# Hamiltonian in non-rectangular coordinates

Shankar, R. (1994), Principles of Quantum Mechanics, Plenum Press. Section 7.4, Exercise 7.4.10.

The standard procedure for quantizing a classical hamiltonian is to write the classical hamiltonian in terms of position and momentum variables in rectangular coordinates and then convert the position and momentum variables to operators satisfying the usual commutation relations. However, in some cases, another coordinate system makes solving the differential equation resulting from the Schrödinger equation easier (as, for example, with the hydrogen atom, where the system has spherical symmetry).

As a 2-d example, suppose we have the classical hamiltonian

$\displaystyle H=\frac{p_{x}^{2}+p_{y}^{2}}{2m}+a\sqrt{x^{2}+y^{2}} \ \ \ \ \ (1)$

for some constant ${a}$. Since the system has radial symmetry, polar coordinates should make things easier. That is, we’d like to transform to

 $\displaystyle \rho$ $\displaystyle =$ $\displaystyle \sqrt{x^{2}+y^{2}}\ \ \ \ \ (2)$ $\displaystyle \phi$ $\displaystyle =$ $\displaystyle \arctan\frac{y}{x} \ \ \ \ \ (3)$

In the rectangular position basis, the quantized operators are

 $\displaystyle P_{x}$ $\displaystyle =$ $\displaystyle -i\hbar\frac{\partial}{\partial x}\ \ \ \ \ (4)$ $\displaystyle P_{y}$ $\displaystyle =$ $\displaystyle -i\hbar\frac{\partial}{\partial y}\ \ \ \ \ (5)$ $\displaystyle X$ $\displaystyle =$ $\displaystyle x\ \ \ \ \ (6)$ $\displaystyle Y$ $\displaystyle =$ $\displaystyle y \ \ \ \ \ (7)$

so the quantum hamiltonian is

$\displaystyle H=-\frac{\hbar^{2}}{2m}\left(\frac{\partial^{2}}{\partial x^{2}}+\frac{\partial^{2}}{\partial y^{2}}\right)+a\sqrt{x^{2}+y^{2}} \ \ \ \ \ (8)$

The first term contains the Laplacian derivative operator, which can be written in polar coordinates as

$\displaystyle \frac{\partial^{2}}{\partial x^{2}}+\frac{\partial^{2}}{\partial y^{2}}=\frac{\partial^{2}}{\partial\rho^{2}}+\frac{1}{\rho}\frac{\partial}{\partial\rho}+\frac{1}{\rho^{2}}\frac{\partial^{2}}{\partial\phi^{2}} \ \ \ \ \ (9)$

Thus the quantum hamiltonian in polar coordinates is

$\displaystyle H=-\frac{\hbar^{2}}{2m}\left(\frac{\partial^{2}}{\partial\rho^{2}}+\frac{1}{\rho}\frac{\partial}{\partial\rho}+\frac{1}{\rho^{2}}\frac{\partial^{2}}{\partial\phi^{2}}\right)+a\rho \ \ \ \ \ (10)$

The question is: can we instead convert the hamiltonian 1 to polar coordinates and then quantize the result, rather than converting the rectangular coordinates after the hamiltonian is written? The answer turns out to be surprisingly complicated, and I’m not sure I follow everything Shankar says, but here’s the argument anyway. Comments, as usual, are welcome.

We first convert the rectangular momentum coordinates to polar momentum coordinates by means of the substitutions

 $\displaystyle p_{\rho}$ $\displaystyle =$ $\displaystyle \hat{\mathbf{r}}\cdot\mathbf{p}=\frac{xp_{x}+yp_{y}}{\sqrt{x^{2}+y^{2}}}\ \ \ \ \ (11)$ $\displaystyle p_{\phi}=\ell_{z}$ $\displaystyle =$ $\displaystyle xp_{y}-yp_{x} \ \ \ \ \ (12)$

Note that the two components of polar momentum have different units: ${p_{\rho}}$ has the dimensions of linear momentum while ${p_{\phi}}$ is actually the angular momentum about the ${z}$ axis ${\ell_{z}}$. In terms of these new momenta, the classical hamiltonian 1 becomes

$\displaystyle H=\frac{p_{\rho}^{2}}{2m}+\frac{p_{\phi}^{2}}{2m\rho^{2}}+a\rho \ \ \ \ \ (13)$

This can be verified either by inverting equations 11 and 12 to get ${p_{x}}$ and ${p_{y}}$ in terms of ${p_{\rho}}$ and ${p_{\phi}}$ and then plugging these into 1 (very messy), or else just starting with 13 and showing it reduces to 1. We’ll do the latter.

 $\displaystyle p_{\rho}^{2}+\frac{p_{\phi}^{2}}{\rho^{2}}$ $\displaystyle =$ $\displaystyle \left[\frac{xp_{x}+yp_{y}}{\sqrt{x^{2}+y^{2}}}\right]^{2}+\frac{\left(xp_{y}-yp_{x}\right)^{2}}{x^{2}+y^{2}}\ \ \ \ \ (14)$ $\displaystyle$ $\displaystyle =$ $\displaystyle \frac{1}{\rho^{2}}\left(x^{2}p_{x}^{2}+y^{2}p_{y}^{2}+2xyp_{x}p_{y}+x^{2}p_{y}^{2}+y^{2}p_{x}^{2}-2xyp_{x}p_{y}\right)\ \ \ \ \ (15)$ $\displaystyle$ $\displaystyle =$ $\displaystyle \frac{1}{\rho^{2}}\left(x^{2}+y^{2}\right)\left(p_{x}^{2}+p_{y}^{2}\right)\ \ \ \ \ (16)$ $\displaystyle$ $\displaystyle =$ $\displaystyle p_{x}^{2}+p_{y}^{2} \ \ \ \ \ (17)$

We can now try quantizing 13 by creating a couple of quantum momentum operators according to the standard rule:

 $\displaystyle P_{\rho}$ $\displaystyle =$ $\displaystyle -i\hbar\frac{\partial}{\partial\rho}\ \ \ \ \ (18)$ $\displaystyle P_{\phi}$ $\displaystyle =$ $\displaystyle -i\hbar\frac{\partial}{\partial\phi} \ \ \ \ \ (19)$

These operators satisfy the usual commutation rule, in the sense that

$\displaystyle \left[\rho,P_{\rho}\right]=\left[\phi,P_{\phi}\right]=i\hbar \ \ \ \ \ (20)$

However, substituting them into 13 gives

$\displaystyle H=-\frac{\hbar^{2}}{2m}\left(\frac{\partial^{2}}{\partial\rho^{2}}+\frac{1}{\rho^{2}}\frac{\partial^{2}}{\partial\phi^{2}}\right)+a\rho \ \ \ \ \ (21)$

Comparing with 10 we see that the middle term with the first order derivative is missing. The problem is due to the fact that 18 is actually not a hermitian operator, which we can see by calculating the bracket as follows:

$\displaystyle \left\langle \psi_{1}\left|P_{\rho}\right|\psi_{2}\right\rangle =-i\hbar\int_{0}^{2\pi}\int_{0}^{\infty}\psi_{1}^*\frac{\partial\psi_{2}}{\partial\rho}\rho\;d\rho\;d\phi \ \ \ \ \ (22)$

We can do the ${\rho}$ integral by parts and, assuming that ${\rho\psi_{1}^*\psi_{2}\rightarrow0}$ at both ${\rho\rightarrow0}$ and ${\rho\rightarrow\infty}$, we have

 $\displaystyle \int_{0}^{\infty}\psi_{1}^*\frac{\partial\psi_{2}}{\partial\rho}\rho\;d\rho$ $\displaystyle =$ $\displaystyle \left.\rho\psi_{1}^*\psi_{2}\right|_{0}^{\infty}-\int_{0}^{\infty}\psi_{2}\frac{\partial\left(\rho\psi_{1}^*\right)}{\partial\rho}\;d\rho\ \ \ \ \ (23)$ $\displaystyle$ $\displaystyle =$ $\displaystyle -\int_{0}^{\infty}\psi_{2}\frac{\partial\psi_{1}^*}{\partial\rho}\rho\;d\rho-\int_{0}^{\infty}\psi_{1}^*\psi_{2}d\rho \ \ \ \ \ (24)$

Substituting back into 22 we get

$\displaystyle \left\langle \psi_{1}\left|P_{\rho}\right|\psi_{2}\right\rangle =i\hbar\int_{0}^{2\pi}\int_{0}^{\infty}\left[\psi_{2}\frac{\partial\psi_{1}^*}{\partial\rho}\rho+\psi_{1}^*\psi_{2}\right]d\rho\;d\phi \ \ \ \ \ (25)$

If ${P_{\rho}}$ is to be hermitian, we need to satisfy

 $\displaystyle \left\langle \psi_{1}\left|P_{\rho}\right|\psi_{2}\right\rangle$ $\displaystyle =$ $\displaystyle \left\langle P_{\rho}\psi_{1}\left|\psi_{2}\right.\right\rangle \ \ \ \ \ (26)$ $\displaystyle$ $\displaystyle =$ $\displaystyle i\hbar\int_{0}^{2\pi}\int_{0}^{\infty}\psi_{2}\frac{\partial\psi_{1}^*}{\partial\rho}\rho\;d\rho\;d\phi \ \ \ \ \ (27)$

We can see that the presence of the second term in the integrand of 25 messes things up. This term arises from the presence of the extra factor of ${\rho}$ that is present in a polar area integral.

We can, in fact, attempt to fix this by defining the radial momentum operator to be, instead of 18:

$\displaystyle P_{\rho}=-i\hbar\left(\frac{\partial}{\partial\rho}+\frac{1}{2\rho}\right) \ \ \ \ \ (28)$

We first verify that this is hermitian:

 $\displaystyle \left\langle \psi_{1}\left|P_{\rho}\right|\psi_{2}\right\rangle$ $\displaystyle =$ $\displaystyle -i\hbar\int_{0}^{2\pi}\int_{0}^{\infty}\psi_{1}^*\left[\frac{\partial\psi_{2}}{\partial\rho}\rho+\frac{\psi_{2}}{2}\right]\;d\rho\;d\phi\ \ \ \ \ (29)$ $\displaystyle$ $\displaystyle =$ $\displaystyle i\hbar\int_{0}^{2\pi}\int_{0}^{\infty}\left[\psi_{2}\frac{\partial\psi_{1}^*}{\partial\rho}\rho+\psi_{1}^*\psi_{2}-\frac{1}{2}\psi_{1}^*\psi_{2}\right]d\rho\;d\phi\ \ \ \ \ (30)$ $\displaystyle$ $\displaystyle =$ $\displaystyle i\hbar\int_{0}^{2\pi}\int_{0}^{\infty}\left[\psi_{2}\frac{\partial\psi_{1}^*}{\partial\rho}\rho+\frac{1}{2}\psi_{1}^*\psi_{2}\right]d\rho\;d\phi\ \ \ \ \ (31)$ $\displaystyle$ $\displaystyle =$ $\displaystyle \left\langle P_{\rho}\psi_{1}\left|\psi_{2}\right.\right\rangle \ \ \ \ \ (32)$

In the second line we did the same integration by parts on the first term and used the result in 24. Thus this new ${P_{\rho}}$ is indeed hermitian. If we now insert this along with the old ${P_{\phi}}$ from 19 into 13 we get

$\displaystyle H=-\frac{\hbar^{2}}{2m}\left(\frac{\partial}{\partial\rho}+\frac{1}{2\rho}\right)^{2}+a\rho \ \ \ \ \ (33)$

To work out the differential part of the hamiltonian we can apply it to a test function.

