ${x^{0}>y^{0}}$ we close the contour using a large semicircle in the lower half plane, which excludes the ${-E_{\mathbf{p}}}$ pole and includes the ${+E_{\mathbf{p}}}$ pole. If ${x^{0}, we close the contour in the upper half plane, which includes the ${-E_{\mathbf{p}}}$ pole and excludes the ${+E_{\mathbf{p}}}$ pole. In either case, the integral includes only one pole, and the result is a propagator that we met when discussing causality.

$\displaystyle D\left(x-y\right)=\left\langle 0\left|\phi\left(x\right)\phi\left(y\right)\right|0\right\rangle =\frac{1}{\left(2\pi\right)^{3}}\int d^{3}p\;\left.\frac{e^{-ip\left(x-y\right)}}{2E_{\mathbf{p}}}\right|_{p^{0}=E_{\mathbf{p}}} \ \ \ \ \ (2)$

Take ${x^{0}>y^{0}}$ first. Then the contour is in the lower half plane, and is clockwise. The residue at ${p^{0}=+E_{\mathbf{p}}}$ is

$\displaystyle \mbox{Res}\left(E_{\mathbf{p}}\right)=\left.\frac{e^{-ip\left(x-y\right)}}{2E_{\mathbf{p}}}\right|_{p^{0}=E_{\mathbf{p}}} \ \ \ \ \ (3)$

The integral is the same as 1, but with a different contour, so we’ll call it ${D_{F}}$. Doing the ${p^{0}}$ integral around this contour gives

 $\displaystyle D_{F}\left(x-y\right)$ $\displaystyle =$ $\displaystyle \frac{1}{\left(2\pi\right)^{3}}\int d^{3}p\int dp^{0}\left(\frac{-1}{2\pi i}\right)\frac{e^{-ip\left(x-y\right)}}{p^{2}-m^{2}}\ \ \ \ \ (4)$ $\displaystyle$ $\displaystyle =$ $\displaystyle \frac{1}{\left(2\pi\right)^{3}}\int d^{3}p\left(\frac{-1}{2\pi i}\right)\left(-2\pi i\right)\mbox{Res}\left(E_{\mathbf{p}}\right)\ \ \ \ \ (5)$ $\displaystyle$ $\displaystyle =$ $\displaystyle \frac{1}{\left(2\pi\right)^{3}}\int d^{3}p\;\left.\frac{e^{-ip\left(x-y\right)}}{2E_{\mathbf{p}}}\right|_{p^{0}=E_{\mathbf{p}}}\ \ \ \ \ (6)$ $\displaystyle$ $\displaystyle =$ $\displaystyle D\left(x-y\right) \ \ \ \ \ (7)$

For ${y^{0}>x^{0}}$, the contour is in the upper half plane and is now counterclockwise and the residue is at ${p^{0}=-E_{\mathbf{p}}}$:

$\displaystyle \mbox{Res}\left(-E_{\mathbf{p}}\right)=\left.-\frac{e^{-ip\left(x-y\right)}}{2E_{\mathbf{p}}}\right|_{p^{0}=-E_{\mathbf{p}}} \ \ \ \ \ (8)$

The integral to be done is now

 $\displaystyle D_{F}\left(x-y\right)$ $\displaystyle =$ $\displaystyle \frac{1}{\left(2\pi\right)^{3}}\int d^{3}p\int dp^{0}\left(\frac{-1}{2\pi i}\right)\frac{e^{-ip\left(x-y\right)}}{p^{2}-m^{2}}\ \ \ \ \ (9)$ $\displaystyle$ $\displaystyle =$ $\displaystyle \frac{1}{\left(2\pi\right)^{3}}\int d^{3}p\left(\frac{-1}{2\pi i}\right)\left(2\pi i\right)\mbox{Res}\left(-E_{\mathbf{p}}\right)\ \ \ \ \ (10)$ $\displaystyle$ $\displaystyle =$ $\displaystyle \frac{1}{\left(2\pi\right)^{3}}\int d^{3}p\;\left.\frac{e^{-ip\left(x-y\right)}}{2E_{\mathbf{p}}}\right|_{p^{0}=-E_{\mathbf{p}}} \ \ \ \ \ (11)$

Note that we now multiply the residue by ${+2\pi i}$ because the contour is now counterclockwise. We can now write for the exponent

$\displaystyle -ip^{0}\left(x^{0}-y^{0}\right)=iE_{\mathbf{p}}\left(x^{0}-y^{0}\right)=-iE_{\mathbf{p}}\left(y^{0}-x^{0}\right) \ \ \ \ \ (12)$

Since the spatial part of the exponent is integrated over all ${\mathbf{p}}$, we can replace ${\mathbf{p}}$ by ${\mathbf{-p}}$, or equivalently, ${\mathbf{x-y}}$ by ${\mathbf{y-x}}$, to get

 $\displaystyle D_{F}\left(x-y\right)$ $\displaystyle =$ $\displaystyle \frac{1}{\left(2\pi\right)^{3}}\int d^{3}p\;\frac{e^{-ip\left(y-x\right)}}{2E_{\mathbf{p}}}\ \ \ \ \ (13)$ $\displaystyle$ $\displaystyle =$ $\displaystyle D\left(y-x\right) \ \ \ \ \ (14)$

Combining the two results 7 and 14 with step functions, we get

 $\displaystyle D_{F}\left(x-y\right)$ $\displaystyle =$ $\displaystyle \theta\left(x^{0}-y^{0}\right)\left\langle 0\left|\phi\left(x\right)\phi\left(y\right)\right|0\right\rangle +\theta\left(y^{0}-x^{0}\right)\left\langle 0\left|\phi\left(y\right)\phi\left(x\right)\right|0\right\rangle \ \ \ \ \ (15)$ $\displaystyle$ $\displaystyle =$ $\displaystyle \left\langle 0\left|T\phi\left(x\right)\phi\left(y\right)\right|0\right\rangle \ \ \ \ \ (16)$

where ${T}$ is the time-ordering symbol which places the field with the later time first.

The Feynman propagator 15 is still a Green’s function for the Klein-Gordon operator, as we can show by following through the same steps we did earlier for ${D_{R}}$. Applying the Klein-Gordon operator to the first term in 15 we get (remember that all derivatives are with respect to ${x}$, not ${y}$):

 $\displaystyle \left(\partial^{2}+m^{2}\right)\theta\left(x^{0}-y^{0}\right)\left\langle 0\left|\phi\left(x\right)\phi\left(y\right)\right|0\right\rangle$ $\displaystyle =$ $\displaystyle -\delta\left(x^{0}-y^{0}\right)\left\langle 0\left|\pi\left(x\right)\phi\left(y\right)\right|0\right\rangle +\ \ \ \ \ (17)$ $\displaystyle$ $\displaystyle$ $\displaystyle 2\delta\left(x^{0}-y^{0}\right)\left\langle 0\left|\pi\left(x\right)\phi\left(y\right)\right|0\right\rangle +0\ \ \ \ \ (18)$ $\displaystyle$ $\displaystyle =$ $\displaystyle \delta\left(x^{0}-y^{0}\right)\left\langle 0\left|\pi\left(x\right)\phi\left(y\right)\right|0\right\rangle \ \ \ \ \ (19)$

Doing the same to the second term gives the same result with opposite signs on the delta functions because

$\displaystyle \frac{d\theta\left(y^{0}-x^{0}\right)}{dx^{0}}=-\frac{d\theta\left(x^{0}-y^{0}\right)}{dx^{0}}=-\delta\left(x^{0}-y^{0}\right) \ \ \ \ \ (20)$

Thus we get

 $\displaystyle \left(\partial^{2}+m^{2}\right)\theta\left(y^{0}-x^{0}\right)\left\langle 0\left|\phi\left(y\right)\phi\left(x\right)\right|0\right\rangle$ $\displaystyle =$ $\displaystyle \delta\left(x^{0}-y^{0}\right)\left\langle 0\left|\phi\left(y\right)\pi\left(x\right)\right|0\right\rangle +\ \ \ \ \ (21)$ $\displaystyle$ $\displaystyle$ $\displaystyle -2\delta\left(x^{0}-y^{0}\right)\left\langle 0\left|\phi\left(y\right)\pi\left(x\right)\right|0\right\rangle +0\ \ \ \ \ (22)$ $\displaystyle$ $\displaystyle =$ $\displaystyle -\delta\left(x^{0}-y^{0}\right)\left\langle 0\left|\phi\left(y\right)\pi\left(x\right)\right|0\right\rangle \ \ \ \ \ (23)$

Combining the two gives

 $\displaystyle \left(\partial^{2}+m^{2}\right)D_{F}\left(x-y\right)$ $\displaystyle =$ $\displaystyle \delta\left(x^{0}-y^{0}\right)\left\langle 0\left|\left[\pi\left(x\right),\phi\left(y\right)\right]\right|0\right\rangle \ \ \ \ \ (24)$ $\displaystyle$ $\displaystyle =$ $\displaystyle -i\delta^{\left(4\right)}\left(x-y\right) \ \ \ \ \ (25)$

which is the same as the result we got for ${D_{R}}$.

# Klein-Gordon Green’s function as a Fourier transform

Michael E. Peskin & Daniel V. Schroeder, An Introduction to Quantum Field Theory, (Perseus Books, 1995) – Chapter 2.

We saw that the function

$\displaystyle D_{R}\left(x-y\right)=\theta\left(x^{0}-y^{0}\right)\left\langle 0\left|\left[\phi\left(x\right),\phi\left(y\right)\right]\right|0\right\rangle \ \ \ \ \ (1)$

where ${\theta\left(x^{0}-y^{0}\right)}$ is the Heaviside step function, is a Green’s function for the Klein-Gordon operator ${\partial^{2}+m^{2}}$, in the sense that

$\displaystyle \left(\partial^{2}+m^{2}\right)D_{R}\left(x-y\right)=-i\delta^{\left(4\right)}\left(x-y\right) \ \ \ \ \ (2)$

We also saw that ${D_{R}}$ can be written as an integral

$\displaystyle D_{R}\left(x-y\right)=\frac{\theta\left(x^{0}-y^{0}\right)}{\left(2\pi\right)^{3}}\int d^{3}p\int dp^{0}\left(\frac{-1}{2\pi i}\right)\frac{e^{-ip\left(x-y\right)}}{p^{2}-m^{2}} \ \ \ \ \ (3)$

where the ${p^{0}}$ integral is done as a contour integral in the complex ${p^{0}}$ plane, in which the contour runs along the real axis with small semicircular arcs going round the poles at ${p^{0}=\pm E_{\mathbf{p}}=\pm\left(\mathbf{p}^{2}+m^{2}\right)}$, and a large semicircular arc in the lower half plane that tends to infinity (over which the integral goes to zero). This integral form was obtained by substituting the explict integral form for the fields into the commutator in 1.

However, the derivation of the Green’s function 2 from the definition 1 doesn’t require the explicit integral form of the fields, and it turns out that 3 can actually be derived from 2 by using a Fourier transform. Suppose we write

$\displaystyle D_{R}\left(x-y\right)=\frac{1}{\left(2\pi\right)^{4}}\int d^{4}p\;e^{-ip\left(x-y\right)}\tilde{D}_{R}\left(p\right) \ \ \ \ \ (4)$

where ${\tilde{D}_{R}\left(p\right)}$ is the Fourier transform of ${D_{R}\left(x-y\right)}$ in 4-momentum space. [This is the standard definition of a Fourier transform.] Now apply the Klein-Gordon operator to this equation

 $\displaystyle \left(\partial^{2}+m^{2}\right)D_{R}\left(x-y\right)$ $\displaystyle =$ $\displaystyle \frac{1}{\left(2\pi\right)^{4}}\int d^{4}p\;\left(\partial^{2}+m^{2}\right)e^{-ip\left(x-y\right)}\tilde{D}_{R}\left(p\right)\ \ \ \ \ (5)$ $\displaystyle$ $\displaystyle =$ $\displaystyle \frac{1}{\left(2\pi\right)^{4}}\int d^{4}p\;\left(-p^{2}+m^{2}\right)e^{-ip\left(x-y\right)}\tilde{D}_{R}\left(p\right) \ \ \ \ \ (6)$

From 2, this must equal ${-i\delta^{\left(4\right)}\left(x-y\right)}$, and since one definition of the 4-d delta function is

$\displaystyle \delta^{\left(4\right)}\left(x-y\right)=\frac{1}{\left(2\pi\right)^{4}}\int d^{4}p\;e^{-ip\left(x-y\right)} \ \ \ \ \ (7)$

we must have

 $\displaystyle \left(-p^{2}+m^{2}\right)\tilde{D}_{R}\left(p\right)$ $\displaystyle =$ $\displaystyle -i\ \ \ \ \ (8)$ $\displaystyle \tilde{D}_{R}\left(p\right)$ $\displaystyle =$ $\displaystyle \frac{i}{p^{2}-m^{2}} \ \ \ \ \ (9)$

From this we get

$\displaystyle D_{R}\left(x-y\right)=\frac{i}{\left(2\pi\right)^{4}}\int\frac{d^{4}p}{\left(p^{2}-m^{2}\right)}\;e^{-ip\left(x-y\right)} \ \ \ \ \ (10)$

which is the same as 3. [In all these calculations, we’ve assumed implicitly that ${x^{0}>y^{0}}$ so the ${\theta\left(x^{0}-y^{0}\right)}$ is implied.]

# Green’s function for Klein-Gordon equation

Michael E. Peskin & Daniel V. Schroeder, An Introduction to Quantum Field Theory, (Perseus Books, 1995) – Chapter 2.

