Tag Archives: momentum space

Harmonic oscillator – eigenfunctions in momentum space

Shankar, R. (1994), Principles of Quantum Mechanics, Plenum Press. Section 7.3, Exercise 7.3.7.

We’ve seen how to solve the Schrödinger equation for the harmonic oscillator in the position basis, where the independent variable is {x}. It’s actually fairly easy to adapt this solution to find the wave functions in momentum space. (We’ve also found these functions by using the Fourier transform of the position functions, but the present post shows an easier way.)

The Schrödinger equation for the stationary states of the harmonic oscillator is, in operator form:

\displaystyle  \frac{P^{2}}{2m}\psi+\frac{1}{2}m\omega^{2}X^{2}\psi=E\psi \ \ \ \ \ (1)

To work in momentum space, we use the results

\displaystyle   P \displaystyle  = \displaystyle  p\ \ \ \ \ (2)
\displaystyle  X \displaystyle  = \displaystyle  i\hbar\frac{\partial}{\partial p} \ \ \ \ \ (3)

This gives

\displaystyle  \frac{p^{2}}{2m}\psi-\frac{1}{2}\hbar^{2}m\omega^{2}\frac{d^{2}\psi}{dp^{2}}=E\psi \ \ \ \ \ (4)

Dividing through by {\left(m\omega\right)^{2}} we get

\displaystyle  -\frac{\hbar^{2}}{2m}\psi^{\prime\prime}+\frac{p^{2}}{2m^{3}\omega^{2}}=\frac{E}{\left(m\omega\right)^{2}}\psi \ \ \ \ \ (5)

where a prime on {\psi} indicates a derivative with respect to {p}.

This is similar to the Schrödinger equation in position space:

\displaystyle  -\frac{\hbar^{2}}{2m}\psi^{\prime\prime}+\frac{1}{2}m\omega^{2}x^{2}\psi=E\psi \ \ \ \ \ (6)

(where a prime here indicates a derivative with respect to {x}). When we solved the position space equation, we introduced a dimensionless variable

\displaystyle  y\equiv\sqrt{\frac{m\omega}{\hbar}}x \ \ \ \ \ (7)

Using this technique to solve 5, we try a definition for {y} of

\displaystyle  y\equiv\frac{p}{\sqrt{\hbar m\omega}} \ \ \ \ \ (8)

(You can check the units of {\sqrt{\hbar m\omega}} to see they are the units of momentum, so {y} is indeed dimensionless here.) Making this substitution, we get

\displaystyle   \frac{d^{2}\psi}{dp^{2}} \displaystyle  = \displaystyle  \frac{1}{\hbar m\omega}\frac{d^{2}\psi}{dy^{2}}\ \ \ \ \ (9)
\displaystyle  \frac{p^{2}}{2m^{3}\omega^{2}} \displaystyle  = \displaystyle  \frac{\hbar y^{2}}{2m^{2}\omega} \ \ \ \ \ (10)

Thus 5 becomes

\displaystyle   -\frac{\hbar^{2}}{2m}\frac{1}{\hbar m\omega}\frac{d^{2}\psi}{dy^{2}}+\frac{\hbar y^{2}}{2m^{2}\omega}\psi \displaystyle  = \displaystyle  \frac{E}{\left(m\omega\right)^{2}}\psi\ \ \ \ \ (11)
\displaystyle  \frac{\hbar^{2}}{2m}\left[-\frac{d^{2}\psi}{dy^{2}}+y^{2}\psi\right] \displaystyle  = \displaystyle  \frac{\hbar E}{\omega}\psi \ \ \ \ \ (12)

We can now use the same dimensionless parameter we used in the earlier derivation:

\displaystyle  \varepsilon\equiv\frac{E}{\hbar\omega} \ \ \ \ \ (13)

This results in the differential equation

\displaystyle  \psi^{\prime\prime}+\left(2\varepsilon-y^{2}\right)\psi=0 \ \ \ \ \ (14)

where a prime now indicates a derivative with respect to {y}. This is exactly the same differential equation that we got for the position basis, except that the independent variable {y} is now defined in terms of {p} by 8 instead of {x}. We can solve it in the same way, which results in the same quantization condition on the allowable energies of {E_{n}=\left(n+\frac{1}{2}\right)\hbar\omega}. The eigenfunctions look the same when expressed in terms of {y}:

\displaystyle  \psi_{n}(y)=A\frac{1}{\sqrt{2^{n}n!}}H_{n}(y)e^{-y^{2}/2} \ \ \ \ \ (15)

where {A} is a normalization constant with the value in the position basis of

\displaystyle  A=\left(\frac{m\omega}{\pi\hbar}\right)^{1/4} \ \ \ \ \ (16)

and {H_{n}} is a Hermite polynomial. We can get the eigenfunctions in momentum space by replacing {y} by 8. We can see that this amounts to replacing {x\rightarrow p} and {m\omega\rightarrow\frac{1}{m\omega}}, so we get

\displaystyle  \psi_{n}(p)=\frac{1}{\left(\pi\hbar m\omega\right)^{1/4}}\frac{1}{\sqrt{2^{n}n!}}H_{n}\left(\frac{p}{\sqrt{\hbar m\omega}}\right)e^{-p^{2}/2\hbar m\omega} \ \ \ \ \ (17)

In particular, the ground state is

\displaystyle  \psi_{0}\left(p\right)=\frac{1}{\left(\pi\hbar m\omega\right)^{1/4}}e^{-p^{2}/2\hbar m\omega} \ \ \ \ \ (18)

Momentum space in 3-d

Required math: calculus

Required physics: 3-d Schrödinger equation

Reference: Griffiths, David J. (2005), Introduction to Quantum Mechanics, 2nd Edition; Pearson Education – Problem 4.42.