 $\displaystyle \left(\frac{\partial}{\partial\rho}+\frac{1}{2\rho}\right)^{2}\psi$ $\displaystyle =$ $\displaystyle \left(\frac{\partial}{\partial\rho}+\frac{1}{2\rho}\right)\left(\frac{\partial\psi}{\partial\rho}+\frac{\psi}{2\rho}\right)\ \ \ \ \ (34)$ $\displaystyle$ $\displaystyle =$ $\displaystyle \frac{\partial^{2}\psi}{\partial\rho^{2}}+\frac{1}{2\rho}\frac{\partial\psi}{\partial\rho}-\frac{\psi}{2\rho^{2}}+\frac{1}{2\rho}\frac{\partial\psi}{\partial\rho}+\frac{\psi}{4\rho^{2}}\ \ \ \ \ (35)$ $\displaystyle$ $\displaystyle =$ $\displaystyle \frac{\partial^{2}\psi}{\partial\rho^{2}}+\frac{1}{\rho}\frac{\partial\psi}{\partial\rho}-\frac{\psi}{4\rho^{2}} \ \ \ \ \ (36)$

The hamiltonian then becomes

$\displaystyle H=-\frac{\hbar^{2}}{2m}\left(\frac{\partial^{2}}{\partial\rho^{2}}+\frac{1}{\rho}\frac{\partial}{\partial\rho}-\frac{1}{4\rho^{2}}+\frac{1}{\rho^{2}}\frac{\partial^{2}}{\partial\phi^{2}}\right)+a\rho \ \ \ \ \ (37)$

Comparing this with 10 we see that now we have an extra term ${-\frac{1}{4\rho^{2}}}$. Shankar doesn’t really explain in detail what the problem is, except to state that when converting from a classical to a quantum hamiltonian, terms of order ${\hbar}$ or higher may be present in the quantum version that are absent in the classical version. Presumably he means terms of order ${\hbar}$ that don’t involve derivatives, since the entire momentum-dependent part of the hamiltonian is multiplied by a factor of ${\hbar^{2}}$. In any case, we’ll have to leave it at that.

# Relation between action and energy

References: Shankar, R. (1994), Principles of Quantum Mechanics, Plenum Press. Section 2.8; Exercises 2.8.6 – 2.8.7.

Here we’ll examine an interesting relation between the action ${S}$ and the total energy of a system, as given by the Hamiltonian ${H}$. Suppose a single particle moving in one dimension follows a classical path given by ${x_{cl}\left(t\right)}$, and moves from an initial position at time ${t_{i}}$ of ${x_{cl}\left(t_{i}\right)=x_{i}}$ to a final position at time ${t_{f}}$ of ${x_{cl}\left(t_{f}\right)=x_{f}}$. The action ${S_{cl}}$ of this classical path is given by the integral of the Lagrangian

$\displaystyle S_{cl}=\int_{t_{i}}^{t_{f}}L\left(x,\dot{x}\right)dt \ \ \ \ \ (1)$

What can we say about the rate of change of the action with respect to the final time ${t_{f}}$? That is, we want to calculate ${\partial S_{cl}/\partial t_{f}}$, where all other parameters ${t_{i},x_{i}}$and ${x_{f}}$ are held constant. The situation can be illustrated as shown:

Since the only thing that is changing is ${t_{f}}$, the particle starts at the same initial time (which we’ve taken to be ${t_{i}=0}$ in the diagram) and moves to the same location ${x_{f}}$, but at a different time (in the diagram, later time). This means that the particle must follow a different path, possibly over its entire trajectory. This path, which we’ll call ${x\left(t\right)}$, is related to the original path ${x_{cl}\left(t\right)}$ by perturbing the original path by an amount ${\eta\left(t\right)}$:

$\displaystyle x\left(t\right)=x_{cl}\left(t\right)+\eta\left(t\right) \ \ \ \ \ (2)$

In the diagram, the original path ${x_{cl}}$ is shown in red and the perturbed path ${x}$ in blue. The amount ${\eta}$ is seen to be the vertical distance between these two curves at each time, and in the case of the paths shown in the diagram, ${\eta\left(t\right)<0}$.

The difference in the action between the two paths is due to two contributions: first, there is the contribution due to the extra time, from ${t_{f}}$ to ${t_{f}+\Delta t}$, that the particle takes to complete its path. Second, there is the difference in the two actions over the path from ${t_{i}}$ to ${t_{f}}$. The first contribution is entirely new and, for an infinitesimal extra time ${\Delta t}$, it is given by

$\displaystyle \delta S_{1}=L\left(t_{f}\right)\Delta t \ \ \ \ \ (3)$

where ${L\left(t_{f}\right)}$ is the Lagrangian evaluated at time ${t_{f}}$. The other contribution can be obtained by varying the action over the path from ${t_{i}=0}$ to ${t_{f}}$:

$\displaystyle \delta S_{2}=\int_{0}^{t_{f}}\delta L\;dt \ \ \ \ \ (4)$

Since ${L}$ depends on ${x}$ and ${\dot{x}}$, we have

$\displaystyle \delta L=\frac{\partial L}{\partial x}\delta x+\frac{\partial L}{\partial\dot{x}}\delta\dot{x} \ \ \ \ \ (5)$

For infinitesimally different trajectories, we can see from the diagram above that ${\delta x=\eta\left(t\right)}$ at each point on the curve, so ${\delta\dot{x}=\dot{\eta}\left(t\right)}$, so we get

 $\displaystyle \delta S_{2}$ $\displaystyle =$ $\displaystyle \int_{0}^{t_{f}}\left[\frac{\partial L}{\partial x}\eta\left(t\right)+\frac{\partial L}{\partial\dot{x}}\dot{\eta}\left(t\right)\right]\;dt\ \ \ \ \ (6)$ $\displaystyle$ $\displaystyle =$ $\displaystyle \int_{0}^{t_{f}}\left[-\frac{d}{dt}\frac{\partial L}{\partial\dot{x}}+\frac{\partial L}{\partial x}\right]\eta\left(t\right)dt+\int_{0}^{t_{f}}\frac{d}{dt}\left(\frac{\partial L}{\partial\dot{x}}\eta\left(t\right)\right)dt\ \ \ \ \ (7)$ $\displaystyle$ $\displaystyle =$ $\displaystyle 0+\left.\frac{\partial L}{\partial\dot{x}}\eta\left(t\right)\right|_{t_{f}} \ \ \ \ \ (8)$

In these equations, the derivatives of ${L}$ are evaluated on the original curve ${x_{cl}}$. To verify the second line, use the product rule on the second integrand and cancel terms to get the first line. The second term in the last is evaluated at ${t=t_{f}}$ only since we’re assuming that ${\eta\left(0\right)=0}$.

The quantity in brackets in the first integral is zero, because of the Euler-Lagrange equations which are valid on the original curve ${x_{cl}}$:

$\displaystyle \frac{d}{dt}\frac{\partial L}{\partial\dot{x}}-\frac{\partial L}{\partial x}=0 \ \ \ \ \ (9)$

Putting everything together, we get for the total variation in the action:

 $\displaystyle \delta S_{cl}$ $\displaystyle =$ $\displaystyle \delta S_{1}+\delta S_{2}\ \ \ \ \ (10)$ $\displaystyle$ $\displaystyle =$ $\displaystyle \left[\frac{\partial L}{\partial\dot{x}}\eta\left(t\right)+L\Delta t\right]_{t_{f}} \ \ \ \ \ (11)$

Looking at the diagram above, the slope of the blue curve ${x\left(t_{f}\right)}$ at the time ${t_{f}}$ is given by

$\displaystyle \dot{x}\left(t_{f}\right)=\frac{\left|\eta\left(t_{f}\right)\right|}{\Delta t} \ \ \ \ \ (12)$

From the definition 2 of ${\eta}$ we see that ${\eta\left(t_{f}\right)<0}$, so

$\displaystyle \eta\left(t_{f}\right)=-\dot{x}\left(t_{f}\right)\Delta t \ \ \ \ \ (13)$

This gives the final equation for the variation of the action:

 $\displaystyle \delta S_{cl}$ $\displaystyle =$ $\displaystyle \left[-\frac{\partial L}{\partial\dot{x}}\dot{x}+L\right]_{t_{f}}\Delta t\ \ \ \ \ (14)$ $\displaystyle$ $\displaystyle =$ $\displaystyle \left(-p\dot{x}+L\right)\Delta t\ \ \ \ \ (15)$ $\displaystyle$ $\displaystyle =$ $\displaystyle -H\Delta t \ \ \ \ \ (16)$

where the second line follows from the definition of the canonical momentum ${p=\partial L/\partial\dot{x}}$.

The required derivative is

$\displaystyle \boxed{\frac{\partial S_{cl}}{\partial t_{f}}=-H\left(t_{f}\right)} \ \ \ \ \ (17)$

Using a similar technique, we can work out ${\partial S_{cl}/\partial x_{f}}$. In this case, the situation is as shown in this diagram:

The two trajectories now take the same time, but in the modified trajectory, the particle moves a distance ${\Delta x}$ further. Since both paths take the same time, there is no extra contribution ${L\Delta t}$. In this case ${\eta\left(t\right)>0}$, since the new (blue) curve ${x\left(t\right)}$ is above the old (red) one ${x_{cl}\left(t\right)}$. The derivation is the same as above up to 8, and the total variation in the action is now

$\displaystyle \delta S_{cl}=\left.\frac{\partial L}{\partial\dot{x}}\eta\left(t\right)\right|_{t_{f}} \ \ \ \ \ (18)$

At ${t=t_{f}}$, ${\eta\left(t_{f}\right)=\Delta x}$, so we get

 $\displaystyle \delta S_{cl}$ $\displaystyle =$ $\displaystyle \left.\frac{\partial L}{\partial\dot{x}}\right|_{t_{f}}\Delta x\ \ \ \ \ (19)$ $\displaystyle \frac{\partial S_{cl}}{\partial x_{f}}$ $\displaystyle =$ $\displaystyle \left.\frac{\partial L}{\partial\dot{x}}\right|_{t_{f}}=p\left(t_{f}\right) \ \ \ \ \ (20)$

Example We can verify 17 for the case of the one-dimensional harmonic oscillator. The general solution for the position is given by

 $\displaystyle x\left(t\right)$ $\displaystyle =$ $\displaystyle A\cos\omega t+B\sin\omega t\ \ \ \ \ (21)$ $\displaystyle \dot{x}\left(t\right)$ $\displaystyle =$ $\displaystyle -A\omega\sin\omega t+B\omega\cos\omega t \ \ \ \ \ (22)$

The total energy is given by

 $\displaystyle E$ $\displaystyle =$ $\displaystyle \frac{1}{2}m\dot{x}^{2}+\frac{1}{2}m\omega^{2}x^{2}\ \ \ \ \ (23)$ $\displaystyle$ $\displaystyle =$ $\displaystyle \frac{m}{2}\left(\left(-A\omega\sin\omega t+B\omega\cos\omega t\right)^{2}+\omega^{2}\left(A\cos\omega t+B\sin\omega t\right)^{2}\right)\ \ \ \ \ (24)$ $\displaystyle$ $\displaystyle =$ $\displaystyle \frac{m\omega^{2}}{2}\left(A^{2}+B^{2}\right) \ \ \ \ \ (25)$

where we just multiplied out the second line, cancelled terms and used ${\cos^{2}x+\sin^{2}x=1}$.