We saw that for the real Klein-Gordon field ${\phi\left(x\right)}$, the commutator ${\left[\phi\left(x\right),\phi\left(y\right)\right]}$ if the spacetime interval ${x-y}$ is spacelike, indicating that a measurement at event ${x}$ cannot influence a measurement at event ${y}$. This commutator has the form

$\displaystyle \left[\phi\left(x\right),\phi\left(y\right)\right]=\frac{1}{\left(2\pi\right)^{3}}\int\frac{d^{3}p}{2E_{\mathbf{p}}}\left[e^{-ip\left(x-y\right)}-e^{ip\left(x-y\right)}\right] \ \ \ \ \ (1)$

When evaluated, this integral is a numerical function of ${x}$ and ${y}$ which are, in field theory, just labels for points in spacetime; that is, they are not operators. Therefore ${\left[\phi\left(x\right),\phi\left(y\right)\right]}$ is a numerical quantity (P&S call it a ‘c-number’, which is a somewhat obsolete term for a numerical quantity, as opposed to an operator quantity which can be called a ‘q-number’). As such, when applied to a state such as the vacuum state ${\left|0\right\rangle }$, it just multiplies the state by a numerical quantity. Since ${\left\langle 0\left|0\right.\right\rangle =1}$ by definition, we can write

 $\displaystyle \left[\phi\left(x\right),\phi\left(y\right)\right]$ $\displaystyle =$ $\displaystyle \left\langle 0\left|\left[\phi\left(x\right),\phi\left(y\right)\right]\right|0\right\rangle \ \ \ \ \ (2)$ $\displaystyle$ $\displaystyle =$ $\displaystyle \frac{1}{\left(2\pi\right)^{3}}\int\frac{d^{3}p}{2E_{\mathbf{p}}}\left[e^{-ip\left(x-y\right)}-e^{ip\left(x-y\right)}\right] \ \ \ \ \ (3)$

The ${p^{0}}$ component in both exponents is ${p^{0}=E_{\mathbf{p}}}$. Since we’re integrating over all ${\mathbf{p}}$ we can change the integration variable from ${\mathbf{p}}$ to ${-\mathbf{p}}$ in the second term. The integral then becomes

$\displaystyle \left[\phi\left(x\right),\phi\left(y\right)\right]=\frac{1}{\left(2\pi\right)^{3}}\int d^{3}p\left[\left.\frac{1}{2E_{\mathbf{p}}}e^{-ip\left(x-y\right)}\right|_{p^{0}=E_{\mathbf{p}}}+\left.\frac{1}{-2E_{\mathbf{p}}}e^{-ip\left(x-y\right)}\right|_{p^{0}=-E_{\mathbf{p}}}\right] \ \ \ \ \ (4)$

We can convert this to a 4-d integral by integrating over a contour in the ${p^{0}}$ plane. If we use the integrand

$\displaystyle \frac{e^{-ip\left(x-y\right)}}{p^{2}-m^{2}}=\frac{e^{-ip\left(x-y\right)}}{\left(p^{0}\right)^{2}-\left(\mathbf{p}^{2}+m^{2}\right)}=\frac{e^{-ip\left(x-y\right)}}{\left(p^{0}\right)^{2}-E_{\mathbf{p}}^{2}} \ \ \ \ \ (5)$

we observe that it has poles at ${p^{0}=\pm E_{\mathbf{p}}}$, and the residues at these poles are

 $\displaystyle \mbox{Res}\left(+E_{\mathbf{p}}\right)$ $\displaystyle =$ $\displaystyle \left.\frac{1}{2E_{\mathbf{p}}}e^{-ip\left(x-y\right)}\right|_{p^{0}=E_{\mathbf{p}}}\ \ \ \ \ (6)$ $\displaystyle \mbox{Res}\left(-E_{\mathbf{p}}\right)$ $\displaystyle =$ $\displaystyle \left.-\frac{1}{2E_{\mathbf{p}}}e^{-ip\left(x-y\right)}\right|_{p^{0}=-E_{\mathbf{p}}} \ \ \ \ \ (7)$

For ${x^{0}>y^{0}}$, the exponential ${e^{-ip^{0}\left(x^{0}-y^{0}\right)}}$ tends to zero around a semicircular contour for ${p^{0}}$ in the lower half-plane. Therefore if we use a contour which runs along the real axis from left to right and loops around the two poles with a small semicircular arc that goes above each pole, and then close the contour with an infinite semicircle in the lower half plane, we can apply the residue theorem to get, for ${x^{0}>y^{0}}$

$\displaystyle \left[\phi\left(x\right),\phi\left(y\right)\right]=\frac{1}{\left(2\pi\right)^{3}}\int d^{3}p\int dp^{0}\left(\frac{-1}{2\pi i}\right)\frac{e^{-ip\left(x-y\right)}}{p^{2}-m^{2}} \ \ \ \ \ (8)$

[The ${-1}$ inside the integral is because the contour we’re using goes clockwise around the poles rather than counterclockwise as specified in the residue theorem.]

For ${x^{0}, if we use the same contour along the real ${p^{0}}$ axis but a semicircular arc in the upper half-plane, then ${e^{-ip^{0}\left(x^{0}-y^{0}\right)}}$ tends to zero on this upper arc. However, this contour now excludes both poles so the integral is zero for ${x^{0}. We can combine these two results using the step function ${\theta\left(x^{0}-y^{0}\right)\equiv1}$ for ${x^{0}>y^{0}}$ and 0 for ${x^{0}:

$\displaystyle D_{R}\left(x-y\right)=\theta\left(x^{0}-y^{0}\right)\left\langle 0\left|\left[\phi\left(x\right),\phi\left(y\right)\right]\right|0\right\rangle \ \ \ \ \ (9)$

${D_{R}}$ turns out to be a Green’s function for the Klein-Gordon operator ${\partial^{2}+m^{2}}$. To see this, we can apply the operator directly to ${D_{R}}$ and see what happens (the derivatives are all with respect to ${x}$, so ${y}$ is being held constant here). Since ${\theta\left(x^{0}-y^{0}\right)}$ is a function of ${x^{0}}$ only we are dealing with a derivative of the form

 $\displaystyle \left(\partial^{2}+m^{2}\right)f\left(x^{0}\right)g\left(x\right)$ $\displaystyle =$ $\displaystyle \partial_{\mu}\partial^{\mu}\left[f\left(x^{0}\right)g\left(x\right)\right]+m^{2}f\left(x^{0}\right)g\left(x\right)\ \ \ \ \ (10)$ $\displaystyle$ $\displaystyle =$ $\displaystyle \partial_{0}\left(g\left(x\right)\partial^{0}f\left(x^{0}\right)\right)+\partial_{\mu}\left(f\left(x^{0}\right)\partial^{\mu}g\left(x\right)\right)+m^{2}f\left(x^{0}\right)g\left(x\right)\ \ \ \ \ (11)$ $\displaystyle$ $\displaystyle =$ $\displaystyle g\left(x\right)\partial_{0}\partial^{0}f\left(x^{0}\right)+2\partial_{0}f\left(x^{0}\right)\partial^{0}g\left(x\right)+f\left(x^{0}\right)\left(\partial_{\mu}\partial^{\mu}+m^{2}\right)g\left(x\right) \ \ \ \ \ (12)$

In this case,

 $\displaystyle f\left(x^{0}\right)$ $\displaystyle =$ $\displaystyle \theta\left(x^{0}-y^{0}\right)\ \ \ \ \ (13)$ $\displaystyle g\left(x\right)$ $\displaystyle =$ $\displaystyle \left\langle 0\left|\left[\phi\left(x\right),\phi\left(y\right)\right]\right|0\right\rangle \ \ \ \ \ (14)$

and we’re trying to calculate

$\displaystyle \left(\partial^{2}+m^{2}\right)f\left(x^{0}\right)g\left(x\right)=\left(\partial^{2}+m^{2}\right)D_{R}\left(x-y\right) \ \ \ \ \ (15)$

The derivative of the step function is a delta function

$\displaystyle \partial^{0}\theta\left(x^{0}-y^{0}\right)=\delta\left(x^{0}-y^{0}\right) \ \ \ \ \ (16)$

and from the definition of the conjugate momentum ${\pi=\dot{\phi}}$:

 $\displaystyle \partial^{0}\left\langle 0\left|\left[\phi\left(x\right),\phi\left(y\right)\right]\right|0\right\rangle$ $\displaystyle =$ $\displaystyle \left\langle 0\left|\left[\dot{\phi}\left(x\right),\phi\left(y\right)\right]\right|0\right\rangle \ \ \ \ \ (17)$ $\displaystyle$ $\displaystyle =$ $\displaystyle \left\langle 0\left|\left[\pi\left(x\right),\phi\left(y\right)\right]\right|0\right\rangle \ \ \ \ \ (18)$

To take the second derivative, we need the derivative of the delta function. In particular

 $\displaystyle \phi\left(x\right)\partial^{0}\delta\left(x^{0}-y^{0}\right)$ $\displaystyle =$ $\displaystyle -\dot{\phi}\left(x\right)\delta\left(x^{0}-y^{0}\right)\ \ \ \ \ (19)$ $\displaystyle$ $\displaystyle =$ $\displaystyle -\pi\left(x\right)\delta\left(x^{0}-y^{0}\right) \ \ \ \ \ (20)$

Thus the first term in 12 comes out to

$\displaystyle g\left(x\right)\partial_{0}\partial^{0}f\left(x^{0}\right)=-\delta\left(x^{0}-y^{0}\right)\left\langle 0\left|\left[\pi\left(x\right),\phi\left(y\right)\right]\right|0\right\rangle \ \ \ \ \ (21)$

The second term in 12 is

$\displaystyle 2\partial_{0}f\left(x^{0}\right)\partial^{0}g\left(x\right)=2\delta\left(x^{0}-y^{0}\right)\left\langle 0\left|\left[\pi\left(x\right),\phi\left(y\right)\right]\right|0\right\rangle \ \ \ \ \ (22)$

The last term in 12 is zero, because ${\phi\left(x\right)}$ satisfies the Klein-Gordon equation

$\displaystyle \left(\partial_{\mu}\partial^{\mu}+m^{2}\right)\left\langle 0\left|\left[\phi\left(x\right),\phi\left(y\right)\right]\right|0\right\rangle =\left\langle 0\left|\left[\left(\partial_{\mu}\partial^{\mu}+m^{2}\right)\phi\left(x\right),\phi\left(y\right)\right]\right|0\right\rangle =0 \ \ \ \ \ (23)$

We therefore get

$\displaystyle \left(\partial^{2}+m^{2}\right)D_{R}\left(x-y\right)=\delta\left(x^{0}-y^{0}\right)\left\langle 0\left|\left[\pi\left(x\right),\phi\left(y\right)\right]\right|0\right\rangle \ \ \ \ \ (24)$

Using the commutation relation

$\displaystyle \left[\pi\left(x\right),\phi\left(y\right)\right]=-i\delta^{\left(3\right)}\left(\mathbf{x-y}\right) \ \ \ \ \ (25)$

we get the final form

 $\displaystyle \left(\partial^{2}+m^{2}\right)D_{R}\left(x-y\right)$ $\displaystyle =$ $\displaystyle -i\delta^{\left(3\right)}\left(\mathbf{x-y}\right)\delta\left(x^{0}-y^{0}\right)\ \ \ \ \ (26)$ $\displaystyle$ $\displaystyle =$ $\displaystyle -i\delta^{\left(4\right)}\left(x-y\right) \ \ \ \ \ (27)$

Thus apart from the ${-i}$, ${D_{R}\left(x-y\right)}$ is the Green’s function for the Klein-Gordon operator.

# Causality in the Klein-Gordon field: commutators and measurements

Michael E. Peskin & Daniel V. Schroeder, An Introduction to Quantum Field Theory, (Perseus Books, 1995) – Chapter 2.

We investigated causality in the Klein-Gordon field by calculating the quantity

 $\displaystyle D\left(x-y\right)$ $\displaystyle \equiv$ $\displaystyle \left\langle 0\left|\phi\left(x\right)\phi\left(y\right)\right|0\right\rangle \ \ \ \ \ (1)$ $\displaystyle$ $\displaystyle =$ $\displaystyle \frac{1}{\left(2\pi\right)^{3}}\int d^{3}p\;\frac{1}{2E_{\mathbf{p}}}e^{-ip\left(x-y\right)} \ \ \ \ \ (2)$

It turns out that ${D\left(x-y\right)\ne0}$ for any pair of events ${x}$ and ${y}$. In particular, it’s non-zero for timelike events (which we’d expect, since one timelike event in a pair can affect the other, as it’s possible for a light signal to travel between the two events), but also for spacelike events, where according to relativity, it is impossible for either event to affect the other since this would require faster than light travel between the events.

What matters in quantum theory, however, is what can be measured (as opposed to what can merely be calculated). Quantum mechanics tells us that two quantities can be independently measured precisely only if the operators corresponding to these quantities commute. This condition is the origin of the uncertainty principle and its most famous prediction: that position and momentum operators do not commute and thus these two quantities cannot be measured independently to arbitrarily precise accuracy. Another way of looking at it is that measuring either position or momentum affects the other quantity, so we can’t get separate independent measurements of both.

Getting back to the Klein-Gordon field, what this means is that if two events ${x}$ and ${y}$ in 4-d spacetime are separated by a spacelike interval (that is, ${\left(x-y\right)^{2}<0}$), then the field operators ${\phi\left(x\right)}$ and ${\phi\left(y\right)}$ should commute, indicating that finding a particle at event ${x}$ cannot affect the existence of a particle at event ${y}$. On the other hand, if ${x}$ and ${y}$ are separated by a timelike interval, then ${\phi\left(x\right)}$ and ${\phi\left(y\right)}$ should not commute, since a light signal could travel between the two events causing one to affect the other. Using the real field version in the Heisenberg picture, we have

$\displaystyle \phi\left(x\right)=\frac{1}{\left(2\pi\right)^{3}}\int d^{3}p\;\frac{1}{\sqrt{2E_{\mathbf{p}}}}\left(a_{\mathbf{p}}e^{-ipx}+a_{\mathbf{p}}^{\dagger}e^{ipx}\right) \ \ \ \ \ (3)$

So the commutator is

 $\displaystyle \left[\phi\left(x\right),\phi\left(y\right)\right]$ $\displaystyle =$ $\displaystyle \frac{1}{\left(2\pi\right)^{3}}\int d^{3}p\;\frac{1}{\sqrt{2E_{\mathbf{p}}}}\frac{1}{\left(2\pi\right)^{3}}\int d^{3}q\;\frac{1}{\sqrt{2E_{\mathbf{q}}}}\times\nonumber$ $\displaystyle$ $\displaystyle$ $\displaystyle \left[\left(a_{\mathbf{p}}e^{-ipx}+a_{\mathbf{p}}^{\dagger}e^{ipx}\right),\left(a_{\mathbf{q}}e^{-iqy}+a_{\mathbf{q}}^{\dagger}e^{iqy}\right)\right] \ \ \ \ \ (4)$

The commutation relations for ${a_{\mathbf{p}}}$ and ${a_{\mathbf{p}}^{\dagger}}$ are

 $\displaystyle \left[a_{\mathbf{p}},a_{\mathbf{q}}^{\dagger}\right]$ $\displaystyle =$ $\displaystyle \left(2\pi\right)^{3}\delta^{\left(3\right)}\left(\mathbf{p}-\mathbf{q}\right)\ \ \ \ \ (5)$ $\displaystyle \left[a_{\mathbf{p}},a_{\mathbf{q}}\right]$ $\displaystyle =$ $\displaystyle 0\ \ \ \ \ (6)$ $\displaystyle \left[a_{\mathbf{p}}^{\dagger},a_{\mathbf{q}}^{\dagger}\right]$ $\displaystyle =$ $\displaystyle 0 \ \ \ \ \ (7)$

So plugging these into the integrals gives

 $\displaystyle \left[\phi\left(x\right),\phi\left(y\right)\right]$ $\displaystyle =$ $\displaystyle \frac{1}{\left(2\pi\right)^{3}}\int d^{3}p\;\frac{1}{\sqrt{2E_{\mathbf{p}}}}\frac{1}{\left(2\pi\right)^{3}}\int d^{3}q\;\frac{1}{\sqrt{2E_{\mathbf{q}}}}\times\nonumber$ $\displaystyle$ $\displaystyle$ $\displaystyle \left\{ \left(2\pi\right)^{3}\delta^{\left(3\right)}\left(\mathbf{p}-\mathbf{q}\right)e^{i\left(qy-px\right)}-\left(2\pi\right)^{3}\delta^{\left(3\right)}\left(\mathbf{p}-\mathbf{q}\right)e^{-i\left(qy-px\right)}\right\} \ \ \ \ \ (8)$ $\displaystyle$ $\displaystyle =$ $\displaystyle \frac{1}{\left(2\pi\right)^{3}}\int\frac{d^{3}p}{2E_{\mathbf{p}}}\left[e^{-ip\left(x-y\right)}-e^{ip\left(x-y\right)}\right]\ \ \ \ \ (9)$ $\displaystyle$ $\displaystyle =$ $\displaystyle D\left(x-y\right)-D\left(y-x\right) \ \ \ \ \ (10)$

This integral is of the Lorentz invariant form that we derived earlier; in fact the original integral 2 is Lorentz invariant, since ${p\left(x-y\right)}$ is a scalar product of two four-vectors, so ${e^{-ip\left(x-y\right)}}$ is a Lorentz invariant function.

At this point, P&S use the argument that, for a spacelike interval it is always possible to transform ${x-y}$ to ${-\left(x-y\right)=y-x}$ by a Lorentz transformation and give a diagram with minimal information to support this. I can sort of see what they mean, in that ${x-y}$ and ${y-x}$ are located on the same hyperboloid that defines locations with the same separation ${\left(x-y\right)^{2}}$, so it should be possible to find a transformation that converts one into the other. I made a few half-hearted attempts at finding such a transformation but couldn’t do it, so I will take their word for it.

A more convincing argument (at least for me) goes like this. For a spacelike interval ${x-y}$, it’s always possible to find an inertial frame in which ${x}$ and ${y}$ are simultaneous, so that in that frame ${t_{x}=t_{y}}$ and ${\left(x-y\right)^{2}=-\left(\mathbf{x}-\mathbf{y}\right)^{2}<0}$. In that frame, we have

$\displaystyle \left[\phi\left(x\right),\phi\left(y\right)\right]=\frac{1}{\left(2\pi\right)^{3}}\int\frac{d^{3}p}{2E_{\mathbf{p}}}\left[e^{-i\mathbf{p}\cdot\left(\mathbf{x-y}\right)}-e^{i\mathbf{p}\cdot\left(\mathbf{x-y}\right)}\right] \ \ \ \ \ (11)$

Now, the ${\mathbf{p}}$ in the exponents is the vector over which the integration is being done, so it’s a dummy variable. Since ${E_{\mathbf{p}}=E_{-\mathbf{p}}=\sqrt{\mathbf{p}^{2}+m^{2}}}$, we can replace ${\mathbf{p}}$ by ${\mathbf{-}\mathbf{p}}$ in the second term without changing the integral, with the result that the two terms cancel each other and we’re left with

$\displaystyle \left[\phi\left(x\right),\phi\left(y\right)\right]=\frac{1}{\left(2\pi\right)^{3}}\int\frac{d^{3}p}{2E_{\mathbf{p}}}\left[e^{-i\mathbf{p}\cdot\left(\mathbf{x-y}\right)}-e^{-i\mathbf{p}\cdot\left(\mathbf{x-y}\right)}\right]=0 \ \ \ \ \ (12)$

Because the original integral 9 is Lorentz invariant, this result must be true when we transform to any other frame with the same value of ${\left(x-y\right)^{2}}$, even one in which the two events are not simultaneous. Therefore, ${\left[\phi\left(x\right),\phi\left(y\right)\right]=0}$ for all spacelike intervals.

For timelike intervals, ${\left(x-y\right)^{2}>0}$ and it’s always possible to find a frame in which the two events occur at the same spatial location ${\mathbf{x}=\mathbf{y}}$, but separated by a time interval ${\Delta t>0}$. In this case,

 $\displaystyle \left[\phi\left(x\right),\phi\left(y\right)\right]$ $\displaystyle =$ $\displaystyle \frac{1}{\left(2\pi\right)^{3}}\int\frac{d^{3}p}{2E_{\mathbf{p}}}\left[e^{iE_{\mathbf{p}}\Delta t}-e^{-iE_{\mathbf{p}}\Delta t}\right]\ \ \ \ \ (13)$ $\displaystyle$ $\displaystyle =$ $\displaystyle \frac{i}{\left(2\pi\right)^{3}}\int\frac{d^{3}p}{E_{\mathbf{p}}}\sin\left(E_{\mathbf{p}}\Delta t\right) \ \ \ \ \ (14)$

which is not zero. Most of the contribution to this integral occurs near ${\mathbf{p}=0}$ where ${E_{\mathbf{p}}\approx m}$ so

$\displaystyle \left[\phi\left(x\right),\phi\left(y\right)\right]\sim\frac{i}{\left(2\pi\right)^{3}m}\sin\left(m\Delta t\right) \ \ \ \ \ (15)$

Thus for timelike intervals ${\left[\phi\left(x\right),\phi\left(y\right)\right]\ne0}$ and the measurement of a particle at event ${x}$ can affect the measurement of the particle at a later event ${y}$.

From the point of view of the measurements of particles at spacetime locations ${x}$ and ${y}$, the Klein-Gordon field does preserve causality, even though the amplitude for the existence of a particle at these two locations is non-zero for all intervals.

# Causality in the Klein-Gordon field

Michael E. Peskin & Daniel V. Schroeder, An Introduction to Quantum Field Theory, (Perseus Books, 1995) – Chapter 2.

In old quantum theory, relativistic causality is violated, which is one of the reasons quantum field theory was created. For the real Klein-Gordon field, we can now investigate whether causality is preserved.

In the Heisenberg picture, the time-dependent field is

$\displaystyle \phi\left(x\right)=\frac{1}{\left(2\pi\right)^{3}}\int d^{3}p\;\frac{1}{\sqrt{2E_{\mathbf{p}}}}\left(a_{\mathbf{p}}e^{-ipx}+a_{\mathbf{p}}^{\dagger}e^{ipx}\right) \ \ \ \ \ (1)$

The state representing a single particle at position ${\mathbf{x}}$ at time ${t_{x}}$ is ${\phi\left(x\right)\left|0\right\rangle }$ (where ${x}$ is the four-vector ${x=\left(\mathbf{x},t_{x}\right)}$) and for a particle at spacetime location ${y}$ is ${\phi\left(y\right)\left|0\right\rangle }$. The amplitude for the particle propagating from ${y}$ to ${x}$ is therefore (remember we’re considering real fields so that ${\phi^{\dagger}=\phi}$):

$\displaystyle D\left(x-y\right)\equiv\left\langle 0\left|\phi\left(x\right)\phi\left(y\right)\right|0\right\rangle \ \ \ \ \ (2)$

Because ${a_{\mathbf{q}}\left|0\right\rangle =0}$, only the ${a_{\mathbf{q}}^{\dagger}}$ terms in ${\phi\left(y\right)}$ will give non-zero results when acting on ${\left|0\right\rangle }$. Since ${\left\langle 0\left|\mathbf{q}\right.\right\rangle =0}$, only the ${a_{\mathbf{p}}}$ terms in ${\phi\left(x\right)}$ will give a non-zero result, and then only when ${\mathbf{q}=\mathbf{p}}$ since we must annihilate a particle with the same momentum as the particle we created in order to restore the vacuum state. From the Lorentz invariant normalization

 $\displaystyle \left|\mathbf{p}\right\rangle$ $\displaystyle =$ $\displaystyle \sqrt{2E_{\mathbf{p}}}a_{\mathbf{p}}^{\dagger}\left|0\right\rangle \ \ \ \ \ (3)$ $\displaystyle \left\langle \mathbf{p}\left|\mathbf{q}\right.\right\rangle$ $\displaystyle =$ $\displaystyle \left(2\pi\right)^{3}2E_{\mathbf{p}}\delta^{\left(3\right)}\left(\mathbf{p}-\mathbf{q}\right) \ \ \ \ \ (4)$

we get

$\displaystyle \left\langle 0\left|a_{\mathbf{p}}a_{\mathbf{q}}^{\dagger}\right|0\right\rangle =\left(2\pi\right)^{3}\delta^{\left(3\right)}\left(\mathbf{p}-\mathbf{q}\right) \ \ \ \ \ (5)$

Therefore

 $\displaystyle D\left(x-y\right)$ $\displaystyle =$ $\displaystyle \left\langle 0\left|\frac{1}{\left(2\pi\right)^{3}}\int d^{3}p\;\frac{1}{\sqrt{2E_{\mathbf{p}}}}\left(a_{\mathbf{p}}e^{-ipx}+a_{\mathbf{p}}^{\dagger}e^{ipx}\right)\frac{1}{\left(2\pi\right)^{3}}\int d^{3}q\;\frac{1}{\sqrt{2E_{\mathbf{q}}}}\left(a_{\mathbf{q}}e^{-iqy}+a_{\mathbf{q}}^{\dagger}e^{iqy}\right)\right|0\right\rangle \ \ \ \ \ (6)$ $\displaystyle$ $\displaystyle =$ $\displaystyle \frac{1}{\left(2\pi\right)^{6}}\int d^{3}p\int d^{3}q\frac{1}{\sqrt{4E_{\mathbf{p}}E_{\mathbf{q}}}}\left\langle 0\left|a_{\mathbf{p}}a_{\mathbf{q}}^{\dagger}\right|0\right\rangle e^{-ipx+iqy}\ \ \ \ \ (7)$ $\displaystyle$ $\displaystyle =$ $\displaystyle \frac{1}{\left(2\pi\right)^{3}}\int d^{3}p\int d^{3}q\frac{1}{\sqrt{4E_{\mathbf{p}}E_{\mathbf{q}}}}\delta^{\left(3\right)}\left(\mathbf{p}-\mathbf{q}\right)e^{-ipx+iqy}\ \ \ \ \ (8)$ $\displaystyle$ $\displaystyle =$ $\displaystyle \frac{1}{\left(2\pi\right)^{3}}\int d^{3}p\;\frac{1}{2E_{\mathbf{p}}}e^{-ip\left(x-y\right)} \ \ \ \ \ (9)$

P&S calculate this propagator ${D}$ for two cases, which we’ll review here to fill in the details. First, for a timelike separation (that is, where there exists a frame in which ${x}$ and ${y}$ can appear at the same location but separated in time), ${x^{0}-y^{0}=t\ne0}$ and ${\mathbf{x}-\mathbf{y}=0}$. In that case

$\displaystyle D\left(x-y\right)=\frac{1}{\left(2\pi\right)^{3}}\int d^{3}p\;\frac{1}{2\sqrt{p^{2}+m^{2}}}e^{-i\sqrt{p^{2}+m^{2}}t} \ \ \ \ \ (10)$

We can convert this to spherical coordinates and since the integrand is spherically symmetric the integrals over ${\phi}$ and ${\theta}$ give ${4\pi}$, so we get

$\displaystyle D\left(x-y\right)=\frac{1}{4\pi^{2}}\int_{0}^{\infty}dp\;\frac{p^{2}}{\sqrt{p^{2}+m^{2}}}e^{-i\sqrt{p^{2}+m^{2}}t} \ \ \ \ \ (11)$

Substituting

 $\displaystyle E$ $\displaystyle =$ $\displaystyle \sqrt{p^{2}+m^{2}}\ \ \ \ \ (12)$ $\displaystyle dE$ $\displaystyle =$ $\displaystyle \frac{p\;dp}{\sqrt{p^{2}+m^{2}}}\ \ \ \ \ (13)$ $\displaystyle dp$ $\displaystyle =$ $\displaystyle \frac{\sqrt{p^{2}+m^{2}}}{p}dE \ \ \ \ \ (14)$

and using ${E=m}$ when ${p=0}$ we get

 $\displaystyle D\left(x-y\right)$ $\displaystyle =$ $\displaystyle \frac{1}{4\pi^{2}}\int_{m}^{\infty}dE\;pe^{-iEt}\ \ \ \ \ (15)$ $\displaystyle$ $\displaystyle =$ $\displaystyle \frac{1}{4\pi^{2}}\int_{m}^{\infty}dE\;\sqrt{E^{2}-m^{2}}e^{-iEt} \ \ \ \ \ (16)$

For very large ${t}$ the exponential oscillates very rapidly so contributions to the integral will be very small. The largest contribution will come from the smallest value of ${E}$ so for large ${t}$ we have

$\displaystyle D\left(x-y\right)\sim e^{-imt} \ \ \ \ \ (17)$

In this case, the amplitude for propagating between ${y}$ and ${x}$ is non-zero and roughly constant as ${t\rightarrow\infty}$. As there is always a frame at which both events occur at the same place, this is fine (a light signal can always connect the two events) so this doesn’t violate causality.