To get the action, we need the Lagrangian:

 $\displaystyle L$ $\displaystyle =$ $\displaystyle T-V\ \ \ \ \ (26)$ $\displaystyle$ $\displaystyle =$ $\displaystyle \frac{1}{2}m\dot{x}^{2}-\frac{1}{2}m\omega^{2}x^{2}\ \ \ \ \ (27)$ $\displaystyle$ $\displaystyle =$ $\displaystyle \frac{m}{2}\left(\left(-A\omega\sin\omega t+B\omega\cos\omega t\right)^{2}-\omega^{2}\left(A\cos\omega t+B\sin\omega t\right)^{2}\right)\ \ \ \ \ (28)$ $\displaystyle$ $\displaystyle =$ $\displaystyle \frac{m\omega^{2}}{2}\left[A^{2}\left(\sin^{2}\omega t-\cos^{2}\omega t\right)+B^{2}\left(\cos^{2}\omega t-\sin^{2}\omega t\right)-4AB\sin\omega t\cos\omega t\right]\ \ \ \ \ (29)$ $\displaystyle$ $\displaystyle =$ $\displaystyle \frac{m\omega^{2}}{2}\left(\left(B^{2}-A^{2}\right)\cos2\omega t-2AB\sin2\omega t\right) \ \ \ \ \ (30)$

The action for a trajectory from ${t=0}$ to ${t=T}$ is then

 $\displaystyle S$ $\displaystyle =$ $\displaystyle \int_{0}^{T}Ldt\ \ \ \ \ (31)$ $\displaystyle$ $\displaystyle =$ $\displaystyle \frac{m\omega}{4}\left[\left(B^{2}-A^{2}\right)\sin2\omega t+2AB\cos2\omega t\right]_{0}^{T}\ \ \ \ \ (32)$ $\displaystyle$ $\displaystyle =$ $\displaystyle \frac{m\omega}{4}\left[\left(B^{2}-A^{2}\right)\sin2\omega T+2AB\left(\cos2\omega T-1\right)\right]\ \ \ \ \ (33)$ $\displaystyle$ $\displaystyle =$ $\displaystyle \frac{m\omega}{2}\left[\left(B^{2}-A^{2}\right)\sin\omega T\cos\omega T+AB\left(\cos^{2}\omega T-\sin^{2}\omega T-1\right)\right]\ \ \ \ \ (34)$ $\displaystyle$ $\displaystyle =$ $\displaystyle \frac{m\omega}{2}\left[\left(B^{2}-A^{2}\right)\sin\omega T\cos\omega T-2AB\sin^{2}\omega T\right] \ \ \ \ \ (35)$

To proceed further, we need to specify ${A}$ and ${B}$, since these depend on the boundary conditions (that is, on where we require the mass to be at ${t=0}$ and ${t=T}$). If we require ${x\left(0\right)=x_{1}}$ and ${x\left(T\right)=x_{2}}$, then

 $\displaystyle A$ $\displaystyle =$ $\displaystyle x_{1}\ \ \ \ \ (36)$ $\displaystyle x_{1}\cos\omega T+B\sin\omega T$ $\displaystyle =$ $\displaystyle x_{2}\ \ \ \ \ (37)$ $\displaystyle B$ $\displaystyle =$ $\displaystyle \frac{x_{2}-x_{1}\cos\omega T}{\sin\omega T} \ \ \ \ \ (38)$

Plugging these into 25 gives the energy as

 $\displaystyle E$ $\displaystyle =$ $\displaystyle \frac{m\omega^{2}}{2}\left(x_{1}^{2}+\left(\frac{x_{2}-x_{1}\cos\omega T}{\sin\omega T}\right)^{2}\right)\ \ \ \ \ (39)$ $\displaystyle$ $\displaystyle =$ $\displaystyle \frac{m\omega^{2}}{2\sin^{2}\omega T}\left(x_{1}^{2}+x_{2}^{2}-2x_{1}x_{2}\cos\omega T\right) \ \ \ \ \ (40)$

Plugging ${A}$ and ${B}$ into 35, we get (using ${c\equiv\cos\omega T}$ and ${s\equiv\sin\omega T}$, so that ${s^{2}+c^{2}=1}$):

 $\displaystyle S$ $\displaystyle =$ $\displaystyle \frac{m\omega}{2s}\left[\left(x_{2}-x_{1}c\right)^{2}c-x_{1}s^{2}c-2x_{1}s^{2}\left(x_{2}-x_{1}c\right)\right]\ \ \ \ \ (41)$ $\displaystyle$ $\displaystyle =$ $\displaystyle \frac{m\omega}{2s}\left[\left(x_{2}^{2}-2x_{1}x_{2}c+x_{1}^{2}c^{2}\right)c-x_{1}^{2}s^{2}c-2x_{1}x_{2}s^{2}+2x_{1}s^{2}c\right]\ \ \ \ \ (42)$ $\displaystyle$ $\displaystyle =$ $\displaystyle \frac{m\omega}{2s}\left[\left(x_{1}^{2}+x_{2}^{2}\right)c-2x_{1}x_{2}\right]\ \ \ \ \ (43)$ $\displaystyle$ $\displaystyle =$ $\displaystyle \frac{m\omega}{2\sin\omega T}\left[\left(x_{1}^{2}+x_{2}^{2}\right)\cos\omega T-2x_{1}x_{2}\right] \ \ \ \ \ (44)$

Taking the derivative, we get

 $\displaystyle \frac{\partial S}{\partial T}$ $\displaystyle =$ $\displaystyle \frac{m\omega}{2s^{2}}\left[-\omega\left(x_{1}^{2}+x_{2}^{2}\right)s^{2}-\left(\left(x_{1}^{2}+x_{2}^{2}\right)c-2x_{1}x_{2}\right)\omega c\right]\ \ \ \ \ (45)$ $\displaystyle$ $\displaystyle =$ $\displaystyle \frac{m\omega^{2}}{2s^{2}}\left[-\left(x_{1}^{2}+x_{2}^{2}\right)+2x_{1}x_{2}c\right]\ \ \ \ \ (46)$ $\displaystyle$ $\displaystyle =$ $\displaystyle -\frac{m\omega^{2}}{2\sin^{2}\omega T}\left(x_{1}^{2}+x_{2}^{2}-2x_{1}x_{2}\cos\omega T\right)\ \ \ \ \ (47)$ $\displaystyle$ $\displaystyle =$ $\displaystyle -E \ \ \ \ \ (48)$

Thus the result is verified for the harmonic oscillator.

# Hamiltonian for the electromagnetic force

References: Shankar, R. (1994), Principles of Quantum Mechanics, Plenum Press. Section 2.6.

Here we derive the equations of motion for the electromagnetic force using the Hamiltonian formalism.

The Hamiltonian is given by

$\displaystyle H\left(q,p\right)=\sum_{i}p_{i}\dot{q}_{i}-L\left(q,\dot{q}\right) \ \ \ \ \ (1)$

where the velocities ${\dot{q}_{i}}$ are expressed in terms of the positions ${q_{i}}$ and momenta ${p_{i}}$. The electromagnetic Lagrangian is

$\displaystyle L=\frac{1}{2}m\mathbf{v}\cdot\mathbf{v}-q\phi+\frac{q}{c}\mathbf{v}\cdot\mathbf{A} \ \ \ \ \ (2)$

where ${\phi}$ is the electric potential and ${\mathbf{A}}$ is the magnetic potential, with ${\mathbf{v}}$ the velocity of the charge ${q}$ with mass ${m}$. To convert to the Hamiltonian, we need the momentum, defined as

$\displaystyle p_{i}=\frac{\partial L}{\partial\dot{q}_{i}}$

In this case, the generalized velocity is given by

$\displaystyle \dot{q}_{i}=v_{i} \ \ \ \ \ (3)$

so we have

$\displaystyle p_{i}=mv_{i}+\frac{q}{c}A_{i} \ \ \ \ \ (4)$

or, in vector notation

 $\displaystyle \mathbf{p}$ $\displaystyle =$ $\displaystyle m\mathbf{v}+\frac{q}{c}\mathbf{A}\ \ \ \ \ (5)$ $\displaystyle \mathbf{v}$ $\displaystyle =$ $\displaystyle \frac{\mathbf{p}}{m}-\frac{q}{mc}\mathbf{A} \ \ \ \ \ (6)$

The Lagrangian is therefore

$\displaystyle L=\frac{\left|\mathbf{p}-q\mathbf{A}/c\right|^{2}}{2m}-q\phi+\frac{q}{c}\left(\frac{\mathbf{p}}{m}-\frac{q}{mc}\mathbf{A}\right)\cdot\mathbf{A} \ \ \ \ \ (7)$

The first sum in the Hamiltonian is

$\displaystyle \sum_{i}p_{i}\dot{q}_{i}=\mathbf{p}\cdot\mathbf{v}=\mathbf{p}\cdot\left(\frac{\mathbf{p}}{m}-\frac{q}{mc}\mathbf{A}\right) \ \ \ \ \ (8)$

The Hamiltonian is then

 $\displaystyle H$ $\displaystyle =$ $\displaystyle \mathbf{p}\cdot\left(\frac{\mathbf{p}}{m}-\frac{q}{mc}\mathbf{A}\right)-\frac{\left|\mathbf{p}-q\mathbf{A}/c\right|^{2}}{2m}+q\phi-\frac{q}{c}\left(\frac{\mathbf{p}}{m}-\frac{q}{mc}\mathbf{A}\right)\cdot\mathbf{A}\ \ \ \ \ (9)$ $\displaystyle$ $\displaystyle =$ $\displaystyle \left(\frac{\mathbf{p}}{m}-\frac{q}{mc}\mathbf{A}\right)\left(\mathbf{p}-\frac{q}{c}\mathbf{A}\right)-\frac{\left|\mathbf{p}-q\mathbf{A}/c\right|^{2}}{2m}+q\phi\ \ \ \ \ (10)$ $\displaystyle$ $\displaystyle =$ $\displaystyle \frac{\left|\mathbf{p}-q\mathbf{A}/c\right|^{2}}{2m}+q\phi \ \ \ \ \ (11)$

# Hamiltonian for the two-body problem

References: Shankar, R. (1994), Principles of Quantum Mechanics, Plenum Press. Section 2.5; Exercise 2.5.4.

Here we derive the equations of motion of the two-body problem using the Hamiltonian formalism.

The Hamiltonian is given by

$\displaystyle H\left(q,p\right)=\sum_{i}p_{i}\dot{q}_{i}-L\left(q,\dot{q}\right) \ \ \ \ \ (1)$

where the velocities ${\dot{q}_{i}}$ are expressed in terms of the positions ${q_{i}}$ and momenta ${p_{i}}$. In this case, we start with the Lagrangian in terms of the centre of mass position ${\mathbf{r}_{CM}}$ and the relative position ${\mathbf{r}}$ of mass 2 to mass 1.

 $\displaystyle L$ $\displaystyle =$ $\displaystyle \frac{1}{2}\left(m_{1}+m_{2}\right)\left|\dot{\mathbf{r}}_{CM}\right|^{2}+\frac{1}{2}\frac{m_{1}m_{2}}{m_{1}+m_{2}}\left|\dot{\mathbf{r}}\right|^{2}-V\left(\mathbf{r}\right)\ \ \ \ \ (2)$ $\displaystyle$ $\displaystyle =$ $\displaystyle \frac{M}{2}\left|\dot{\mathbf{r}}_{CM}\right|^{2}+\frac{\mu}{2}\left|\dot{\mathbf{r}}\right|^{2}-V\left(\mathbf{r}\right) \ \ \ \ \ (3)$

where ${M=m_{1}+m_{2}}$ is the total mass and ${\mu=\frac{m_{1}m_{2}}{m_{1}+m_{2}}}$ is the reduced mass.

There are potentially 6 velocity components and 6 coordinate components in the Lagrangian, but the 3 components of ${\mathbf{r}_{CM}}$ do not appear, which simplifies things a bit. To convert to a Hamiltonian, we need the momenta

$\displaystyle p_{i}=\frac{\partial L}{\partial\dot{q}_{i}} \ \ \ \ \ (4)$

The ${x}$ component of momentum of the centre of mass is

$\displaystyle p_{CM,x}=\frac{\partial L}{\partial\dot{r}_{CM,x}}=M\dot{r}_{CM,x} \ \ \ \ \ (5)$

The other two components of the centre of mass velocity, and of the relative velocity, have a similar form, and in general we can write

 $\displaystyle p_{CM,i}$ $\displaystyle =$ $\displaystyle M\dot{r}_{CM,i}\ \ \ \ \ (6)$ $\displaystyle p_{i}$ $\displaystyle =$ $\displaystyle \mu\dot{r}_{i} \ \ \ \ \ (7)$

In vector notation, this becomes

 $\displaystyle \dot{\mathbf{r}}_{CM}$ $\displaystyle =$ $\displaystyle \frac{\mathbf{p}_{CM}}{M}\ \ \ \ \ (8)$ $\displaystyle \dot{\mathbf{r}}$ $\displaystyle =$ $\displaystyle \frac{\mathbf{p}}{\mu}\ \ \ \ \ (9)$ $\displaystyle \left|\dot{\mathbf{r}}_{CM}\right|^{2}$ $\displaystyle =$ $\displaystyle \frac{\left|\mathbf{p}_{CM}\right|^{2}}{M^{2}}\ \ \ \ \ (10)$ $\displaystyle \left|\dot{\mathbf{r}}\right|^{2}$ $\displaystyle =$ $\displaystyle \frac{\left|\mathbf{p}\right|^{2}}{\mu^{2}} \ \ \ \ \ (11)$

The Lagrangian thus becomes

$\displaystyle L=\frac{\left|\mathbf{p}_{CM}\right|^{2}}{2M}+\frac{\left|\mathbf{p}\right|^{2}}{2\mu}-V\left(\mathbf{r}\right) \ \ \ \ \ (12)$

The Hamiltonian is

 $\displaystyle H$ $\displaystyle =$ $\displaystyle \mathbf{p}\cdot\dot{\mathbf{r}}+\mathbf{p}_{CM}\cdot\dot{\mathbf{r}}_{CM}-L\ \ \ \ \ (13)$ $\displaystyle$ $\displaystyle =$ $\displaystyle \frac{\left|\mathbf{p}\right|^{2}}{\mu}+\frac{\left|\mathbf{p}_{CM}\right|^{2}}{M}-\left[\frac{\left|\mathbf{p}_{CM}\right|^{2}}{2M}+\frac{\left|\mathbf{p}\right|^{2}}{2\mu}-V\left(\mathbf{r}\right)\right]\ \ \ \ \ (14)$ $\displaystyle$ $\displaystyle =$ $\displaystyle \frac{\left|\mathbf{p}_{CM}\right|^{2}}{2M}+\frac{\left|\mathbf{p}\right|^{2}}{2\mu}+V\left(\mathbf{r}\right) \ \ \ \ \ (15)$

Once we’ve got the Hamiltonian, we can apply Hamilton’s canonical equations to get the equations of motion.

 $\displaystyle \frac{\partial H}{\partial p_{i}}$ $\displaystyle =$ $\displaystyle \dot{r}_{i}\ \ \ \ \ (16)$ $\displaystyle -\frac{\partial H}{\partial r_{i}}$ $\displaystyle =$ $\displaystyle \dot{p}_{i} \ \ \ \ \ (17)$

Since ${\mathbf{r}_{CM}}$ does not appear in the Hamiltonian, we have

 $\displaystyle \dot{\mathbf{p}}_{CM}$ $\displaystyle =$ $\displaystyle 0\ \ \ \ \ (18)$ $\displaystyle \mathbf{p}_{CM}$ $\displaystyle =$ $\displaystyle \mbox{constant} \ \ \ \ \ (19)$

so the momentum of the centre of mass does not change, as expected.

For ${\mathbf{r}}$, we have

 $\displaystyle \frac{\partial H}{\partial p_{i}}$ $\displaystyle =$ $\displaystyle \frac{p_{i}}{\mu}=\dot{r}_{i}\ \ \ \ \ (20)$ $\displaystyle \frac{\partial H}{\partial r_{i}}$ $\displaystyle =$ $\displaystyle \frac{\partial V}{\partial r_{i}}=-\dot{p}_{i} \ \ \ \ \ (21)$

The first equation tells us nothing new, while the second is just Newton’s law for a central force: ${\mathbf{\dot{p}}=-\nabla V}$.

# Hamiltonians for harmonic oscillators

References: Shankar, R. (1994), Principles of Quantum Mechanics, Plenum Press. Section 2.5; Exercises 2.5.2 – 2.5.3.

Here are a couple of examples of equations of motion using the Hamiltonian formalism. First, we look at the simple harmonic oscillator, in which we have a mass ${m}$ sliding on a frictionless horizontal surface. The mass is connected to a spring with constant ${k}$, with the other end of the spring connected to a fixed support.

The Hamiltonian is given by

$\displaystyle H\left(q,p\right)=\sum_{i}p_{i}\dot{q}_{i}-L\left(q,\dot{q}\right) \ \ \ \ \ (1)$

where the velocities ${\dot{q}_{i}}$ are expressed in terms of the positions ${q_{i}}$ and momenta ${p_{i}}$. In this case, we have, using the coordinate ${x}$ as the displacement from equilibrium

 $\displaystyle L\left(x,\dot{x}\right)$ $\displaystyle =$ $\displaystyle \frac{1}{2}m\dot{x}^{2}-\frac{1}{2}kx^{2}\ \ \ \ \ (2)$ $\displaystyle p$ $\displaystyle =$ $\displaystyle \frac{\partial L}{\partial\dot{x}}=m\dot{x}\ \ \ \ \ (3)$ $\displaystyle \dot{x}$ $\displaystyle =$ $\displaystyle \frac{p}{m}\ \ \ \ \ (4)$ $\displaystyle L\left(x,\dot{x}\left(x,p\right)\right)$ $\displaystyle =$ $\displaystyle \frac{p^{2}}{2m}-\frac{1}{2}kx^{2}\ \ \ \ \ (5)$ $\displaystyle H$ $\displaystyle =$ $\displaystyle \frac{p^{2}}{m}-\left(\frac{p^{2}}{2m}-\frac{1}{2}kx^{2}\right)\ \ \ \ \ (6)$ $\displaystyle$ $\displaystyle =$ $\displaystyle \frac{p^{2}}{2m}+\frac{1}{2}kx^{2} \ \ \ \ \ (7)$

We can now apply Hamilton’s canonical equations:

 $\displaystyle \frac{\partial H}{\partial p}$ $\displaystyle =$ $\displaystyle \dot{x}\ \ \ \ \ (8)$ $\displaystyle -\frac{\partial H}{\partial x}$ $\displaystyle =$ $\displaystyle \dot{p} \ \ \ \ \ (9)$

We get

 $\displaystyle \frac{\partial H}{\partial p}$ $\displaystyle =$ $\displaystyle \frac{p}{m}=\dot{x}\ \ \ \ \ (10)$ $\displaystyle -\frac{\partial H}{\partial x}$ $\displaystyle =$ $\displaystyle -kx=\dot{p} \ \ \ \ \ (11)$

We thus get a pair of first order ODEs which can be solved in the usual way, given ${x\left(0\right)}$ and ${p\left(0\right)}$. The second order ODE that we got by using the Lagrangian method can be obtained by differentiating the first equation and plugging it into the second:

 $\displaystyle \ddot{x}$ $\displaystyle =$ $\displaystyle \frac{\dot{p}}{m}\ \ \ \ \ (12)$ $\displaystyle$ $\displaystyle =$ $\displaystyle -\frac{k}{m}x \ \ \ \ \ (13)$

From 7 we see that, since in the absence of external force, the total energy ${H=T+V=E}$ is a constant,

$\displaystyle \frac{p^{2}}{2m}+\frac{1}{2}kx^{2}=E=\mbox{constant} \ \ \ \ \ (14)$

This can be written as the equation of an ellipse:

$\displaystyle \frac{p^{2}}{b^{2}}+\frac{x^{2}}{a^{2}}=1 \ \ \ \ \ (15)$

where

 $\displaystyle a^{2}$ $\displaystyle =$ $\displaystyle \frac{2E}{k}\ \ \ \ \ (16)$ $\displaystyle b^{2}$ $\displaystyle =$ $\displaystyle 2mE \ \ \ \ \ (17)$

We can use the Hamiltonian formalism to get the equations of motion of the coupled harmonic oscillator. From our Lagrangian treatment, we had

$\displaystyle L=\frac{1}{2}m\left(\dot{x}_{1}^{2}+\dot{x}_{2}^{2}\right)-k\left(x_{1}^{2}+x_{2}^{2}-x_{1}x_{2}\right) \ \ \ \ \ (18)$

Converting to coordinates and momenta, we have

 $\displaystyle p_{i}$ $\displaystyle =$ $\displaystyle \frac{\partial L}{\partial\dot{x}_{i}}=m\dot{x}_{i}\ \ \ \ \ (19)$ $\displaystyle \dot{x}_{i}$ $\displaystyle =$ $\displaystyle \frac{p_{i}}{m}\ \ \ \ \ (20)$ $\displaystyle H$ $\displaystyle =$ $\displaystyle \sum_{i}p_{i}\dot{x}_{i}-L\left(x,\dot{x}\right)\ \ \ \ \ (21)$ $\displaystyle$ $\displaystyle =$ $\displaystyle \frac{1}{m}\left(p_{1}^{2}+p_{2}^{2}\right)-\left[\frac{1}{2m}m\left(p_{1}^{2}+p_{2}^{2}\right)-k\left(x_{1}^{2}+x_{2}^{2}-x_{1}x_{2}\right)\right]\ \ \ \ \ (22)$ $\displaystyle$ $\displaystyle =$ $\displaystyle \frac{1}{2m}\left(p_{1}^{2}+p_{2}^{2}\right)+k\left(x_{1}^{2}+x_{2}^{2}-x_{1}x_{2}\right) \ \ \ \ \ (23)$

Applying the canonical equations gives

 $\displaystyle \frac{\partial H}{\partial p_{i}}$ $\displaystyle =$ $\displaystyle \frac{p_{i}}{m}=\dot{x}_{i}\ \ \ \ \ (24)$ $\displaystyle -\frac{\partial H}{\partial x_{1}}$ $\displaystyle =$ $\displaystyle -2kx_{1}+kx_{2}=\dot{p}_{1}\ \ \ \ \ (25)$ $\displaystyle -\frac{\partial H}{\partial x_{2}}$ $\displaystyle =$ $\displaystyle -2kx_{2}+kx_{1}=\dot{p}_{2} \ \ \ \ \ (26)$

Again, by taking the derivative of the first line and substituting into the last two lines, we get back the previous equations of motion:

 $\displaystyle \ddot{x}_{1}$ $\displaystyle =$ $\displaystyle -2\frac{k}{m}x_{1}+\frac{k}{m}x_{2}\ \ \ \ \ (27)$ $\displaystyle \ddot{x}_{2}$ $\displaystyle =$ $\displaystyle \frac{k}{m}x_{1}-2\frac{k}{m}x_{2} \ \ \ \ \ (28)$

# Lagrangian for the Schrödinger equation

References: W. Greiner & J. Reinhardt, Field Quantization, Springer-Verlag (1996), Chapter 3, Section 3.1.