For the second example, P&S choose a spacelike interval, that is, ${x^{0}-y^{0}=0}$ and ${\mathbf{x}-\mathbf{y}=\mathbf{r}\ne0}$. In this case, we have

$\displaystyle D\left(x-y\right)=\frac{1}{\left(2\pi\right)^{3}}\int d^{3}p\;\frac{1}{2E_{\mathbf{p}}}e^{i\mathbf{p}\cdot\mathbf{r}} \ \ \ \ \ (18)$

Again switching to spherical coordinates helps. Remember that for the purposes of the integral, ${\mathbf{r}}$ is a constant, so we can choose the polar axis to be along ${\mathbf{r}}$. Then we get

 $\displaystyle D\left(x-y\right)$ $\displaystyle =$ $\displaystyle \frac{1}{\left(2\pi\right)^{3}}\int_{0}^{\infty}dp\int_{0}^{\pi}d\theta\int_{0}^{2\pi}d\phi\frac{p^{2}\sin\theta}{2\sqrt{p^{2}+m^{2}}}e^{ipr\cos\theta}\ \ \ \ \ (19)$ $\displaystyle$ $\displaystyle =$ $\displaystyle \frac{2\pi}{\left(2\pi\right)^{3}}\int_{0}^{\infty}dp\frac{p^{2}}{2\sqrt{p^{2}+m^{2}}}\frac{e^{ipr}-e^{-ipr}}{ipr}\ \ \ \ \ (20)$ $\displaystyle$ $\displaystyle =$ $\displaystyle \frac{-i}{8\pi^{2}r}\int_{-\infty}^{\infty}dp\frac{p}{\sqrt{p^{2}+m^{2}}}e^{ipr} \ \ \ \ \ (21)$

The integral can be treated as a contour integral in the complex ${p}$ plane. To do this, however, we need to recognize that ${\sqrt{p^{2}+m^{2}}}$ is a multivalued function with branch points at ${p=\pm im}$, that is, on the imaginary axis, and also a branch point at ${p=\infty}$. A suitable branch cut consists of lines extending from ${p=+im}$ to ${+i\infty}$ and from ${p=-im}$ to ${-i\infty}$. We can do the integral over the contour shown, which avoids the branch cut by taking a dip around it.

We can write the integration variable as

$\displaystyle p=Re^{i\alpha} \ \ \ \ \ (22)$

and let ${R\rightarrow\infty}$. In that case, the integrand is

$\displaystyle \frac{p}{\sqrt{p^{2}+m^{2}}}e^{ipr}=\frac{Re^{i\alpha}}{\sqrt{R^{2}e^{2i\alpha}+m^{2}}}e^{irR\cos\alpha}e^{-rR\sin\alpha} \ \ \ \ \ (23)$

The integral over the circular arc in the first quadrant is (using ${dp=iRe^{i\alpha}d\alpha}$ and integrating over ${\alpha}$):

$\displaystyle \int_{0}^{\pi/2}\frac{Re^{i\alpha}}{\sqrt{R^{2}e^{2i\alpha}+m^{2}}}e^{irR\cos\alpha}e^{-rR\sin\alpha}iRe^{i\alpha}d\alpha \ \ \ \ \ (24)$

Because ${\sin\alpha>0}$, the last exponential term tends to zero as ${R\rightarrow\infty}$, so the integral is zero. The same is true for the integral over the circular arc in the second quadrant, since ${\sin\alpha>0}$ here too. Thus the integrals over both arcs vanish for large ${R}$ and we’re left with the integral along the real axis (which is what we want) and the integral around the branch cut.

The point where the contour wraps around the branch point ${p=+im}$ is actually a little circle of radius ${\epsilon}$ which we let go to zero in the limit. This part of the contour can be modelled by letting ${p=im+\epsilon e^{i\beta}}$ so that the integrand becomes

$\displaystyle dp\frac{p}{\sqrt{p^{2}+m^{2}}}e^{ipr}=\left(i\epsilon e^{i\beta}d\beta\right)\frac{im+\epsilon e^{i\beta}}{\sqrt{2m\epsilon e^{i\beta}+\epsilon^{2}e^{2i\beta}}}e^{ipr} \ \ \ \ \ (25)$

In the limit of small ${\epsilon}$, the integrand goes to zero like ${\sqrt{\epsilon}}$, so it is only the two vertical sections of the integral around the branch cut that are left. Because ${\sqrt{p^{2}+m^{2}}}$ has opposite signs on each side of the cut, and the direction of integration is also opposite (down in the first quadrant and up in the second), these two effects cancel each other, and the integral around the cut is

$\displaystyle \frac{i}{4\pi^{2}r}\int_{im}^{\infty}dp\frac{p}{\sqrt{p^{2}+m^{2}}}e^{ipr} \ \ \ \ \ (26)$

and this must be the negative of the integral ${D\left(x-y\right)}$ that we’re trying to find, so that

$\displaystyle D\left(x-y\right)=\frac{-i}{4\pi^{2}r}\int_{im}^{\infty}dp\frac{p}{\sqrt{p^{2}+m^{2}}}e^{ipr} \ \ \ \ \ (27)$

With the substitution ${\rho=-ip}$, this converts into the real integral

$\displaystyle D\left(x-y\right)=\frac{1}{4\pi^{2}r}\int_{m}^{\infty}d\rho\frac{\rho}{\sqrt{\rho^{2}-m^{2}}}e^{-\rho r} \ \ \ \ \ (28)$

For large ${r}$, the main contribution to the integral comes from the smallest values of ${\rho}$ so the asymptotic behaviour of the integral is

$\displaystyle D\left(x-y\right)\sim e^{-mr} \ \ \ \ \ (29)$

Thus even for spacelike intervals, the amplitude for particle propagation is still non-zero so causality appears to be violated. However, in quantum theory, it is what can be measured that matters, rather than what the amplitudes are. We’ll look at that later.

# Klein-Gordon field in the Heisenberg picture; time dependence

Michael E. Peskin & Daniel V. Schroeder, An Introduction to Quantum Field Theory, (Perseus Books, 1995) – Chapter 2.

Although we’ve already ground through the derivation of the Klein-Gordon equation in the Heisenberg picture, it’s useful to review the process as given in P&S’s chapter 2, which omits many of the steps in the derivation. We start by converting the field operator ${\phi\left(\mathbf{x}\right)}$ and conjugate momentum ${\pi\left(\mathbf{x}\right)}$ to the Heisenberg picture using the unitary transformation

 $\displaystyle \phi\left(x\right)$ $\displaystyle =$ $\displaystyle \phi\left(\mathbf{x},t\right)=e^{iHt}\phi\left(\mathbf{x}\right)e^{-iHt}\ \ \ \ \ (1)$ $\displaystyle \pi\left(x\right)$ $\displaystyle =$ $\displaystyle \pi\left(\mathbf{x},t\right)=e^{iHt}\pi\left(\mathbf{x}\right)e^{-iHt} \ \ \ \ \ (2)$

The equation of motion for a Heisenberg operator is

$\displaystyle i\frac{\partial\mathcal{O}}{\partial t}=\left[\mathcal{O},H\right] \ \ \ \ \ (3)$

We can then plug in the integral of the Hamiltonian density ${\mathcal{H}}$ (remember we’re dealing with a real field, so ${\phi^{\dagger}=\phi}$):

$\displaystyle H=\frac{1}{2}\int d^{3}x^{\prime}\left\{ \pi^{2}\left(\mathbf{x}^{\prime},t\right)+\left(\nabla\phi\left(\mathbf{x}^{\prime},t\right)\right)^{2}+m^{2}\phi^{2}\left(\mathbf{x}^{\prime},t\right)\right\} \ \ \ \ \ (4)$

We can now calculate the commutators with ${\phi}$ and ${\pi}$ by using the commutation relations

$\displaystyle \left[\phi\left(\mathbf{x},t\right),\pi\left(\mathbf{x}^{\prime},t\right)\right]=i\delta^{\left(3\right)}\left(\mathbf{x}-\mathbf{x}^{\prime}\right) \ \ \ \ \ (5)$

For the field

 $\displaystyle i\frac{\partial\phi\left(\mathbf{x},t\right)}{\partial t}$ $\displaystyle =$ $\displaystyle \left[\phi\left(\mathbf{x},t\right),H\right]\ \ \ \ \ (6)$ $\displaystyle$ $\displaystyle =$ $\displaystyle \left[\phi\left(\mathbf{x},t\right),\frac{1}{2}\int d^{3}x^{\prime}\left\{ \pi^{2}\left(\mathbf{x}^{\prime},t\right)+\left(\nabla\phi\left(\mathbf{x}^{\prime},t\right)\right)^{2}+m^{2}\phi^{2}\left(\mathbf{x}^{\prime},t\right)\right\} \right] \ \ \ \ \ (7)$

The key point to notice here is that the coordinate ${\mathbf{x}}$ in ${\phi\left(\mathbf{x},t\right)}$ is a constant relative to the ${\mathbf{x}^{\prime}}$ used as an integration variable, so the ${\phi\left(\mathbf{x},t\right)}$ can be taken inside the integral and all commutators evaluated under the integral sign. We find that ${\phi\left(\mathbf{x},t\right)}$ commutes with ${\left(\nabla\phi\left(\mathbf{x}^{\prime},t\right)\right)^{2}+m^{2}\phi^{2}\left(\mathbf{x}^{\prime},t\right)}$ so we’re left with

 $\displaystyle i\frac{\partial\phi\left(\mathbf{x},t\right)}{\partial t}$ $\displaystyle =$ $\displaystyle \frac{1}{2}\int d^{3}x^{\prime}\left[\phi\left(\mathbf{x},t\right),\pi^{2}\left(\mathbf{x}^{\prime},t\right)\right]\ \ \ \ \ (8)$ $\displaystyle$ $\displaystyle =$ $\displaystyle \frac{1}{2}\int d^{3}x^{\prime}\left(i\delta^{\left(3\right)}\left(\mathbf{x}-\mathbf{x}^{\prime}\right)\pi\left(\mathbf{x}^{\prime},t\right)+\pi\left(\mathbf{x}^{\prime},t\right)\phi\left(\mathbf{x},t\right)\pi\left(\mathbf{x}^{\prime},t\right)-\pi^{2}\left(\mathbf{x}^{\prime},t\right)\phi\left(\mathbf{x},t\right)\right)\ \ \ \ \ (9)$ $\displaystyle$ $\displaystyle =$ $\displaystyle \int d^{3}x^{\prime}i\delta^{\left(3\right)}\left(\mathbf{x}-\mathbf{x}^{\prime}\right)\pi\left(\mathbf{x}^{\prime},t\right)\ \ \ \ \ (10)$ $\displaystyle$ $\displaystyle =$ $\displaystyle i\pi\left(\mathbf{x},t\right) \ \ \ \ \ (11)$

The equation for the conjugate momentum is

 $\displaystyle i\frac{\partial\pi\left(\mathbf{x},t\right)}{\partial t}$ $\displaystyle =$ $\displaystyle \left[\pi\left(\mathbf{x},t\right),H\right]\ \ \ \ \ (12)$ $\displaystyle$ $\displaystyle =$ $\displaystyle \left[\pi\left(\mathbf{x},t\right),\frac{1}{2}\int d^{3}x^{\prime}\left\{ \pi^{2}\left(\mathbf{x}^{\prime},t\right)+\left(\nabla\phi\left(\mathbf{x}^{\prime},t\right)\right)^{2}+m^{2}\phi^{2}\left(\mathbf{x}^{\prime},t\right)\right\} \right] \ \ \ \ \ (13)$

Since ${\pi\left(\mathbf{x},t\right)}$ commutes with the ${\pi^{2}\left(\mathbf{x}^{\prime},t\right)}$ in the integral we’re left with

$\displaystyle i\frac{\partial\pi\left(\mathbf{x},t\right)}{\partial t}=\left[\pi\left(\mathbf{x},t\right),\frac{1}{2}\int d^{3}x^{\prime}\left\{ \left(\nabla\phi\left(\mathbf{x}^{\prime},t\right)\right)^{2}+m^{2}\phi^{2}\left(\mathbf{x}^{\prime},t\right)\right\} \right] \ \ \ \ \ (14)$

The gradient term is transformed using

 $\displaystyle \nabla\left(\phi\nabla\phi\right)$ $\displaystyle =$ $\displaystyle \left(\nabla\phi\right)^{2}+\phi\nabla^{2}\phi\ \ \ \ \ (15)$ $\displaystyle \left(\nabla\phi\right)^{2}$ $\displaystyle =$ $\displaystyle \nabla\left(\phi\nabla\phi\right)-\phi\nabla^{2}\phi \ \ \ \ \ (16)$

The first term on the RHS is a divergence and is converted to a surface integral which goes to zero using Gauss’s theorem, so we’re left with

$\displaystyle i\frac{\partial\pi\left(\mathbf{x},t\right)}{\partial t}=\left[\pi\left(\mathbf{x},t\right),\frac{1}{2}\int d^{3}x^{\prime}\left\{ -\phi\left(\mathbf{x}^{\prime},t\right)\nabla^{2}\phi\left(\mathbf{x}^{\prime},t\right)+m^{2}\phi^{2}\left(\mathbf{x}^{\prime},t\right)\right\} \right] \ \ \ \ \ (17)$

Using the commutator 5 in a similar manner to the calculation for ${\phi}$ above, this integral gets reduced to

$\displaystyle i\frac{\partial\pi\left(\mathbf{x},t\right)}{\partial t}=-i\left(\nabla^{2}+m^{2}\right)\phi\left(\mathbf{x},t\right) \ \ \ \ \ (18)$

Taking the time derivative of 11 and using 18 gives the Klein-Gordon equation

$\displaystyle \frac{\partial^{2}\phi}{\partial t^{2}}=\left(\nabla^{2}-m^{2}\right)\phi \ \ \ \ \ (19)$

We can also write the Heisenberg field and conjugate momentum in terms of creation and annihilation operators. In this form, the Hamiltonian is

$\displaystyle H=\int\frac{d^{3}p}{\left(2\pi\right)^{3}}\omega_{\mathbf{p}}\left(a_{\mathbf{p}}^{\dagger}a_{\mathbf{p}}+\frac{1}{2}\left[a_{\mathbf{p}},a_{\mathbf{p}}^{\dagger}\right]\right) \ \ \ \ \ (20)$