As a prelude to ‘proper’ quantum field theory, we’ll look first at turning the non-relativistic quantum theory based on the Schrödinger equation into a field theory. Before we develop a quantum field theory of the Schrödinger equation, we’ll first look at this equation treating the wave function ${\psi\left(\mathbf{x},t\right)}$ as a classical (that is, non-quantum) field. The Schrödinger equation is

$\displaystyle i\hbar\frac{\partial\psi}{\partial t}=-\frac{\hbar^{2}}{2m}\nabla^{2}\psi+V\left(\mathbf{x},t\right)\psi \ \ \ \ \ (1)$

where ${V\left(\mathbf{x},t\right)}$ is, as usual, the potential function.

In order to apply the techniques of classical field theory, we need a Lagrangian density ${\mathcal{L}}$. There doesn’t seem to be any way of actually deriving Lagrangian densities; presumably they are found through trial and error, with perhaps a bit of physical intuition. In any case, the Lagrangian density for the Schrödinger equation turns out to be

$\displaystyle \mathcal{L}\left(\psi,\nabla\psi,\dot{\psi}\right)=i\hbar\psi^*\dot{\psi}-\frac{\hbar^{2}}{2m}\nabla\psi^*\cdot\nabla\psi-V\left(\mathbf{x},t\right)\psi^*\psi \ \ \ \ \ (2)$

As ${\psi}$ is a complex function, it has real and imaginary parts, so we can treat ${\psi}$ and ${\psi^*}$ as independent fields. As we saw earlier, we can derive the Euler-Lagrange equations for multiple fields from the principle of least action and end up with

$\displaystyle \frac{\partial\mathcal{L}}{\partial\phi^{r}}-\frac{\partial}{\partial q^{\mu}}\left(\frac{\partial\mathcal{L}}{\partial\phi_{,\mu}^{r}}\right)=0 \ \ \ \ \ (3)$

where the ${\phi^{r}}$ are the fields and ${q^{\mu}=\left(\mathbf{x},t\right)}$. In this case, the two fields are ${\psi}$ and ${\psi^*}$ and we get the two equations

 $\displaystyle \frac{\partial\mathcal{L}}{\partial\psi}-\frac{\partial}{\partial x^{i}}\frac{\partial\mathcal{L}}{\partial\psi_{,i}}-\frac{\partial}{\partial t}\frac{\partial\mathcal{L}}{\partial\dot{\psi}}$ $\displaystyle =$ $\displaystyle \frac{\partial\mathcal{L}}{\partial\psi}-\nabla\cdot\frac{\partial\mathcal{L}}{\partial\nabla\psi}-\frac{\partial}{\partial t}\frac{\partial\mathcal{L}}{\partial\dot{\psi}}=0\ \ \ \ \ (4)$ $\displaystyle \frac{\partial\mathcal{L}}{\partial\psi^*}-\frac{\partial}{\partial x^{i}}\frac{\partial\mathcal{L}}{\partial\psi_{,i}^*}-\frac{\partial}{\partial t}\frac{\partial\mathcal{L}}{\partial\dot{\psi}^*}$ $\displaystyle =$ $\displaystyle \frac{\partial\mathcal{L}}{\partial\psi^*}-\nabla\cdot\frac{\partial\mathcal{L}}{\partial\nabla\psi^*}-\frac{\partial}{\partial t}\frac{\partial\mathcal{L}}{\partial\dot{\psi}^*}=0 \ \ \ \ \ (5)$

The second term in each row just introduces the gradient sign ${\nabla}$ as a shorthand for the ${\frac{\partial}{\partial x^{i}}\frac{\partial\mathcal{L}}{\partial\psi_{,i}}}$ and ${\frac{\partial}{\partial x^{i}}\frac{\partial\mathcal{L}}{\partial\psi_{,i}^*}}$ terms.

We can plug 2 into these two equations to verify that we recover the original Schrödinger equation 1 and its complex conjugate. From 4 we have

 $\displaystyle \frac{\partial\mathcal{L}}{\partial\psi}$ $\displaystyle =$ $\displaystyle -V\left(\mathbf{x},t\right)\psi^*\ \ \ \ \ (6)$ $\displaystyle \nabla\cdot\frac{\partial\mathcal{L}}{\partial\nabla\psi}$ $\displaystyle =$ $\displaystyle -\frac{\hbar^{2}}{2m}\nabla^{2}\psi^*\ \ \ \ \ (7)$ $\displaystyle \frac{\partial}{\partial t}\frac{\partial\mathcal{L}}{\partial\dot{\psi}}$ $\displaystyle =$ $\displaystyle i\hbar\dot{\psi}^*\ \ \ \ \ (8)$ $\displaystyle \frac{\partial\mathcal{L}}{\partial\psi}-\nabla\cdot\frac{\partial\mathcal{L}}{\partial\nabla\psi}-\frac{\partial}{\partial t}\frac{\partial\mathcal{L}}{\partial\dot{\psi}}$ $\displaystyle =$ $\displaystyle -i\hbar\dot{\psi}^*+\frac{\hbar^{2}}{2m}\nabla^{2}\psi^*-V\left(\mathbf{x},t\right)\psi^*=0\ \ \ \ \ (9)$ $\displaystyle -i\hbar\dot{\psi}^*$ $\displaystyle =$ $\displaystyle \frac{\hbar^{2}}{2m}\nabla^{2}\psi^*-V\left(\mathbf{x},t\right)\psi^* \ \ \ \ \ (10)$

which is the complex conjugate of 1. Plugging 2 into 5 just reproduces 1.

The conjugate momentum density ${\pi}$ can be calculated for the two fields ${\psi}$ and ${\psi^*}$. We get

 $\displaystyle \pi_{1}\left(\mathbf{x},t\right)$ $\displaystyle =$ $\displaystyle \frac{\partial\mathcal{L}}{\partial\dot{\psi}}=i\hbar\psi^*\left(\mathbf{x},t\right)\ \ \ \ \ (11)$ $\displaystyle \pi_{2}\left(\mathbf{x},t\right)$ $\displaystyle =$ $\displaystyle \frac{\partial\mathcal{L}}{\partial\dot{\psi}^*}=0 \ \ \ \ \ (12)$

The Hamiltonian density is defined as

 $\displaystyle \mathcal{H}$ $\displaystyle =$ $\displaystyle \sum_{r}\pi_{r}\dot{\phi}^{r}-\mathcal{L}\ \ \ \ \ (13)$ $\displaystyle$ $\displaystyle =$ $\displaystyle i\hbar\psi^*\dot{\psi}-\left[i\hbar\psi^*\dot{\psi}-\frac{\hbar^{2}}{2m}\nabla\psi^*\cdot\nabla\psi-V\left(\mathbf{x},t\right)\psi^*\psi\right]\ \ \ \ \ (14)$ $\displaystyle$ $\displaystyle =$ $\displaystyle \frac{\hbar^{2}}{2m}\nabla\psi^*\cdot\nabla\psi+V\left(\mathbf{x},t\right)\psi^*\psi \ \ \ \ \ (15)$

The total Hamiltonian is the integral of this over 3-d space:

 $\displaystyle H$ $\displaystyle =$ $\displaystyle \int d^{3}x\;\mathcal{H}\ \ \ \ \ (16)$ $\displaystyle$ $\displaystyle =$ $\displaystyle \int d^{3}x\left[\frac{\hbar^{2}}{2m}\nabla\psi^*\cdot\nabla\psi+V\left(\mathbf{x},t\right)\psi^*\psi\right] \ \ \ \ \ (17)$

We can integrate the first term by parts, by integrating the ${\nabla\psi^*}$ term and invoking the usual assumption that ${\psi^*\rightarrow0}$ fast enough at infinity that the integrated term is zero. We then get

 $\displaystyle H$ $\displaystyle =$ $\displaystyle \int d^{3}x\left[-\frac{\hbar^{2}}{2m}\psi^*\nabla^{2}\psi+V\left(\mathbf{x},t\right)\psi^*\psi\right]\ \ \ \ \ (18)$ $\displaystyle$ $\displaystyle =$ $\displaystyle \int d^{3}x\;\psi^*\left[-\frac{\hbar^{2}}{2m}\nabla^{2}\psi+V\left(\mathbf{x},t\right)\psi\right] \ \ \ \ \ (19)$

Referring back to quantum mechanics for a moment, we see that this last integral is just ${\left\langle \psi\left|\hat{H}\right|\psi\right\rangle }$, that is, the expectation value of the Hamiltonian operator, which is the total energy of the system.

Finally, we can write down the Poisson brackets, since these are general results for any field ${\psi}$ and its conjugate momentum ${\pi}$:

 $\displaystyle \left\{ \psi\left(\mathbf{x},t\right),\pi\left(\mathbf{x}^{\prime},t\right)\right\} _{PB}$ $\displaystyle =$ $\displaystyle \delta^{3}\left(\mathbf{x}-\mathbf{x}^{\prime}\right)\ \ \ \ \ (20)$ $\displaystyle \left\{ \phi\left(\mathbf{x},t\right),\phi\left(\mathbf{x}^{\prime},t\right)\right\} _{PB}$ $\displaystyle =$ $\displaystyle 0\ \ \ \ \ (21)$ $\displaystyle \left\{ \pi\left(\mathbf{x},t\right),\pi\left(\mathbf{x}^{\prime},t\right)\right\} _{PB}$ $\displaystyle =$ $\displaystyle 0 \ \ \ \ \ (22)$

These brackets will be used later when we quantize the theory.

# 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.

# Hamiltonian for the Dirac equation

Reference: References: Robert D. Klauber, Student Friendly Quantum Field Theory, (Sandtrove Press, 2013) – Chapter 4, Problem 4.24.

Here we continue with the derivation of the total Dirac Hamiltonian. We’ve seen that the Hamiltonian density is given as

$\displaystyle \mathcal{H}_{0}^{1/2}=-i\bar{\psi}\gamma^{j}\partial_{j}\psi+m\bar{\psi}\psi \ \ \ \ \ (1)$

where the general solutions of the Dirac equation are given by

 $\displaystyle \psi$ $\displaystyle =$ $\displaystyle \sum_{r=1}^{2}\sum_{\mathbf{p}}\sqrt{\frac{m}{VE_{\mathbf{p}}}}\left[c_{r}\left(\mathbf{p}\right)u_{r}\left(\mathbf{p}\right)e^{-ipx}+d_{r}^{\dagger}\left(\mathbf{p}\right)v_{r}\left(\mathbf{p}\right)e^{ipx}\right]\ \ \ \ \ (2)$ $\displaystyle \bar{\psi}$ $\displaystyle =$ $\displaystyle \sum_{r=1}^{2}\sum_{\mathbf{p}}\sqrt{\frac{m}{VE_{\mathbf{p}}}}\left[d_{r}\left(\mathbf{p}\right)\bar{v}_{r}\left(\mathbf{p}\right)e^{-ipx}+c_{r}^{\dagger}\left(\mathbf{p}\right)\bar{u}_{r}\left(\mathbf{p}\right)e^{ipx}\right] \ \ \ \ \ (3)$

The full Hamiltonian is the integral of the density over all space, that is

$\displaystyle H_{0}^{1/2}=\int\left[-i\bar{\psi}\gamma^{j}\partial_{j}\psi+m\bar{\psi}\psi\right]d^{3}x \ \ \ \ \ (4)$

From section 4.4.1 in Klauber’s book and the previous post, we see that four of the terms in the expansion of this integral (after substituting in 2 and 3) cancel out, leaving us with (implied sum over ${j=1,2,3}$ in the first two lines):

 $\displaystyle H_{0}^{1/2}$ $\displaystyle =$ $\displaystyle -\int\sum_{r,s,\mathbf{p}}\frac{m}{VE_{\mathbf{p}}}d_{r}\left(\mathbf{p}\right)\bar{v}_{r}\left(\mathbf{p}\right)\gamma^{j}p^{j}v_{s}\left(\mathbf{p}\right)d_{s}^{\dagger}\left(\mathbf{p}\right)d^{3}x+\nonumber$ $\displaystyle$ $\displaystyle$ $\displaystyle \int\sum_{r,s,\mathbf{p}}\frac{m}{VE_{\mathbf{p}}}c_{r}^{\dagger}\left(\mathbf{p}\right)\bar{u}_{r}\left(\mathbf{p}\right)\gamma^{j}p^{j}u_{s}\left(\mathbf{p}\right)c_{s}\left(\mathbf{p}\right)d^{3}x+\nonumber$ $\displaystyle$ $\displaystyle$ $\displaystyle \int\sum_{r,s,\mathbf{p}}\frac{m}{VE_{\mathbf{p}}}d_{r}\left(\mathbf{p}\right)\bar{v}_{r}\left(\mathbf{p}\right)mv_{s}\left(\mathbf{p}\right)d_{s}^{\dagger}\left(\mathbf{p}\right)d^{3}x+\nonumber$ $\displaystyle$ $\displaystyle$ $\displaystyle \int\sum_{r,s,\mathbf{p}}\frac{m}{VE_{\mathbf{p}}}c_{r}^{\dagger}\left(\mathbf{p}\right)\bar{u}_{r}\left(\mathbf{p}\right)mu_{s}\left(\mathbf{p}\right)c_{s}\left(\mathbf{p}\right)d^{3}x \ \ \ \ \ (5)$

Klauber shows that the sum of the first and third lines gives us

$\displaystyle \sum_{r,\mathbf{p}}E_{\mathbf{p}}d^{\dagger}\left(\mathbf{p}\right)d_{r}\left(\mathbf{p}\right)-1\equiv\sum_{r,\mathbf{p}}E_{\mathbf{p}}\bar{N}_{r}\left(\mathbf{p}\right)-1 \ \ \ \ \ (6)$

By following similar steps to Klauber’s equations 4-70 to 4.73, we can work out the sum of the second and fourth lines, as follows. We can use the inner products of the spinors:

$\displaystyle u_{r}^{\dagger}\left(\mathbf{p}\right)u_{s}\left(\mathbf{p}\right)=v_{r}^{\dagger}\left(\mathbf{p}\right)v_{s}\left(\mathbf{p}\right)=\frac{E_{\mathbf{p}}}{m}\delta_{rs}=\frac{p_{0}}{m}\delta_{rs} \ \ \ \ \ (7)$

The adjoint spinors are defined as

 $\displaystyle \bar{u}_{r}$ $\displaystyle =$ $\displaystyle u_{r}^{\dagger}\gamma^{0} \ \ \ \ \ (8)$

so combined with the identity ${\gamma^{0}\gamma^{0}=1}$ we have

 $\displaystyle u_{r}^{\dagger}\left(\mathbf{p}\right)u_{s}\left(\mathbf{p}\right)$ $\displaystyle =$ $\displaystyle \bar{u}_{r}\left(\mathbf{p}\right)\gamma^{0}u_{s}\left(\mathbf{p}\right)\ \ \ \ \ (9)$ $\displaystyle \bar{u}_{r}\left(\mathbf{p}\right)\gamma^{0}p_{0}u_{s}\left(\mathbf{p}\right)$ $\displaystyle =$ $\displaystyle \frac{\left(p_{0}\right)^{2}}{m}\delta_{rs}\ \ \ \ \ (10)$ $\displaystyle$ $\displaystyle =$ $\displaystyle \frac{E_{\mathbf{p}}^{2}}{m}\delta_{rs} \ \ \ \ \ (11)$

The sum of the second and fourth lines of 5 can then be written as

 $\displaystyle \int\sum_{r,s,\mathbf{p}}\frac{m}{VE_{\mathbf{p}}}c_{r}^{\dagger}\left(\mathbf{p}\right)\bar{u}_{r}\left(\mathbf{p}\right)\left[-\gamma^{j}p_{j}+m-\gamma^{0}p_{0}+\gamma^{0}p_{0}\right]u_{s}\left(\mathbf{p}\right)c_{s}\left(\mathbf{p}\right)d^{3}x$ $\displaystyle =$ $\displaystyle \int\frac{d^{3}x}{V}\sum_{r,\mathbf{p}}\frac{m}{E_{\mathbf{p}}}c_{r}^{\dagger}\left(\mathbf{p}\right)\left[\bar{u}_{r}\left(\mathbf{p}\right)\left(-\gamma^{\mu}p_{\mu}+m\right)u_{r}\left(\mathbf{p}\right)+\frac{E_{\mathbf{p}}^{2}}{m}\right]c_{r}\left(\mathbf{p}\right) \ \ \ \ \ (12)$

Using Klauber’s equation B4-3.3:

$\displaystyle \left(\gamma^{\mu}p_{\mu}-m\right)u_{r}\left(\mathbf{p}\right)=0 \ \ \ \ \ (13)$

and the fact that

$\displaystyle \int d^{3}x=V \ \ \ \ \ (14)$

that is, the integral over the entire volume is just the volume ${V}$ itself, we see that 12 reduces to

 $\displaystyle \sum_{r,\mathbf{p}}\frac{m}{E_{\mathbf{p}}}c_{r}^{\dagger}\left(\mathbf{p}\right)\frac{E_{\mathbf{p}}^{2}}{m}c_{r}\left(\mathbf{p}\right)$ $\displaystyle =$ $\displaystyle \sum_{r,\mathbf{p}}E_{\mathbf{p}}c_{r}^{\dagger}\left(\mathbf{p}\right)c_{r}\left(\mathbf{p}\right)\ \ \ \ \ (15)$ $\displaystyle$ $\displaystyle \equiv$ $\displaystyle \sum_{r,\mathbf{p}}E_{\mathbf{p}}N_{r}\left(\mathbf{p}\right) \ \ \ \ \ (16)$

Combining this result with 6 we get the final Hamiltonian as

$\displaystyle H_{0}^{1/2}=\sum_{r,\mathbf{p}}E_{\mathbf{p}}\left(N_{r}\left(\mathbf{p}\right)-\frac{1}{2}+\bar{N}_{r}\left(\mathbf{p}\right)-\frac{1}{2}\right) \ \ \ \ \ (17)$

The operators are interpreted as number operators so that ${N_{r}\left(\mathbf{p}\right)}$ is the operator whose eigenvalue is the number of particles with spin ${r}$ and momentum ${\mathbf{p}}$, and ${\bar{N}_{r}\left(\mathbf{p}\right)}$ is the operator whose eigenvalue is the number of antiparticles with spin ${r}$ and momentum ${\mathbf{p}}$. The two ${\frac{1}{2}}$ terms give rise to an infinite negative vacuum energy.

# Hamiltonian density for the Dirac equation

Reference: References: Robert D. Klauber, Student Friendly Quantum Field Theory, (Sandtrove Press, 2013) – Chapter 4, Problem 4.23.

The Lagrangian density which gives the Dirac equation is

$\displaystyle \mathcal{L}_{0}^{1/2}=\bar{\psi}\left(i\gamma^{\mu}\partial_{\mu}-m\right)\psi \ \ \ \ \ (1)$

We can follow the same procedure that we used for scalar fields to derive the conjugate momenta for the Dirac spin-${\frac{1}{2}}$ field. For the conjugate momenta we have from 1

 $\displaystyle \pi^{1/2}$ $\displaystyle \equiv$ $\displaystyle \frac{\partial\mathcal{L}_{0}^{1/2}}{\partial\psi_{,0}}\ \ \ \ \ (2)$ $\displaystyle$ $\displaystyle =$ $\displaystyle i\bar{\psi}\gamma^{0}\ \ \ \ \ (3)$ $\displaystyle$ $\displaystyle =$ $\displaystyle i\psi^{\dagger}\gamma^{0}\gamma^{0}\ \ \ \ \ (4)$ $\displaystyle$ $\displaystyle =$ $\displaystyle i\psi^{\dagger} \ \ \ \ \ (5)$

where we’ve used the definition of the adjoint field ${\bar{\psi}\equiv\psi^{\dagger}\gamma^{0}}$.

Because the Lagrangian is not symmetric with respect to ${\psi}$ and ${\bar{\psi}}$ (it contains derivatives of ${\psi}$ but not of ${\bar{\psi}}$), the adjoint momentum turns out to be zero:

$\displaystyle \bar{\pi}^{1/2}\equiv\frac{\partial\mathcal{L}_{0}^{1/2}}{\partial\bar{\psi}_{,0}}=0 \ \ \ \ \ (6)$

From the conjugate momenta and the Lagrangian, we can derive the Hamiltonian density:

 $\displaystyle \mathcal{H}_{0}^{1/2}$ $\displaystyle =$ $\displaystyle \pi_{r}^{1/2}\phi_{,0}^{r}\ \ \ \ \ (7)$ $\displaystyle$ $\displaystyle =$ $\displaystyle \pi^{1/2}\dot{\psi}+\bar{\pi}^{1/2}\dot{\bar{\psi}}-\mathcal{L}_{0}^{1/2}\ \ \ \ \ (8)$ $\displaystyle$ $\displaystyle =$ $\displaystyle i\psi^{\dagger}\dot{\psi}+0-\bar{\psi}\left(i\gamma^{\mu}\partial_{\mu}-m\right)\psi\ \ \ \ \ (9)$ $\displaystyle$ $\displaystyle =$ $\displaystyle i\bar{\psi}\gamma^{0}\dot{\psi}-i\bar{\psi}\gamma^{0}\dot{\psi}-i\bar{\psi}\gamma^{j}\partial_{j}\psi+m\bar{\psi}\psi\ \ \ \ \ (10)$ $\displaystyle$ $\displaystyle =$ $\displaystyle -i\bar{\psi}\gamma^{j}\partial_{j}\psi+m\bar{\psi}\psi \ \ \ \ \ (11)$

where in the last 2 lines, the ${j}$ index is summed over spatial coordinates only.