Using the fact that

$\displaystyle Ha_{\mathbf{p}}^{\dagger}\left|0\right\rangle =E_{\mathbf{p}}a_{\mathbf{p}}^{\dagger}\left|0\right\rangle \ \ \ \ \ (21)$

and (ignoring the infinite vacuum energy)

$\displaystyle a_{\mathbf{p}}^{\dagger}H\left|0\right\rangle =0 \ \ \ \ \ (22)$

(since the ${a_{\mathbf{p}}^{\dagger}a_{\mathbf{p}}}$ operating on ${\left|0\right\rangle }$ produces 0), we have

$\displaystyle \left[H,a_{\mathbf{p}}^{\dagger}\right]=E_{\mathbf{p}}a_{\mathbf{p}}^{\dagger} \ \ \ \ \ (23)$

For ${a_{\mathbf{p}}}$ we have

 $\displaystyle Ha_{\mathbf{p}}\left|\mathbf{p}\right\rangle$ $\displaystyle =$ $\displaystyle H\left|0\right\rangle =0\ \ \ \ \ (24)$ $\displaystyle a_{\mathbf{p}}H\left|\mathbf{p}\right\rangle$ $\displaystyle =$ $\displaystyle E_{\mathbf{p}}a_{\mathbf{p}}\left|\mathbf{p}\right\rangle \ \ \ \ \ (25)$

so

$\displaystyle \left[H,a_{\mathbf{p}}\right]=-E_{\mathbf{p}}a_{\mathbf{p}} \ \ \ \ \ (26)$

We can write this as

$\displaystyle Ha_{\mathbf{p}}=a_{\mathbf{p}}\left(H-E_{\mathbf{p}}\right) \ \ \ \ \ (27)$

For higher powers we get

 $\displaystyle H^{2}a_{\mathbf{p}}$ $\displaystyle =$ $\displaystyle H\left(Ha_{\mathbf{p}}\right)\ \ \ \ \ (28)$ $\displaystyle$ $\displaystyle =$ $\displaystyle H\left[a_{\mathbf{p}}\left(H-E_{\mathbf{p}}\right)\right]\ \ \ \ \ (29)$ $\displaystyle$ $\displaystyle =$ $\displaystyle a_{\mathbf{p}}\left(H-E_{\mathbf{p}}\right)^{2} \ \ \ \ \ (30)$

This fairly obviously generalizes to (or you can prove it by induction)

$\displaystyle H^{n}a_{\mathbf{p}}=a_{\mathbf{p}}\left(H-E_{\mathbf{p}}\right)^{n} \ \ \ \ \ (31)$

For ${a_{\mathbf{p}}^{\dagger}}$ we can use the same argument to show that

$\displaystyle H^{n}a_{\mathbf{p}}^{\dagger}=a_{\mathbf{p}}^{\dagger}\left(H+E_{\mathbf{p}}\right)^{n} \ \ \ \ \ (32)$

 $\displaystyle \phi\left(\mathbf{x}\right)$ $\displaystyle =$ $\displaystyle \frac{1}{\left(2\pi\right)^{3}}\int d^{3}p\;e^{i\mathbf{p\cdot x}}\frac{1}{\sqrt{2E_{\mathbf{p}}}}\left(a_{\mathbf{p}}+a_{-\mathbf{p}}^{\dagger}\right)\ \ \ \ \ (33)$ $\displaystyle$ $\displaystyle =$ $\displaystyle \frac{1}{\left(2\pi\right)^{3}}\int d^{3}p\;\frac{1}{\sqrt{2E_{\mathbf{p}}}}\left(a_{\mathbf{p}}e^{i\mathbf{p\cdot x}}+a_{\mathbf{p}}^{\dagger}e^{-i\mathbf{p\cdot x}}\right)\ \ \ \ \ (34)$ $\displaystyle \pi\left(\mathbf{x}\right)$ $\displaystyle =$ $\displaystyle \frac{-i}{\left(2\pi\right)^{3}}\int d^{3}p\;e^{i\mathbf{p\cdot x}}\sqrt{\frac{E_{\mathbf{p}}}{2}}\left(a_{\mathbf{p}}-a_{-\mathbf{p}}^{\dagger}\right)\ \ \ \ \ (35)$ $\displaystyle$ $\displaystyle$ $\displaystyle \frac{-i}{\left(2\pi\right)^{3}}\int d^{3}p\;\sqrt{\frac{E_{\mathbf{p}}}{2}}\left(a_{\mathbf{p}}e^{i\mathbf{p\cdot x}}-a_{\mathbf{p}}^{\dagger}e^{-i\mathbf{p\cdot x}}\right) \ \ \ \ \ (36)$

To apply the conversion 1 to get the Heisenberg field we need the operators ${e^{iHt}}$ and ${e^{-iHt}}$. Since the exponentials expand as a power series in ${\pm iHt}$ we can apply 31 and 32 to get

 $\displaystyle e^{iHt}a_{\mathbf{p}}e^{-iHt}$ $\displaystyle =$ $\displaystyle a_{\mathbf{p}}e^{i\left(H-E_{\mathbf{p}}\right)t}e^{-iHt}\ \ \ \ \ (37)$ $\displaystyle$ $\displaystyle =$ $\displaystyle a_{\mathbf{p}}e^{-iE_{\mathbf{p}}t}\ \ \ \ \ (38)$ $\displaystyle e^{iHt}a_{\mathbf{p}}^{\dagger}e^{-iHt}$ $\displaystyle =$ $\displaystyle a_{\mathbf{p}}^{\dagger}e^{i\left(H+E_{\mathbf{p}}\right)t}e^{-iHt}\ \ \ \ \ (39)$ $\displaystyle$ $\displaystyle =$ $\displaystyle a_{\mathbf{p}}^{\dagger}e^{iE_{\mathbf{p}}t} \ \ \ \ \ (40)$

This gives the Heisenberg fields as

 $\displaystyle \phi\left(\mathbf{x},t\right)$ $\displaystyle =$ $\displaystyle e^{iHt}\phi\left(\mathbf{x}\right)e^{-iHt}\ \ \ \ \ (41)$ $\displaystyle$ $\displaystyle =$ $\displaystyle \frac{1}{\left(2\pi\right)^{3}}\int d^{3}p\;\frac{1}{\sqrt{2E_{\mathbf{p}}}}\left(a_{\mathbf{p}}e^{-iE_{\mathbf{p}}t}e^{i\mathbf{p\cdot x}}+a_{\mathbf{p}}^{\dagger}e^{iE_{\mathbf{p}}t}e^{-i\mathbf{p\cdot x}}\right)\ \ \ \ \ (42)$ $\displaystyle$ $\displaystyle =$ $\displaystyle \frac{1}{\left(2\pi\right)^{3}}\int d^{3}p\;\frac{1}{\sqrt{2E_{\mathbf{p}}}}\left(a_{\mathbf{p}}e^{-ipx}+a_{\mathbf{p}}^{\dagger}e^{ipx}\right) \ \ \ \ \ (43)$

where ${p}$ and ${x}$ are now 4-vectors and ${px\equiv p^{\mu}x_{\mu}=E_{\mathbf{p}}t-\mathbf{p}\cdot\mathbf{x}}$. Similarly

$\displaystyle \pi\left(\mathbf{x},t\right)=\frac{-i}{\left(2\pi\right)^{3}}\int d^{3}p\;\sqrt{\frac{E_{\mathbf{p}}}{2}}\left(a_{\mathbf{p}}e^{-ipx}-a_{\mathbf{p}}^{\dagger}e^{ipx}\right)=\frac{\partial\phi\left(\mathbf{x},t\right)}{\partial t} \ \ \ \ \ (44)$

We can do a similar calculation with the total momentum operator ${\mathbf{P}}$. When ${\mathbf{P}}$ operates on a single-particle state ${\left|\mathbf{p}\right\rangle }$ we get

$\displaystyle \mathbf{P}\left|\mathbf{p}\right\rangle =\mathbf{p}\left|\mathbf{p}\right\rangle \ \ \ \ \ (45)$

where ${\mathbf{p}}$ is the eigenvalue (not an operator) of ${\mathbf{P}}$.

Therefore

 $\displaystyle \left[\mathbf{P},a_{\mathbf{p}}\right]\left|\mathbf{p}\right\rangle$ $\displaystyle =$ $\displaystyle \mathbf{P}a_{\mathbf{p}}\left|\mathbf{p}\right\rangle -a_{\mathbf{p}}\mathbf{P}\left|\mathbf{p}\right\rangle \ \ \ \ \ (46)$ $\displaystyle$ $\displaystyle =$ $\displaystyle 0-\mathbf{p}a_{\mathbf{p}}\left|\mathbf{p}\right\rangle \ \ \ \ \ (47)$ $\displaystyle \left[\mathbf{P},a_{\mathbf{p}}\right]$ $\displaystyle =$ $\displaystyle -\mathbf{p}a_{\mathbf{p}}\ \ \ \ \ (48)$ $\displaystyle \left[\mathbf{P},a_{\mathbf{p}}^{\dagger}\right]\left|0\right\rangle$ $\displaystyle =$ $\displaystyle \mathbf{P}a_{\mathbf{p}}^{\dagger}\left|0\right\rangle -a_{\mathbf{p}}^{\dagger}\mathbf{P}\left|0\right\rangle \ \ \ \ \ (49)$ $\displaystyle$ $\displaystyle =$ $\displaystyle \mathbf{p}\left|\mathbf{p}\right\rangle -0\ \ \ \ \ (50)$ $\displaystyle$ $\displaystyle =$ $\displaystyle \mathbf{p}a_{\mathbf{p}}^{\dagger}\left|0\right\rangle \ \ \ \ \ (51)$ $\displaystyle \left[\mathbf{P},a_{\mathbf{p}}^{\dagger}\right]$ $\displaystyle =$ $\displaystyle \mathbf{p}a_{\mathbf{p}}^{\dagger} \ \ \ \ \ (52)$

Therefore, using the same logic as above for ${H}$, we get

 $\displaystyle a_{\mathbf{p}}\mathbf{P}\cdot\mathbf{x}$ $\displaystyle =$ $\displaystyle \left(\mathbf{p}\cdot\mathbf{x}+\mathbf{P}\cdot\mathbf{x}\right)a_{\mathbf{p}}\ \ \ \ \ (53)$ $\displaystyle a_{\mathbf{p}}\left(\mathbf{P}\cdot\mathbf{x}\right)^{n}$ $\displaystyle =$ $\displaystyle \left(\mathbf{p}\cdot\mathbf{x}+\mathbf{P}\cdot\mathbf{x}\right)^{n}a_{\mathbf{p}}\ \ \ \ \ (54)$ $\displaystyle a_{\mathbf{p}}e^{i\mathbf{P}\cdot\mathbf{x}}$ $\displaystyle =$ $\displaystyle e^{i\left(\mathbf{p}\cdot\mathbf{x}+\mathbf{P}\cdot\mathbf{x}\right)}a_{\mathbf{p}}\ \ \ \ \ (55)$ $\displaystyle e^{-i\mathbf{P}\cdot\mathbf{x}}a_{\mathbf{p}}e^{i\mathbf{P}\cdot\mathbf{x}}$ $\displaystyle =$ $\displaystyle a_{\mathbf{p}}e^{i\mathbf{p}\cdot\mathbf{x}} \ \ \ \ \ (56)$

[Because ${e^{i\mathbf{p}\cdot\mathbf{x}}}$ is a number, not an operator, it commutes with ${a_{\mathbf{p}}}$.]

By the same logic,

$\displaystyle e^{-i\mathbf{P}\cdot\mathbf{x}}a_{\mathbf{p}}^{\dagger}e^{i\mathbf{P}\cdot\mathbf{x}}=a_{\mathbf{p}}^{\dagger}e^{-i\mathbf{p}\cdot\mathbf{x}} \ \ \ \ \ (57)$

Therefore, we can write 34 as

 $\displaystyle \phi\left(\mathbf{x}\right)$ $\displaystyle =$ $\displaystyle e^{-i\mathbf{P}\cdot\mathbf{x}}\left[\frac{1}{\left(2\pi\right)^{3}}\int d^{3}p\;\frac{1}{\sqrt{2E_{\mathbf{p}}}}\left(a_{\mathbf{p}}+a_{\mathbf{p}}^{\dagger}\right)\right]e^{i\mathbf{P}\cdot\mathbf{x}}\ \ \ \ \ (58)$ $\displaystyle$ $\displaystyle =$ $\displaystyle e^{-i\mathbf{P}\cdot\mathbf{x}}\phi\left(0\right)e^{i\mathbf{P}\cdot\mathbf{x}} \ \ \ \ \ (59)$

Or, in 4-vector notation from 43, we get

 $\displaystyle \phi\left(\mathbf{x},t\right)$ $\displaystyle =$ $\displaystyle e^{iHt}\phi\left(\mathbf{x}\right)e^{-iHt}\ \ \ \ \ (60)$ $\displaystyle$ $\displaystyle =$ $\displaystyle e^{iHt}e^{-i\mathbf{P}\cdot\mathbf{x}}\phi\left(0\right)e^{i\mathbf{P}\cdot\mathbf{x}}e^{-iHt}\ \ \ \ \ (61)$ $\displaystyle$ $\displaystyle =$ $\displaystyle e^{iPx}\phi\left(0\right)e^{-iPx} \ \ \ \ \ (62)$

where ${P}$ is the 4-vector operator

$\displaystyle P^{\mu}=\left(H,\mathbf{P}\right) \ \ \ \ \ (63)$

# Klein-Gordon field in position space

Michael E. Peskin & Daniel V. Schroeder, An Introduction to Quantum Field Theory, (Perseus Books, 1995) – Chapter 2.