The general solution of the Dirac equation for discrete momenta is

 $\displaystyle \psi$ $\displaystyle =$ $\displaystyle \sum_{r=1}^{2}\sum_{\mathbf{p}}\sqrt{\frac{m}{VE_{\mathbf{p}}}}\left[c_{r}\left(\mathbf{p}\right)u_{r}\left(\mathbf{p}\right)e^{-ipx}+d_{r}^{\dagger}\left(\mathbf{p}\right)v_{r}\left(\mathbf{p}\right)e^{ipx}\right]\ \ \ \ \ (12)$ $\displaystyle \bar{\psi}$ $\displaystyle =$ $\displaystyle \sum_{r=1}^{2}\sum_{\mathbf{p}}\sqrt{\frac{m}{VE_{\mathbf{p}}}}\left[d_{r}\left(\mathbf{p}\right)\bar{v}_{r}\left(\mathbf{p}\right)e^{-ipx}+c_{r}^{\dagger}\left(\mathbf{p}\right)\bar{u}_{r}\left(\mathbf{p}\right)e^{ipx}\right] \ \ \ \ \ (13)$

We can find the total Hamiltonian ${H_{0}^{1/2}}$ by substituting these into 11 and integrating over space:

 $\displaystyle H_{0}^{1/2}$ $\displaystyle =$ $\displaystyle \int\mathcal{H}_{0}^{1/2}d^{3}x\ \ \ \ \ (14)$ $\displaystyle$ $\displaystyle =$ $\displaystyle \int\left[-i\bar{\psi}\gamma^{j}\partial_{j}\psi+m\bar{\psi}\psi\right]d^{3}x \ \ \ \ \ (15)$

As you might guess, this is a messy operation, and Klauber goes through (most) of the gory details in his section 4.4.1. Basically, we substitute 12 and 13 into 15, using different indices for the sums in each case (${r,\mathbf{p}}$ for ${\psi}$ and ${s,\mathbf{p}^{\prime}}$ for ${\bar{\psi}}$). We then use the fact that the integral of ${e^{\pm i\mathbf{p}\cdot\mathbf{x}}}$ over all space gives zero unless ${\mathbf{p}=0}$ (because we’re assuming that the volume ${V}$ is such that an integral number of wavelengths fits exactly within its boundaries). This allows us to collapse the double sum over ${\mathbf{p}}$ and ${\mathbf{p}^{\prime}}$ to a single sum over ${\mathbf{p}}$, although we must still retain the double sum over the spins ${r}$ and ${s}$. We get a set of four integrals resulting from the derivative term in 15 and another four integrals resulting from the mass term in 15. We need to add each integral in the first set of four to the corresponding integral in the second set. We’ll consider the sum of the fourth integral from each set here (the derivation is similar to that done by Klauber in his Box 4-3, where he adds the first integral from each set). The fourth integral from the derivative term is (implied sum over ${j=1,2,3}$)

$\displaystyle I_{d4}\equiv-\int\sum_{r,s,\mathbf{p}}\frac{m}{VE_{\mathbf{p}}}c_{r}^{\dagger}\left(-\mathbf{p}\right)\bar{u}_{r}\left(-\mathbf{p}\right)\gamma^{j}p^{j}v_{s}\left(\mathbf{p}\right)d_{s}^{\dagger}\left(\mathbf{p}\right)e^{2iE_{\mathbf{p}}t}d^{3}x \ \ \ \ \ (16)$

The fourth integral from the mass term is

$\displaystyle I_{m4}\equiv\int\sum_{r,s,\mathbf{p}}\frac{m}{VE_{\mathbf{p}}}c_{r}^{\dagger}\left(-\mathbf{p}\right)\bar{u}_{r}\left(-\mathbf{p}\right)mv_{s}\left(\mathbf{p}\right)d_{s}^{\dagger}\left(\mathbf{p}\right)e^{2iE_{\mathbf{p}}t}d^{3}x \ \ \ \ \ (17)$

The only differences between the two integrals are the minus sign in ${I_{d4}}$ and the replacement of ${\gamma^{j}p^{j}}$ in ${I_{d4}}$ by ${m}$ in ${I_{m4}}$.

[In Klauber’s Box 4-3, he uses a symbol with a slash through it to indicate a sum of that symbol multiplied by ${\gamma^{\mu}}$. Unfortunately, the wordpress.com Latex server doesn’t support symbols with a slash through them, so I’ll need to resort to the longhand notation.]

Suppose we have a single eigensolution which is a single term from the last term in 12, that is (no sum over ${r}$; we can take ${r}$ to be either spin):

$\displaystyle \psi=\sqrt{\frac{m}{VE_{\mathbf{p}}}}d_{r}^{\dagger}\left(\mathbf{p}\right)v_{r}\left(\mathbf{p}\right)e^{ipx} \ \ \ \ \ (18)$

Since this is a solution of the Dirac equation we must have

 $\displaystyle \left(i\gamma^{\mu}\partial_{\mu}-m\right)\psi$ $\displaystyle =$ $\displaystyle 0\ \ \ \ \ (19)$ $\displaystyle \sqrt{\frac{m}{VE_{\mathbf{p}}}}d_{r}^{\dagger}\left(\mathbf{p}\right)e^{ipx}\left(-\gamma^{\mu}p_{\mu}-m\right)v_{r}\left(\mathbf{p}\right)$ $\displaystyle =$ $\displaystyle 0 \ \ \ \ \ (20)$

Remember that this is a matrix equation, and note that the terms ${\sqrt{\frac{m}{VE_{\mathbf{p}}}}d_{r}^{\dagger}\left(\mathbf{p}\right)e^{ipx}}$ are not matrices, nor are they zero, so we must have

$\displaystyle \left(\gamma^{\mu}p_{\mu}+m\right)v_{r}\left(\mathbf{p}\right)=0 \ \ \ \ \ (21)$

Returning to 16, we can convert the integrand using the inner product of the spinor terms

$\displaystyle u_{r}^{\dagger}\left(-\mathbf{p}\right)v_{s}\left(\mathbf{p}\right)=0 \ \ \ \ \ (22)$

From the definition of the adjoint spinor ${\bar{u}_{r}}$ and the identity ${\gamma^{0}\gamma^{0}=1}$, we have

 $\displaystyle u_{r}^{\dagger}\left(-\mathbf{p}\right)v_{s}\left(\mathbf{p}\right)$ $\displaystyle =$ $\displaystyle u_{r}^{\dagger}\left(-\mathbf{p}\right)\gamma^{0}\gamma^{0}v_{s}\left(\mathbf{p}\right)\ \ \ \ \ (23)$ $\displaystyle$ $\displaystyle =$ $\displaystyle \bar{u}_{r}\left(-\mathbf{p}\right)\gamma^{0}v_{s}\left(\mathbf{p}\right)\ \ \ \ \ (24)$ $\displaystyle$ $\displaystyle =$ $\displaystyle 0 \ \ \ \ \ (25)$

Since this term is zero, we can multiply it by anything without changing it, so we multiply it by ${p_{0}}$ to get

$\displaystyle u_{r}^{\dagger}\left(-\mathbf{p}\right)v_{s}\left(\mathbf{p}\right)=\bar{u}_{r}\left(-\mathbf{p}\right)\gamma^{0}p_{0}v_{s}\left(\mathbf{p}\right)=0 \ \ \ \ \ (26)$

The middle part of 16 can be written as

 $\displaystyle \bar{u}_{r}\left(-\mathbf{p}\right)\gamma^{j}p^{j}v_{s}\left(\mathbf{p}\right)$ $\displaystyle =$ $\displaystyle \bar{u}_{r}\left(-\mathbf{p}\right)\gamma^{j}p_{j}v_{s}\left(\mathbf{p}\right)\ \ \ \ \ (27)$ $\displaystyle$ $\displaystyle =$ $\displaystyle -\bar{u}_{r}\left(-\mathbf{p}\right)\gamma^{0}p_{0}v_{s}\left(\mathbf{p}\right)+\bar{u}_{r}\left(-\mathbf{p}\right)\gamma^{j}p_{j}v_{s}\left(\mathbf{p}\right)\ \ \ \ \ (28)$ $\displaystyle$ $\displaystyle =$ $\displaystyle -\bar{u}_{r}\left(-\mathbf{p}\right)\gamma^{\mu}p_{\mu}v_{s}\left(\mathbf{p}\right)\ \ \ \ \ (29)$ $\displaystyle$ $\displaystyle =$ $\displaystyle \bar{u}_{r}\left(-\mathbf{p}\right)mv_{s}\left(\mathbf{p}\right) \ \ \ \ \ (30)$

where we subtracted 26 (which is zero, so it doesn’t change anything) in the second line and used 21 to get the last line. Putting this final result into 16 we get

$\displaystyle I_{d4}=-\int\sum_{r,s,\mathbf{p}}\frac{m}{VE_{\mathbf{p}}}c_{r}^{\dagger}\left(-\mathbf{p}\right)\bar{u}_{r}\left(-\mathbf{p}\right)mv_{s}\left(\mathbf{p}\right)d_{s}^{\dagger}\left(\mathbf{p}\right)e^{2iE_{\mathbf{p}}t}d^{3}=-I_{m4} \ \ \ \ \ (31)$

so ${I_{d4}+I_{m4}=0}$ and the two integrals cancel each other.

# Free scalar Hamiltonian as an integral of the Hamiltonian density

Reference: References: Robert D. Klauber, Student Friendly Quantum Field Theory, (Sandtrove Press, 2013) – Chapter 3, Problem 3.8.

The free-field Hamiltonian density used in the Klein-Gordon equation is, in terms of the field ${\phi}$ and the mass ${\mu}$:

$\displaystyle \mathcal{H}=\dot{\phi}\dot{\phi}^{\dagger}+\nabla\phi^{\dagger}\cdot\nabla\phi+\mu^{2}\phi^{\dagger}\phi \ \ \ \ \ (1)$

[I’ve dropped the superscript and subscript 0 to save space, but this is still a scalar Hamiltonian density with no interactions.]

The discrete plane-wave solutions of the Klein-Gordon equation are

 $\displaystyle \phi$ $\displaystyle =$ $\displaystyle \sum_{\mathbf{k}}\frac{1}{\sqrt{2V\omega_{\mathbf{k}}}}\left(a\left(\mathbf{k}\right)e^{-ikx}+b^{\dagger}\left(\mathbf{k}\right)e^{ikx}\right)\ \ \ \ \ (2)$ $\displaystyle \dot{\phi}^{\dagger}$ $\displaystyle =$ $\displaystyle \sum_{\mathbf{k}^{\prime}}\frac{1}{\sqrt{2V\omega_{\mathbf{k}^{\prime}}}}\left(a^{\dagger}\left(\mathbf{k}^{\prime}\right)e^{ik^{\prime}x}+b\left(\mathbf{k}^{\prime}\right)e^{-ik^{\prime}x}\right) \ \ \ \ \ (3)$

Using these we can find the total Hamiltonian by integrating over 3-d space:

$\displaystyle H=\int\mathcal{H}d^{3}x \ \ \ \ \ (4)$

In this case, we’re dealing with discrete solutions over a finite volume ${V}$, such that the values of ${\mathbf{k}}$ are determined by the condition that an integral number of wavelengths fits into ${V}$.