When we apply the creation operator ${a_{\mathbf{p}}^{\dagger}}$ to the vacuum ${\left|0\right\rangle }$, we get a single particle state with momentum ${\mathbf{p}}$, normalized so that ${\left\langle \mathbf{p}^{\prime}\left|\mathbf{p}\right.\right\rangle }$ is Lorentz invariant:

$\displaystyle \left|\mathbf{p}\right\rangle =\sqrt{2E_{\mathbf{p}}}a_{\mathbf{p}}^{\dagger}\left|0\right\rangle \ \ \ \ \ (1)$

The Klein-Gordon field operator as a function of position is

$\displaystyle \phi\left(\mathbf{x},t\right)=\frac{1}{\left(2\pi\right)^{3}}\int d^{3}p\;e^{i\mathbf{p\cdot x}}\frac{1}{\sqrt{2E_{\mathbf{p}}}}\left(a_{\mathbf{p}}+a_{-\mathbf{p}}^{\dagger}\right) \ \ \ \ \ (2)$

so it’s interesting to see what happens when we apply this operator to the vacuum: ${\phi\left(\mathbf{x},t\right)\left|0\right\rangle }$. Because ${a_{\mathbf{p}}\left|0\right\rangle =0}$ the ${a_{\mathbf{p}}}$ term in the integrand drops out and we get

 $\displaystyle \phi\left(\mathbf{x},t\right)\left|0\right\rangle$ $\displaystyle =$ $\displaystyle \frac{1}{\left(2\pi\right)^{3}}\int d^{3}p\;e^{i\mathbf{p\cdot x}}\frac{1}{\sqrt{2E_{\mathbf{p}}}}\frac{1}{\sqrt{2E_{-\mathbf{p}}}}\left|-\mathbf{p}\right\rangle \ \ \ \ \ (3)$ $\displaystyle$ $\displaystyle =$ $\displaystyle \frac{1}{\left(2\pi\right)^{3}}\int d^{3}p\;e^{-i\mathbf{p\cdot x}}\frac{1}{2E_{\mathbf{p}}}\left|\mathbf{p}\right\rangle \ \ \ \ \ (4)$

where in the last line we changed the integration variable from ${\mathbf{p}}$ to ${-\mathbf{p}}$ and used ${E_{-\mathbf{p}}=E_{\mathbf{p}}}$.

Apart from the factor of ${1/2E_{\mathbf{p}}}$, this is the expansion of the position state ${\left|\mathbf{x}\right\rangle }$ in terms of momentum operators. For non-relativistic speeds ${E_{\mathbf{p}}=\sqrt{\mathbf{p}^{2}+m^{2}}\rightarrow m}$ so is approximately constant. Thus ${\phi\left(\mathbf{x},t\right)\left|0\right\rangle }$ is interpreted as a free particle located at position ${\mathbf{x}}$. In accordance with the uncertainty principle, a particle whose position is known exactly has a completely indeterminate momentum, as shown by the integral over all momentum values.

We can calculate the matrix element ${\left\langle 0\left|\phi\left(\mathbf{x},t\right)\right|\mathbf{p}\right\rangle }$. Because ${\phi}$ here is real, ${\left\langle 0\right|\phi\left(\mathbf{x},t\right)=\left\langle \phi\left(\mathbf{x},t\right)0\right|}$, so the bra part of the bracket is the complex conjugate of 4, giving

 $\displaystyle \left\langle 0\left|\phi\left(\mathbf{x},t\right)\right|\mathbf{p}\right\rangle$ $\displaystyle =$ $\displaystyle \frac{1}{\left(2\pi\right)^{3}}\int d^{3}p^{\prime}\;e^{i\mathbf{p^{\prime}\cdot x}}\frac{1}{2E_{\mathbf{p}^{\prime}}}\left\langle \mathbf{p}^{\prime}\left|\mathbf{p}\right.\right\rangle \ \ \ \ \ (5)$ $\displaystyle$ $\displaystyle =$ $\displaystyle \frac{1}{\left(2\pi\right)^{3}}\int d^{3}p^{\prime}\;e^{i\mathbf{p^{\prime}\cdot x}}\frac{1}{2E_{\mathbf{p}^{\prime}}}\left(2\pi\right)^{3}2E_{\mathbf{p}}\delta^{\left(3\right)}\left(\mathbf{p}^{\prime}-\mathbf{p}\right)\ \ \ \ \ (6)$ $\displaystyle$ $\displaystyle =$ $\displaystyle e^{i\mathbf{p}\cdot\mathbf{x}} \ \ \ \ \ (7)$

where we used the Lorentz-invariant normalization in the second line. The result is proportional to ${\left\langle \mathbf{x}\left|\mathbf{p}\right.\right\rangle }$, so the state ${\phi\left(\mathbf{x},t\right)\left|0\right\rangle }$ does appear to represent the position state ${\left|\mathbf{x}\right\rangle }$.

# Lorentz invariance in Klein-Gordon momentum states

Michael E. Peskin & Daniel V. Schroeder, An Introduction to Quantum Field Theory, (Perseus Books, 1995) – Chapter 2.

The vacuum state ${\left|0\right\rangle }$ in Klein-Gordon field theory is a postulated state which gives 0 when operated on by any annihilation operator ${a_{\mathbf{p}}}$. Applying a creation operator ${a_{\mathbf{p}}^{\dagger}}$ to the vacuum converts it to a state with a single particle of momentum ${\mathbf{p}}$, that is, the state ${\left|\mathbf{p}\right\rangle }$. The vacuum state is normalized so that ${\left\langle 0\left|0\right.\right\rangle =1}$. If we require all single-particle momentum states to be orthogonal for different momenta then, since ${\mathbf{p}}$ is a continuous variable, we might expect that a suitable normalization would be

$\displaystyle \left\langle \mathbf{p}\left|\mathbf{q}\right.\right\rangle =\left(2\pi\right)^{3}\delta^{\left(3\right)}\left(\mathbf{p}-\mathbf{q}\right) \ \ \ \ \ (1)$

[Again, the factors of ${2\pi}$ depend on how other quantities in the theory are defined, since sometimes these factors turn up elsewhere.] The problem with this normalization is that it’s not Lorentz invariant. That is, if we view the system from a frame moving with velocity ${\beta}$ in the ${x_{3}}$ direction, say, the delta function doesn’t remain invariant. Since 4-momentum is a 4-vector, it transforms under Lorentz transformations, so that

 $\displaystyle E^{\prime}$ $\displaystyle =$ $\displaystyle \gamma\left(E+\beta p_{3}\right)\ \ \ \ \ (2)$ $\displaystyle p_{3}^{\prime}$ $\displaystyle =$ $\displaystyle \gamma\left(p_{3}+\beta E\right) \ \ \ \ \ (3)$

where ${\gamma=1/\sqrt{1-\beta^{2}}}$ and ${E=p_{0}}$ as usual.

How does the delta function transform? We can use the formula

$\displaystyle \delta\left(f\left(x\right)-f\left(x_{0}\right)\right)=\frac{1}{\left|f^{\prime}\left(x_{0}\right)\right|}\delta\left(x\right) \ \ \ \ \ (4)$

We want to find ${\delta^{\left(3\right)}\left(\mathbf{p}^{\prime}-\mathbf{q}^{\prime}\right)}$ when we change from ${p_{3}}$ to ${p_{3}^{\prime}}$ so we need

 $\displaystyle \frac{dp_{3}^{\prime}}{dp_{3}}$ $\displaystyle =$ $\displaystyle \gamma\left(1+\beta\frac{dE}{dp_{3}}\right)\ \ \ \ \ (5)$ $\displaystyle$ $\displaystyle =$ $\displaystyle \gamma\left(1+\beta\frac{d}{dp_{3}}\sqrt{\mathbf{p}\cdot\mathbf{p}+m^{2}}\right)\ \ \ \ \ (6)$ $\displaystyle$ $\displaystyle =$ $\displaystyle \gamma\left(1+\frac{\beta}{E}p_{3}\right)\ \ \ \ \ (7)$ $\displaystyle$ $\displaystyle =$ $\displaystyle \frac{\gamma}{E}\left(E+\beta p_{3}\right)\ \ \ \ \ (8)$ $\displaystyle$ $\displaystyle =$ $\displaystyle \frac{E^{\prime}}{E} \ \ \ \ \ (9)$

So the delta function transforms as

$\displaystyle \delta^{\left(3\right)}\left(\mathbf{p}^{\prime}-\mathbf{q}^{\prime}\right)=\frac{E}{E^{\prime}}\delta^{\left(3\right)}\left(\mathbf{p}-\mathbf{q}\right) \ \ \ \ \ (10)$

and is not invariant. However, multiplying through by ${E^{\prime}}$ shows that

$\displaystyle E^{\prime}\delta^{\left(3\right)}\left(\mathbf{p}^{\prime}-\mathbf{q}^{\prime}\right)=E\delta^{\left(3\right)}\left(\mathbf{p}-\mathbf{q}\right) \ \ \ \ \ (11)$

so the quantity ${E\delta^{\left(3\right)}\left(\mathbf{p}-\mathbf{q}\right)}$ is Lorentz invariant. As a result, the momentum state is usually normalized so that

 $\displaystyle \left|\mathbf{p}\right\rangle$ $\displaystyle =$ $\displaystyle \sqrt{2E_{\mathbf{p}}}a_{\mathbf{p}}^{\dagger}\left|0\right\rangle \ \ \ \ \ (12)$ $\displaystyle \left\langle \mathbf{p}\left|\mathbf{q}\right.\right\rangle$ $\displaystyle =$ $\displaystyle \left(2\pi\right)^{3}2E_{\mathbf{p}}\delta^{\left(3\right)}\left(\mathbf{p}-\mathbf{q}\right) \ \ \ \ \ (13)$

[The extra factor of 2 is inserted to make other calculations easier. It doesn’t affect the Lorentz invariance.]

Since we’re defining states to preserve Lorentz invariance, if we apply a Lorentz transformation ${\Lambda}$ to a state to get a new state ${\left|\Lambda\mathbf{p}\right\rangle }$ we require

$\displaystyle \left\langle \Lambda\mathbf{p}\left|\Lambda\mathbf{p}\right.\right\rangle =\left\langle \mathbf{p}\left|\mathbf{p}\right.\right\rangle \ \ \ \ \ (14)$

This means that a Lorentz transformation is a unitary operator ${U\left(\Lambda\right)}$, since it leaves the bracket unchanged. We can write this as

$\displaystyle U\left(\Lambda\right)\left|\mathbf{p}\right\rangle =\left|\Lambda\mathbf{p}\right\rangle \ \ \ \ \ (15)$

Because of 12 and the fact that operators transform under a unitary transformation according to ${Q^{\prime}=UQU^{-1}}$, we get the transformation rule

 $\displaystyle \sqrt{2E_{\Lambda\mathbf{p}}}a_{\Lambda\mathbf{p}}^{\dagger}$ $\displaystyle =$ $\displaystyle U\left(\Lambda\right)\sqrt{2E_{\mathbf{p}}}a_{\mathbf{p}}^{\dagger}U^{-1}\left(\Lambda\right)\ \ \ \ \ (16)$ $\displaystyle U\left(\Lambda\right)a_{\mathbf{p}}^{\dagger}U^{-1}\left(\Lambda\right)$ $\displaystyle =$ $\displaystyle \sqrt{\frac{E_{\Lambda\mathbf{p}}}{E_{\mathbf{p}}}}a_{\Lambda\mathbf{p}}^{\dagger} \ \ \ \ \ (17)$

Another useful relation can be derived concerning the integration of Lorentz invariant functions. First, consider the four dimensional ‘volume’ element ${d^{4}p=d^{3}pdp_{0}=d^{3}pdE}$. The four-momentum transforms under Lorentz transformations in the same way as the four-vector representing spacetime, with energy playing the role of time and the three components of ${\mathbf{p}}$ playing the role of the components of ${\mathbf{x}}$. Thus an increment of energy ${dE}$ will be dilated by the factor ${\gamma}$ in the same way that time intervals are dilated, and the momentum ‘volume’ element ${d^{3}p}$ will be contracted by the factor ${1/\gamma}$ in the same way that the spatial volume element ${d^{3}x}$ is contracted in the direction of relative motion. Thus in the 4-volume element ${d^{4}p}$ these two factors cancel out, meaning that ${d^{4}p}$ is Lorentz invariant.

When we integrate over the four components of four-momentum, we must constrain the integral so that it satisfies the relativistic energy-momentum formula

$\displaystyle E^{2}=\mathbf{p}^{2}+m^{2} \ \ \ \ \ (18)$

We can do this by means of a delta-function ${\delta\left(p^{2}-m^{2}\right)}$ where ${p^{2}=p^{\mu}p_{\mu}=E^{2}-\mathbf{p}^{2}}$ is the square of a 4-vector and thus is also Lorentz invariant. Now suppose we have some function ${f\left(p\right)}$ (where ${p}$ is the four-momentum) that is also Lorentz invariant. In that case, the integral

$\displaystyle \left.\int\frac{d^{4}p}{\left(2\pi\right)^{4}}\left(2\pi\right)f\left(p\right)\delta\left(p^{2}-m^{2}\right)\right|_{p^{0}>0} \ \ \ \ \ (19)$

is Lorentz invariant because all of the factors in the integrand are invariant. (The factors of ${2\pi}$ are there for consistency with the rest of the theory.) The subscript ${p^{0}>0}$ reminds us that relativistic energy is always positive, so the integral over ${p^{0}}$ is taken only over this interval. Another way of writing this is to use the Heaviside step function ${\theta\left(p^{0}\right)}$ which is 1 for ${p^{0}>0}$ and 0 for ${p^{0}<0}$:

$\displaystyle \int\frac{d^{4}p}{\left(2\pi\right)^{4}}\left(2\pi\right)f\left(p\right)\delta\left(p^{2}-m^{2}\right)\theta\left(p^{0}\right) \ \ \ \ \ (20)$

We can transform the delta function by using 4, where this time the delta function is taken to be a function of ${p^{0}}$:

$\displaystyle \delta\left(\left(p^{0}\right)^{2}-\mathbf{p}^{2}-m^{2}\right)=\frac{1}{2\sqrt{\mathbf{p}^{2}+m^{2}}}\delta\left(p^{0}\right)=\frac{1}{2E_{\mathbf{p}}}\delta\left(p^{0}\right) \ \ \ \ \ (21)$

The integral over ${p^{0}}$ can now be done with the result

$\displaystyle \left.\int\frac{d^{4}p}{\left(2\pi\right)^{4}}\left(2\pi\right)f\left(p\right)\delta\left(p^{2}-m^{2}\right)\right|_{p^{0}>0}=\int\frac{d^{3}p}{\left(2\pi\right)^{3}}\frac{f\left(E_{\mathbf{p}},\mathbf{p}\right)}{2E_{\mathbf{p}}} \ \ \ \ \ (22)$

where ${f}$ on the RHS is now a function of the four-momentum with energy and 3-momentum properly related to each other, rather than the general ${p}$ that appeared in the integral on the LHS. In particular, if ${f=1}$, the integral

$\displaystyle \int\frac{d^{3}p}{\left(2\pi\right)^{3}}\frac{1}{2E_{\mathbf{p}}} \ \ \ \ \ (23)$

is a Lorentz invariant ‘measure’ integral.

One example of this is ${f=\left|\mathbf{p}\right\rangle \left\langle \mathbf{p}\right|}$, for which the Lorentz invariant integral is

$\displaystyle \mathbf{1}=\int\frac{d^{3}p}{\left(2\pi\right)^{3}}\frac{\left|\mathbf{p}\right\rangle \left\langle \mathbf{p}\right|}{2E_{\mathbf{p}}} \ \ \ \ \ (24)$

This is the Lorentz invariant form of the expansion of the unit operator in terms of momentum states. It can be verified that it gives the correct answer by inserting it into 13 above:

 $\displaystyle \left\langle \mathbf{p}\left|\mathbf{q}\right.\right\rangle$ $\displaystyle =$ $\displaystyle \left\langle \mathbf{p}\right|\int\frac{d^{3}p^{\prime}}{\left(2\pi\right)^{3}}\frac{\left|\mathbf{p}^{\prime}\right\rangle \left\langle \mathbf{p}^{\prime}\right|}{2E_{\mathbf{p}^{\prime}}}\left|\mathbf{q}\right\rangle \ \ \ \ \ (25)$ $\displaystyle$ $\displaystyle =$ $\displaystyle \left\langle \mathbf{p}\right|\int\frac{d^{3}p^{\prime}}{\left(2\pi\right)^{3}}\frac{\left|\mathbf{p}^{\prime}\right\rangle \left\langle \mathbf{p}^{\prime}\left|\mathbf{q}\right.\right\rangle }{2E_{\mathbf{p}^{\prime}}}\ \ \ \ \ (26)$ $\displaystyle$ $\displaystyle =$ $\displaystyle \left\langle \mathbf{p}\right|\int\frac{d^{3}p^{\prime}}{\left(2\pi\right)^{3}}\frac{\left|\mathbf{p}^{\prime}\right\rangle 2E_{\mathbf{p}^{\prime}}\left(2\pi\right)^{3}\delta^{\left(3\right)}\left(\mathbf{p}^{\prime}-\mathbf{q}\right)}{2E_{\mathbf{p}^{\prime}}}\ \ \ \ \ (27)$ $\displaystyle$ $\displaystyle =$ $\displaystyle \left\langle \mathbf{p}\left|\mathbf{q}\right.\right\rangle \ \ \ \ \ (28)$

# Klein-Gordon equation from harmonic oscillator: Hamiltonian, creation and annihilation operators

Michael E. Peskin & Daniel V. Schroeder, An Introduction to Quantum Field Theory, (Perseus Books, 1995) – Chapter 2.

Continuing our discussion of the derivation of the Klein-Gordon field by analogy with the harmonic oscillator, we arrived at the field and its conjugate momentum density:

 $\displaystyle \phi\left(\mathbf{x},t\right)$ $\displaystyle =$ $\displaystyle \frac{1}{\left(2\pi\right)^{3}}\int d^{3}p\;e^{i\mathbf{p\cdot x}}\frac{1}{\sqrt{2\omega_{\mathbf{p}}}}\left(a_{\mathbf{p}}+a_{-\mathbf{p}}^{\dagger}\right)\ \ \ \ \ (1)$ $\displaystyle \pi\left(\mathbf{x},t\right)$ $\displaystyle =$ $\displaystyle \frac{-i}{\left(2\pi\right)^{3}}\int d^{3}p\;e^{i\mathbf{p\cdot x}}\sqrt{\frac{\omega_{\mathbf{p}}}{2}}\left(a_{\mathbf{p}}-a_{-\mathbf{p}}^{\dagger}\right) \ \ \ \ \ (2)$

The operators ${a_{\mathbf{p}}}$ and ${a_{\mathbf{p}}^{\dagger}}$ are analogues of the raising and lowering operators ${a}$ and ${a^{\dagger}}$ in the harmonic oscillator, with the extra condition that we have one pair of operators for each momentum ${\mathbf{p}}$. In the harmonic oscillator, the operators satisfied the commutation relation

$\displaystyle \left[a,a^{\dagger}\right]=1 \ \ \ \ \ (3)$

When applied to the Klein-Gordon field, we assume that operators with different momenta commute, but those with the same momenta do not. The assumption is that

 $\displaystyle \left[a_{\mathbf{p}},a_{\mathbf{p}^{\prime}}^{\dagger}\right]$ $\displaystyle =$ $\displaystyle \left(2\pi\right)^{3}\delta^{\left(3\right)}\left(\mathbf{p}-\mathbf{p}^{\prime}\right)\ \ \ \ \ (4)$ $\displaystyle \left[a_{\mathbf{p}},a_{\mathbf{p}^{\prime}}\right]$ $\displaystyle =$ $\displaystyle 0\ \ \ \ \ (5)$ $\displaystyle \left[a_{\mathbf{p}}^{\dagger},a_{\mathbf{p}^{\prime}}^{\dagger}\right]$ $\displaystyle =$ $\displaystyle 0 \ \ \ \ \ (6)$

[These equations differ from those in Klauber’s book as discussed earlier in that the factors of ${2\pi}$ turn up in different places. However, the final result for the commutator ${\left[\phi\left(\mathbf{x},t\right),\pi\left(\mathbf{x}^{\prime},t\right)\right]}$ is the same, which is what matters.]

From here, we can work out the commutator ${\left[\phi\left(\mathbf{x},t\right),\pi\left(\mathbf{x}^{\prime},t\right)\right]}$ by plugging the operator commutators into the integrals for ${\phi}$ and ${\pi}$. This is similar to the derivation we did earlier, except here we’re assuming the commutators for ${a_{\mathbf{p}}}$ and ${a_{\mathbf{p}}^{\dagger}}$ are as given above, rather than deriving them from the assumed commutator ${\left[\phi\left(\mathbf{x},t\right),\pi\left(\mathbf{x}^{\prime},t\right)\right]}$. This calculation proves to be somewhat easier than the previous one, in that we can throw away all integrals containing ${\left[a_{\mathbf{p}},a_{\mathbf{p}^{\prime}}\right]}$ and ${\left[a_{\mathbf{p}}^{\dagger},a_{\mathbf{p}^{\prime}}^{\dagger}\right]}$. We get

 $\displaystyle \left[\phi\left(\mathbf{x},t\right),\pi\left(\mathbf{x}^{\prime},t\right)\right]$ $\displaystyle =$ $\displaystyle \frac{-i}{2\left(2\pi\right)^{6}}\int d^{3}pd^{3}p^{\prime}\sqrt{\frac{\omega_{\mathbf{p}^{\prime}}}{\omega_{\mathbf{p}}}}\left(\left[a_{-\mathbf{p}}^{\dagger},a_{\mathbf{p}^{\prime}}\right]-\left[a_{\mathbf{p}},a_{-\mathbf{p}^{\prime}}^{\dagger}\right]\right)e^{i\left(\mathbf{p}\cdot\mathbf{x}+\mathbf{p}^{\prime}\cdot\mathbf{x}^{\prime}\right)}\ \ \ \ \ (7)$ $\displaystyle$ $\displaystyle =$ $\displaystyle \frac{-i\left(2\pi\right)^{3}}{2\left(2\pi\right)^{6}}\int d^{3}pd^{3}p^{\prime}\sqrt{\frac{\omega_{\mathbf{p}^{\prime}}}{\omega_{\mathbf{p}}}}\left(-\delta^{\left(3\right)}\left(\mathbf{p}^{\prime}+\mathbf{p}\right)-\delta^{\left(3\right)}\left(\mathbf{p}^{\prime}+\mathbf{p}\right)\right)e^{i\left(\mathbf{p}\cdot\mathbf{x}+\mathbf{p}^{\prime}\cdot\mathbf{x}^{\prime}\right)}\ \ \ \ \ (8)$ $\displaystyle$ $\displaystyle =$ $\displaystyle \frac{i}{\left(2\pi\right)^{3}}\int d^{3}pe^{i\mathbf{p}\cdot\left(\mathbf{x}-\mathbf{x}^{\prime}\right)}\ \ \ \ \ (9)$ $\displaystyle$ $\displaystyle =$ $\displaystyle i\delta^{\left(3\right)}\left(\mathbf{x}-\mathbf{x}^{\prime}\right) \ \ \ \ \ (10)$

This is the same result we got earlier using Klauber’s method. Remember that we’re still taking the field ${\phi}$ to be a real field.

P&S now derive the total Hamiltonian in their equation 2.31. The technique is very similar to that used by Klauber. The differences are (i) we take the momentum ${\mathbf{p}}$ to be continuous rather than the discrete ${\mathbf{k}}$ used by Klauber; (ii) the Klein-Gordon field is real, rather than the two complex fields used by Klauber; (iii) the Hamiltonian density has a factor of ${\frac{1}{2}}$ not found in Klabuer; and (iv) the factors of ${2\pi}$ show up in different places.

P&S’s Hamiltonian density is

$\displaystyle \mathcal{H}=\frac{1}{2}\left[\pi^{2}+\left(\nabla\phi\right)^{2}+m^{2}\phi^{2}\right] \ \ \ \ \ (11)$

From 1 we have

$\displaystyle \nabla\phi=\frac{i}{\left(2\pi\right)^{3}}\int d^{3}p\;\mathbf{p}e^{i\mathbf{p\cdot x}}\frac{1}{\sqrt{2\omega_{\mathbf{p}}}}\left(a_{\mathbf{p}}+a_{-\mathbf{p}}^{\dagger}\right) \ \ \ \ \ (12)$

Plugging this and 2 into 11 and integrating over ${d^{3}x}$ gives the total Hamiltonian

$\displaystyle H=\int d^{3}x\int\frac{d^{3}pd^{3}p^{\prime}}{4\left(2\pi\right)^{6}}e^{i\mathbf{x}\cdot\left(\mathbf{p}+\mathbf{p}^{\prime}\right)}\left(A+B\right) \ \ \ \ \ (13)$

where ${A}$ comes from the ${\pi^{2}}$ term in 11 and ${B}$ comes from ${\left(\nabla\phi\right)^{2}+m^{2}\phi^{2}}$:

 $\displaystyle A$ $\displaystyle =$ $\displaystyle -\sqrt{\omega_{\mathbf{p}}\omega_{\mathbf{p}^{\prime}}}\left(a_{\mathbf{p}}-a_{-\mathbf{p}}^{\dagger}\right)\left(a_{\mathbf{p}^{\prime}}-a_{-\mathbf{p}^{\prime}}^{\dagger}\right)\ \ \ \ \ (14)$ $\displaystyle B$ $\displaystyle =$ $\displaystyle \frac{-\mathbf{p}\cdot\mathbf{p}^{\prime}+m^{2}}{\sqrt{\omega_{\mathbf{p}}\omega_{\mathbf{p}^{\prime}}}}\left(a_{\mathbf{p}}+a_{-\mathbf{p}}^{\dagger}\right)\left(a_{\mathbf{p}^{\prime}}+a_{-\mathbf{p}^{\prime}}^{\dagger}\right) \ \ \ \ \ (15)$

Doing the ${x}$ integral first, we have

$\displaystyle \int d^{3}x\frac{e^{i\mathbf{x}\cdot\left(\mathbf{p}+\mathbf{p}^{\prime}\right)}}{\left(2\pi\right)^{3}}=\delta^{\left(3\right)}\left(\mathbf{p}+\mathbf{p}^{\prime}\right) \ \ \ \ \ (16)$

so we can set ${\mathbf{p}^{\prime}=-\mathbf{p}}$ and eliminate the integral over ${\mathbf{p}^{\prime}}$. Further, using ${\omega_{\mathbf{p}}^{2}=p^{2}+m^{2}}$ converts ${A}$ and ${B}$ to

 $\displaystyle A$ $\displaystyle =$ $\displaystyle -\omega_{\mathbf{p}}\left(a_{\mathbf{p}}-a_{-\mathbf{p}}^{\dagger}\right)\left(a_{-\mathbf{p}}-a_{\mathbf{p}}^{\dagger}\right)\ \ \ \ \ (17)$ $\displaystyle B$ $\displaystyle =$ $\displaystyle \omega_{\mathbf{p}}\left(a_{\mathbf{p}}+a_{-\mathbf{p}}^{\dagger}\right)\left(a_{-\mathbf{p}}+a_{\mathbf{p}}^{\dagger}\right)\ \ \ \ \ (18)$ $\displaystyle A+B$ $\displaystyle =$ $\displaystyle 2\omega_{\mathbf{p}}\left(a_{\mathbf{p}}a_{\mathbf{p}}^{\dagger}+a_{-\mathbf{p}}^{\dagger}a_{-\mathbf{p}}\right) \ \ \ \ \ (19)$

Since we’re integrating over all ${\mathbf{p}}$ we can replace ${-\mathbf{p}}$ by ${\mathbf{p}}$ in the last term, and use 4 on the first term:

 $\displaystyle H$ $\displaystyle =$ $\displaystyle \int\frac{d^{3}p}{4\left(2\pi\right)^{3}}\left(A+B\right)=2\int\frac{d^{3}p}{4\left(2\pi\right)^{3}}\omega_{\mathbf{p}}\left(2a_{\mathbf{p}}^{\dagger}a_{\mathbf{p}}+\left[a_{\mathbf{p}},a_{\mathbf{p}}^{\dagger}\right]\right)\ \ \ \ \ (20)$ $\displaystyle$ $\displaystyle =$ $\displaystyle \int\frac{d^{3}p}{\left(2\pi\right)^{3}}\omega_{\mathbf{p}}\left(a_{\mathbf{p}}^{\dagger}a_{\mathbf{p}}+\frac{1}{2}\left[a_{\mathbf{p}},a_{\mathbf{p}}^{\dagger}\right]\right) \ \ \ \ \ (21)$

Since the commutator in the last term has two operators both with suffix ${\mathbf{p}}$, it is an infinite quantity, so its integral is infinite. This is swept under the carpet by saying that since this energy is present in all states and it’s only the difference between a given state and the ground state that can be measured, we can ignore it. In the harmonic oscillator, the ground state ${\left|0\right\rangle }$ has energy ${\frac{1}{2}\omega}$ so this infinite term can be thought of as the sum of this zero-point energy over all momentem states.