Klauber works out the integral of the first two terms in 1 in his section 3.4.1 and asks us to do the third term, but it’s worth looking at the technique used on the first two terms. To do the derivatives, we remember that

$\displaystyle kx=k_{\mu}x^{\mu}=\omega_{\mathbf{k}}t+k_{i}x^{i}=\omega_{\mathbf{k}}t-\mathbf{k}\cdot\mathbf{x} \ \ \ \ \ (5)$

Therefore

 $\displaystyle \dot{\phi}$ $\displaystyle =$ $\displaystyle \sum_{\mathbf{k}}\frac{i\omega_{\mathbf{k}}}{\sqrt{2V\omega_{\mathbf{k}}}}\left(-a\left(\mathbf{k}\right)e^{-ikx}+b^{\dagger}\left(\mathbf{k}\right)e^{ikx}\right)\ \ \ \ \ (6)$ $\displaystyle \dot{\phi}^{\dagger}$ $\displaystyle =$ $\displaystyle \sum_{\mathbf{k}^{\prime}}\frac{i\omega_{\mathbf{k}^{\prime}}}{\sqrt{2V\omega_{\mathbf{k}^{\prime}}}}\left(a^{\dagger}\left(\mathbf{k}^{\prime}\right)e^{ik^{\prime}x}-b\left(\mathbf{k}^{\prime}\right)e^{-ik^{\prime}x}\right) \ \ \ \ \ (7)$

When we calculate ${\int\dot{\phi}\dot{\phi}^{\dagger}d^{3}x}$, we get a collection of integrals of the form

$\displaystyle I_{\mathbf{kk}^{\prime}}=\begin{cases} \int e^{ikx+ik^{\prime}x}d^{3}x\\ \int e^{ikx-ik^{\prime}x}d^{3}x\\ \int e^{-ikx-ik^{\prime}x}d^{3}x\\ \int e^{-ikx+ik^{\prime}x}d^{3}x \end{cases} \ \ \ \ \ (8)$

If we were integrating over an infinite volume, these would give delta functions, but over a finite volume, with the boundary conditions specified above, the functions ${e^{ikx}}$ and ${e^{ik^{\prime}x}}$ are orthogonal. Therefore, these integrals are zero unless the exponent is zero, in which case we integrate 1 over the volume ${V}$, giving ${V}$. This means that in the first and third integrals in 8, we must have ${\mathbf{k}=-\mathbf{k}^{\prime}}$ and in the second and fourth integrals, we have ${\mathbf{k}=+\mathbf{k}^{\prime}}$. Using this condition, and multiplying out 6 and 7 we get

$\displaystyle \int\dot{\phi}\dot{\phi}^{\dagger}d^{3}x=\sum_{\mathbf{k}}\frac{\left(\omega_{\mathbf{k}}\right)^{2}}{2\omega_{\mathbf{k}}}\left[-a_{\mathbf{k}}b_{-\mathbf{k}}e^{-2i\omega_{\mathbf{k}}t}+a_{\mathbf{k}}a_{\mathbf{k}}^{\dagger}+b_{\mathbf{k}}^{\dagger}b_{\mathbf{k}}-b_{\mathbf{k}}^{\dagger}a_{-\mathbf{k}}^{\dagger}e^{2i\omega_{\mathbf{k}}t}\right] \ \ \ \ \ (9)$

Since we’re summing over all ${\mathbf{k}}$, positive and negative, we can flip the sign of ${\mathbf{k}}$ in the first and fourth terms to give (remember that ${\omega_{\mathbf{k}}=\omega_{-\mathbf{k}}}$):

$\displaystyle \int\dot{\phi}\dot{\phi}^{\dagger}d^{3}x=\sum_{\mathbf{k}}\frac{\left(\omega_{\mathbf{k}}\right)^{2}}{2\omega_{\mathbf{k}}}\left[-a_{-\mathbf{k}}b_{\mathbf{k}}e^{-2i\omega_{\mathbf{k}}t}+a_{\mathbf{k}}a_{\mathbf{k}}^{\dagger}+b_{\mathbf{k}}^{\dagger}b_{\mathbf{k}}-b_{-\mathbf{k}}^{\dagger}a_{\mathbf{k}}^{\dagger}e^{2i\omega_{\mathbf{k}}t}\right] \ \ \ \ \ (10)$

To integrate the gradient terms, we have

$\displaystyle \nabla\phi^{\dagger}\cdot\nabla\phi=-\partial_{i}\phi^{\dagger}\partial^{i}\phi=\sum_{i}\partial_{i}\phi^{\dagger}\partial_{i}\phi \ \ \ \ \ (11)$

[The minus sign is because ${\partial_{i}=-\partial^{i}}$ for spatial components.] Klauber gives the details in his equation 3-52, but the procedure is similar to that for calculating ${\dot{\phi}}$ above. We have, using 5 and ${\partial_{i}e^{ikx}=-ik_{i}}$:

 $\displaystyle \partial_{i}\phi^{\dagger}$ $\displaystyle =$ $\displaystyle \sum_{\mathbf{k}^{\prime}}\frac{-ik_{i}^{\prime}}{\sqrt{2V\omega_{\mathbf{k}^{\prime}}}}\left(a^{\dagger}\left(\mathbf{k}^{\prime}\right)e^{ik^{\prime}x}-b\left(\mathbf{k}^{\prime}\right)e^{-ik^{\prime}x}\right)\ \ \ \ \ (12)$ $\displaystyle \partial_{i}\phi$ $\displaystyle =$ $\displaystyle \sum_{\mathbf{k}}\frac{-ik_{i}}{\sqrt{2V\omega_{\mathbf{k}}}}\left(-a\left(\mathbf{k}\right)e^{-ikx}+b^{\dagger}\left(\mathbf{k}\right)e^{ikx}\right) \ \ \ \ \ (13)$

Multiplying the two terms together and integrating, again using the orthonormality of the exponentials so that ${\mathbf{k}=\pm\mathbf{k}^{\prime}}$, depending on the term, gives

 $\displaystyle \int\nabla\phi^{\dagger}\cdot\nabla\phi d^{3}x$ $\displaystyle =$ $\displaystyle \sum_{\mathbf{k}}\frac{1}{2\omega_{\mathbf{k}}}\left[k_{i}\left(-k_{i}\right)\left(-b_{\mathbf{k}}a_{-\mathbf{k}}e^{-2i\omega_{\mathbf{k}}t}-a_{\mathbf{k}}^{\dagger}b_{-\mathbf{k}}^{\dagger}e^{2i\omega_{\mathbf{k}}t}\right)+k_{i}k_{i}\left(a_{\mathbf{k}}^{\dagger}a_{\mathbf{k}}+b_{\mathbf{k}}b_{\mathbf{k}}^{\dagger}\right)\right]\ \ \ \ \ (14)$ $\displaystyle$ $\displaystyle =$ $\displaystyle \sum_{\mathbf{k}}\frac{\mathbf{k}^{2}}{2\omega_{\mathbf{k}}}\left[b_{\mathbf{k}}a_{-\mathbf{k}}e^{-2i\omega_{\mathbf{k}}t}+a_{\mathbf{k}}^{\dagger}b_{-\mathbf{k}}^{\dagger}e^{2i\omega_{\mathbf{k}}t}+a_{\mathbf{k}}^{\dagger}a_{\mathbf{k}}+b_{\mathbf{k}}b_{\mathbf{k}}^{\dagger}\right] \ \ \ \ \ (15)$

Finally, for the third term in 1 we have

 $\displaystyle \mu^{2}\int\phi^{\dagger}\phi d^{3}x$ $\displaystyle =$ $\displaystyle \mu^{2}\int d^{3}x\left[\sum_{\mathbf{k}^{\prime}}\frac{1}{\sqrt{2V\omega_{\mathbf{k}^{\prime}}}}\left(a^{\dagger}\left(\mathbf{k}^{\prime}\right)e^{ik^{\prime}x}+b\left(\mathbf{k}^{\prime}\right)e^{-ik^{\prime}x}\right)\right]\times\nonumber$ $\displaystyle$ $\displaystyle$ $\displaystyle \left[\sum_{\mathbf{k}}\frac{1}{\sqrt{2V\omega_{\mathbf{k}}}}\left(a\left(\mathbf{k}\right)e^{-ikx}+b^{\dagger}\left(\mathbf{k}\right)e^{ikx}\right)\right]\ \ \ \ \ (16)$ $\displaystyle$ $\displaystyle$ $\displaystyle \sum_{\mathbf{k}}\frac{\mu^{2}}{2\omega_{\mathbf{k}}}\left[a_{\mathbf{k}}^{\dagger}a_{\mathbf{k}}+b_{\mathbf{k}}b_{\mathbf{k}}^{\dagger}+b_{\mathbf{k}}a_{-\mathbf{k}}e^{-2i\omega_{\mathbf{k}}t}+a_{\mathbf{k}}^{\dagger}b_{-\mathbf{k}}^{\dagger}e^{2i\omega_{\mathbf{k}}t}\right] \ \ \ \ \ (17)$

From the relativistic energy formula (in natural units)

$\displaystyle \omega_{\mathbf{k}}^{2}=\mathbf{k}^{2}+\mu^{2} \ \ \ \ \ (18)$

we can add 15 and 17 to get

$\displaystyle \mu^{2}\int\phi^{\dagger}\phi d^{3}x\int\nabla\phi^{\dagger}\cdot\nabla\phi d^{3}x+=\sum_{\mathbf{k}}\frac{\omega_{\mathbf{k}}^{2}}{2\omega_{\mathbf{k}}}\left[a_{\mathbf{k}}^{\dagger}a_{\mathbf{k}}+b_{\mathbf{k}}b_{\mathbf{k}}^{\dagger}+b_{\mathbf{k}}a_{-\mathbf{k}}e^{-2i\omega_{\mathbf{k}}t}+a_{\mathbf{k}}^{\dagger}b_{-\mathbf{k}}^{\dagger}e^{2i\omega_{\mathbf{k}}t}\right] \ \ \ \ \ (19)$

To add this to 10, we need to use the commutators

$\displaystyle \left[a_{\mathbf{k}},a_{\mathbf{k}^{\prime}}^{\dagger}\right]=\left[b_{\mathbf{k}},b_{\mathbf{k}^{\prime}}^{\dagger}\right]=\delta_{\mathbf{kk^{\prime\mbox{ }}}} \ \ \ \ \ (20)$

Then 10 and 19 become

 $\displaystyle \int\dot{\phi}\dot{\phi}^{\dagger}d^{3}x$ $\displaystyle =$ $\displaystyle \sum_{\mathbf{k}}\frac{\left(\omega_{\mathbf{k}}\right)^{2}}{2\omega_{\mathbf{k}}}\left[-a_{-\mathbf{k}}b_{\mathbf{k}}e^{-2i\omega_{\mathbf{k}}t}+a_{\mathbf{k}}^{\dagger}a_{\mathbf{k}}+1+b_{\mathbf{k}}^{\dagger}b_{\mathbf{k}}-b_{-\mathbf{k}}^{\dagger}a_{\mathbf{k}}^{\dagger}e^{2i\omega_{\mathbf{k}}t}\right]\ \ \ \ \ (21)$ $\displaystyle \mu^{2}\int\phi^{\dagger}\phi d^{3}x\int\nabla\phi^{\dagger}\cdot\nabla\phi d^{3}x+$ $\displaystyle =$ $\displaystyle \sum_{\mathbf{k}}\frac{\omega_{\mathbf{k}}^{2}}{2\omega_{\mathbf{k}}}\left[a_{\mathbf{k}}^{\dagger}a_{\mathbf{k}}+b_{\mathbf{k}}^{\dagger}b_{\mathbf{k}}+1+b_{\mathbf{k}}a_{-\mathbf{k}}e^{-2i\omega_{\mathbf{k}}t}+a_{\mathbf{k}}^{\dagger}b_{-\mathbf{k}}^{\dagger}e^{2i\omega_{\mathbf{k}}t}\right] \ \ \ \ \ (22)$

 $\displaystyle H$ $\displaystyle =$ $\displaystyle \int\mathcal{H}d^{3}x=\sum_{\mathbf{k}}\omega_{\mathbf{k}}\left[a_{\mathbf{k}}^{\dagger}a_{\mathbf{k}}+\frac{1}{2}+b_{\mathbf{k}}^{\dagger}b_{\mathbf{k}}+\frac{1}{2}\right]\ \ \ \ \ (23)$ $\displaystyle$ $\displaystyle =$ $\displaystyle \sum_{\mathbf{k}}\omega_{\mathbf{k}}\left[N_{a}\left(\mathbf{k}\right)+\frac{1}{2}+N_{b}\left(\mathbf{k}\right)+\frac{1}{2}\right] \ \ \ \ \ (24)$
where ${N_{a}}$ and ${N_{b}}$ are the number operators that we met earlier, although in a continuous system.