In field theory, a state ${\left|0\right\rangle }$ is postulated such that ${a_{\mathbf{p}}\left|0\right\rangle =0}$ for all ${\mathbf{p}}$. This is the vacuum state, which has the infinite zero-point energy. If we operate on ${\left|0\right\rangle }$ with ${a_{\mathbf{p}}^{\dagger}}$ this produces an eigenstate of ${H}$ as we can show using the commutator 4 and ignoring the infinite energy term:

 $\displaystyle Ha_{\mathbf{p}}^{\dagger}\left|0\right\rangle$ $\displaystyle =$ $\displaystyle \int\frac{d^{3}p^{\prime}}{\left(2\pi\right)^{3}}\omega_{\mathbf{p}^{\prime}}a_{\mathbf{\mathbf{p}^{\prime}}}^{\dagger}a_{\mathbf{p}^{\prime}}a_{\mathbf{p}}^{\dagger}\left|0\right\rangle \ \ \ \ \ (22)$ $\displaystyle$ $\displaystyle =$ $\displaystyle \int\frac{d^{3}p^{\prime}}{\left(2\pi\right)^{3}}\omega_{\mathbf{p}^{\prime}}a_{\mathbf{\mathbf{p}^{\prime}}}^{\dagger}\left(a_{\mathbf{p}}^{\dagger}a_{\mathbf{p}^{\prime}}+\left(2\pi\right)^{3}\delta\left(\mathbf{p}-\mathbf{p}^{\prime}\right)\right)\left|0\right\rangle \ \ \ \ \ (23)$ $\displaystyle$ $\displaystyle =$ $\displaystyle \left[\int\frac{d^{3}p^{\prime}}{\left(2\pi\right)^{3}}\omega_{\mathbf{p}^{\prime}}a_{\mathbf{\mathbf{p}^{\prime}}}^{\dagger}a_{\mathbf{p}}^{\dagger}a_{\mathbf{p}^{\prime}}\left|0\right\rangle \right]+\omega_{\mathbf{p}}a_{\mathbf{p}}^{\dagger}\left|0\right\rangle \ \ \ \ \ (24)$ $\displaystyle$ $\displaystyle =$ $\displaystyle 0+\omega_{\mathbf{p}}a_{\mathbf{p}}^{\dagger}\left|0\right\rangle \ \ \ \ \ (25)$ $\displaystyle$ $\displaystyle =$ $\displaystyle \omega_{\mathbf{p}}a_{\mathbf{p}}^{\dagger}\left|0\right\rangle \ \ \ \ \ (26)$

The operator ${a_{\mathbf{p}}^{\dagger}}$ acts on the vacuum to create an excited state with energy ${\omega_{\mathbf{p}}}$ in the same way that the ${a^{\dagger}}$ operator in the harmonic oscillator operates on the ground state to produce an oscillator in the next highest energy state. In field theory, this excitation is called a particle, so ${a_{\mathbf{p}}^{\dagger}}$ becomes a creation operator that creates a particle with energy ${\omega_{\mathbf{p}}}$. A similar calculation shows that ${a_{\mathbf{p}}}$, acting on a state containing a particle of energy ${\omega_{\mathbf{p}}}$, removes this particle from the state and produces an eigenstate with energy lowered by ${\omega_{\mathbf{p}}}$, so ${a_{\mathbf{p}}}$ is called an annihilation operator. These operators can produce and destroy multiple particles within a single state.

# Klein-Gordon solutions from harmonic oscillator

Michael E. Peskin & Daniel V. Schroeder, An Introduction to Quantum Field Theory, (Perseus Books, 1995) – Chapter 2.

Another way of deriving the Klein-Gordon field and its conjugate momentum density is by drawing an analogy with the harmonic oscillator. The idea is to start with the classical Klein-Gordon field ${\phi\left(\mathbf{x},t\right)}$ and expand it by using a Fourier transform:

$\displaystyle \phi\left(\mathbf{x},t\right)=\frac{1}{\left(2\pi\right)^{3}}\int d^{3}p\;e^{i\mathbf{p\cdot x}}\phi\left(\mathbf{p},t\right) \ \ \ \ \ (1)$

Since ${\phi\left(\mathbf{x},t\right)}$ is a classical field, it must be real, which we can ensure by requiring that ${\phi^*\left(\mathbf{p},t\right)=\phi\left(-\mathbf{p},t\right)}$. Plugging this into the Klein-Gordon equation gives

 $\displaystyle \partial_{\mu}\partial^{\mu}\phi\left(\mathbf{x},t\right)+m^{2}\left(\mathbf{x},t\right)$ $\displaystyle =$ $\displaystyle 0\ \ \ \ \ (2)$ $\displaystyle \frac{1}{\left(2\pi\right)^{3}}\int d^{3}p\;\left[\frac{\partial^{2}}{\partial t^{2}}-\nabla^{2}+m^{2}\right]e^{i\mathbf{p\cdot x}}\phi\left(\mathbf{p},t\right)$ $\displaystyle =$ $\displaystyle 0\ \ \ \ \ (3)$ $\displaystyle \frac{1}{\left(2\pi\right)^{3}}\int d^{3}p\;e^{i\mathbf{p\cdot x}}\left[\frac{\partial^{2}}{\partial t^{2}}+\left|\mathbf{p}\right|^{2}+m^{2}\right]\phi\left(\mathbf{p},t\right)$ $\displaystyle =$ $\displaystyle 0 \ \ \ \ \ (4)$

Since this must be true for all ${\phi\left(\mathbf{p},t\right)}$, the integrand must be zero so we get

$\displaystyle \left[\frac{\partial^{2}}{\partial t^{2}}+\left|\mathbf{p}\right|^{2}+m^{2}\right]\phi\left(\mathbf{p},t\right)=0 \ \ \ \ \ (5)$

The classical equation of motion for a harmonic oscillator in one dimension is

 $\displaystyle F$ $\displaystyle =$ $\displaystyle -kx\ \ \ \ \ (6)$ $\displaystyle m_{0}\frac{d^{2}x}{dt^{2}}+kx$ $\displaystyle =$ $\displaystyle 0 \ \ \ \ \ (7)$

The solution is an oscillator with frequency

$\displaystyle \omega=\sqrt{\frac{k}{m_{0}}} \ \ \ \ \ (8)$

The corresponding Hamiltonian is

$\displaystyle H=\frac{p^{2}}{2m_{0}}+\frac{1}{2}kx^{2} \ \ \ \ \ (9)$

Comparing this with 5, we see that the Klein-Gordon equation in momentum space has the same form as a harmonic oscillator with ${x}$ replaced by ${\mathbf{p}}$, ${k=\left|\mathbf{p}\right|^{2}+m^{2}}$ and ${m_{0}=1}$, so its solution is an oscillator with frequency

$\displaystyle \omega_{\mathbf{p}}=\sqrt{\left|\mathbf{p}\right|^{2}+m^{2}} \ \ \ \ \ (10)$

In the algebraic solution of the harmonic oscillator in non-relativistic quantum mechanics, we introduced the raising and lowering operators (renaming ${a_{+}=a^{\dagger}}$ and ${a_{-}=a}$ to be consistent with P&S’s notation):

 $\displaystyle a^{\dagger}$ $\displaystyle =$ $\displaystyle \frac{1}{\sqrt{2\hbar m_{0}\omega}}\left[-ip+m_{0}\omega x\right]\ \ \ \ \ (11)$ $\displaystyle a$ $\displaystyle =$ $\displaystyle \frac{1}{\sqrt{2\hbar m_{0}\omega}}\left[ip+m_{0}\omega x\right] \ \ \ \ \ (12)$

In natural units ${\hbar=1}$ and with ${m_{0}=1}$ this gives

 $\displaystyle a^{\dagger}$ $\displaystyle =$ $\displaystyle \frac{1}{\sqrt{2\omega}}\left[-ip+\omega x\right]\ \ \ \ \ (13)$ $\displaystyle a$ $\displaystyle =$ $\displaystyle \frac{1}{\sqrt{2\omega}}\left[ip+\omega x\right] \ \ \ \ \ (14)$

which can be inverted to give

 $\displaystyle x$ $\displaystyle =$ $\displaystyle \frac{1}{\sqrt{2\omega}}\left(a+a^{\dagger}\right)\ \ \ \ \ (15)$ $\displaystyle p$ $\displaystyle =$ $\displaystyle -i\sqrt{\frac{\omega}{2}}\left(a-a^{\dagger}\right) \ \ \ \ \ (16)$

Using the commutation relation

$\displaystyle \left[x,p\right]=i \ \ \ \ \ (17)$

we get

$\displaystyle \left[a,a^{\dagger}\right]=1 \ \ \ \ \ (18)$

and the Hamiltonian comes out to

$\displaystyle H_{SHO}=\omega\left(a^{\dagger}a+\frac{1}{2}\right) \ \ \ \ \ (19)$

The effects of ${a}$ and ${a^{\dagger}}$ are to lower and raise a state by an energy quantum ${\omega}$. The wave functions are

 $\displaystyle a\left|n\right\rangle$ $\displaystyle =$ $\displaystyle \sqrt{n}\left|n-1\right\rangle \ \ \ \ \ (20)$ $\displaystyle a^{\dagger}\left|n\right\rangle$ $\displaystyle =$ $\displaystyle \sqrt{n+1}\left|n+1\right\rangle \ \ \ \ \ (21)$

To apply all this to the quantum Klein-Gordon equation, we want to interpret ${x}$ in 15 as a quantum field ${\phi\left(\mathbf{p}\right)}$, and leave ${p}$ in 16 as it is. The field ${\phi\left(\mathbf{x}\right)}$ in position space is the integral 1 over ${\phi\left(\mathbf{p}\right)}$ in momentum space, with each momentum ${\mathbf{p}}$ contributing its own raising and lowering operators ${a_{\mathbf{p}}}$ and ${a_{-\mathbf{p}}^{\dagger}}$ from 13 and 14. That is, we get

 $\displaystyle \phi\left(\mathbf{x}\right)$ $\displaystyle =$ $\displaystyle \frac{1}{\left(2\pi\right)^{3}}\int d^{3}p\;e^{i\mathbf{p\cdot x}}\frac{1}{\sqrt{2\omega_{\mathbf{p}}}}\left(a_{\mathbf{p}}+a_{-\mathbf{p}}^{\dagger}\right)\ \ \ \ \ (22)$ $\displaystyle \pi\left(\mathbf{x}\right)$ $\displaystyle =$ $\displaystyle \frac{-i}{\left(2\pi\right)^{3}}\int d^{3}p\;e^{i\mathbf{p\cdot x}}\sqrt{\frac{\omega_{\mathbf{p}}}{2}}\left(a_{\mathbf{p}}-a_{-\mathbf{p}}^{\dagger}\right) \ \ \ \ \ (23)$

We can invert these equations to get expressions for ${a}$ and ${a^{\dagger}}$. We multiply both sides by ${e^{-i\mathbf{p^{\prime}\cdot x}}}$ and integrate over ${d^{3}x}$:

 $\displaystyle \int d^{3}xe^{-i\mathbf{p^{\prime}\cdot x}}\phi\left(\mathbf{x}\right)$ $\displaystyle =$ $\displaystyle \int d^{3}p\left(\frac{1}{\left(2\pi\right)^{3}}\int d^{3}xe^{i\mathbf{\left(p-p^{\prime}\right)\cdot x}}\right)\frac{1}{\sqrt{2\omega_{\mathbf{p}}}}\left(a_{\mathbf{p}}+a_{-\mathbf{p}}^{\dagger}\right)\ \ \ \ \ (24)$ $\displaystyle$ $\displaystyle =$ $\displaystyle \int d^{3}p\delta\left(\mathbf{p-p}^{\prime}\right)\frac{1}{\sqrt{2\omega_{\mathbf{p}}}}\left(a_{\mathbf{p}}+a_{-\mathbf{p}}^{\dagger}\right)\ \ \ \ \ (25)$ $\displaystyle$ $\displaystyle =$ $\displaystyle \frac{1}{\sqrt{2\omega_{\mathbf{p}^{\prime}}}}\left(a_{\mathbf{p}^{\prime}}+a_{-\mathbf{p}^{\prime}}^{\dagger}\right)\ \ \ \ \ (26)$ $\displaystyle \int d^{3}xe^{-i\mathbf{p^{\prime}\cdot x}}\pi\left(\mathbf{x}\right)$ $\displaystyle =$ $\displaystyle -i\sqrt{\frac{\omega_{\mathbf{p}^{\prime}}}{2}}\left(a_{\mathbf{p}^{\prime}}-a_{-\mathbf{p}^{\prime}}^{\dagger}\right) \ \ \ \ \ (27)$

Dropping the primes on ${\mathbf{p}^{\prime}}$ and solving for ${a}$ and ${a^{\dagger}}$ we get (since ${\omega_{\mathbf{p}}=\omega_{-\mathbf{p}}}$):

 $\displaystyle a_{\mathbf{p}}$ $\displaystyle =$ $\displaystyle \int d^{3}x\frac{1}{\sqrt{2\omega_{\mathbf{p}}}}e^{-i\mathbf{p\cdot x}}\left[i\pi\left(\mathbf{x}\right)+\omega_{\mathbf{p}}\phi\left(\mathbf{x}\right)\right]\ \ \ \ \ (28)$ $\displaystyle a_{\mathbf{p}}^{\dagger}$ $\displaystyle =$ $\displaystyle \int d^{3}x\frac{1}{\sqrt{2\omega_{\mathbf{p}}}}e^{i\mathbf{p\cdot x}}\left[-i\pi\left(\mathbf{x}\right)+\omega_{\mathbf{p}}\phi\left(\mathbf{x}\right)\right] \ \ \ \ \ (29)$