We can generalize the 1-d definition of momentum space to 3-d, so the momentum space wave function is

\displaystyle  \phi\left(\mathbf{p}\right)=\frac{1}{\left(2\pi\hbar\right)^{3/2}}\int e^{-i\mathbf{p\cdot r}/\hbar}\psi\left(\mathbf{r}\right)d^{3}\mathbf{r} \ \ \ \ \ (1)

As an example, we can calculate this function for the ground state of hydrogen:

\displaystyle  \psi_{100}=\frac{1}{\sqrt{\pi}a^{3/2}}e^{-r/a} \ \ \ \ \ (2)

Since {\psi_{100}} is spherically symmetric, the integral above should be the same for a given magnitude {p} regardless of the direction of {\mathbf{p}}. Taking {\mathbf{p}} along the polar axis means that {\mathbf{p}\cdot\mathbf{r}=pr\cos\theta}, so

\displaystyle   \phi(\mathbf{p}) \displaystyle  = \displaystyle  \frac{1}{\sqrt{\pi a^{3}}(2\pi\hbar)^{3/2}}\int_{0}^{2\pi}\int_{0}^{\infty}\int_{0}^{\pi}e^{-ipr\cos\theta/\hbar}e^{-r/a}r^{2}\sin\theta d\theta drd\phi\ \ \ \ \ (3)
\displaystyle  \displaystyle  = \displaystyle  \frac{1}{\pi}\left(\frac{2a}{\hbar}\right)^{3/2}\frac{1}{(1+a^{2}p^{2}/\hbar^{2})^{2}} \ \ \ \ \ (4)

where we used Maple to do the integration. It would be nice to check the independence of the result on the direction of {\mathbf{p}}, but unfortunately, choosing any other direction gives an intractable integral. For example, if we took {\mathbf{p}} to be along the {x} axis, so that {\mathbf{p}=\left[p,0,0\right]}, then since {\mathbf{r}=\left[x,y,z\right]} we have {\mathbf{r}\cdot\mathbf{p}=xp=rp\sin\theta\cos\phi}. Since this factor appears in the exponent, there’s no analytic integral that I know of as a solution.

Also, since any hydrogen wave function with a non-zero {l} number has a dependence on the angles {\theta} and {\phi} via the spherical harmonic, the direction of {\mathbf{p}} does matter, so the calculation above doesn’t apply in those cases. The integrals, even in the case where we take {\mathbf{p}} along the polar axis, are all horrible anyway, so it’s doubtful we could get a closed solution even in that special case.

To check normalization, we integrate {\phi^{2}(\mathbf{p})} over {p}-space in three dimensions.

\displaystyle   \int_{0}^{2\pi}\int_{0}^{\infty}\int_{0}^{\pi}\phi^{2}(\mathbf{p})p^{2}\sin\theta d\theta dpd\phi \displaystyle  = \displaystyle  \frac{1}{\pi^{2}}\left(\frac{2a}{\hbar}\right)^{3}\int_{0}^{2\pi}\int_{0}^{\infty}\int_{0}^{\pi}\frac{p^{2}\sin\theta d\theta dpd\phi}{(1+a^{2}p^{2}/\hbar^{2})^{4}}\ \ \ \ \ (5)
\displaystyle  \displaystyle  = \displaystyle  1 \ \ \ \ \ (6)

Again, Maple was used for the integral.

To get the mean kinetic energy, we first calculate:

\displaystyle   \langle p^{2}\rangle \displaystyle  = \displaystyle  \int_{0}^{2\pi}\int_{0}^{\infty}\int_{0}^{\pi}\phi^{2}(\mathbf{p})p^{4}\sin\theta d\theta dpd\phi\ \ \ \ \ (7)
\displaystyle  \displaystyle  = \displaystyle  \frac{1}{\pi^{2}}\left(\frac{2a}{\hbar}\right)^{3}\int_{0}^{2\pi}\int_{0}^{\infty}\int_{0}^{\pi}\frac{p^{4}\sin\theta d\theta dpd\phi}{(1+a^{2}p^{2}/\hbar^{2})^{4}}\ \ \ \ \ (8)
\displaystyle  \displaystyle  = \displaystyle  \frac{\hbar^{2}}{a^{2}} \ \ \ \ \ (9)

The kinetic energy is therefore

\displaystyle   \langle T\rangle \displaystyle  = \displaystyle  \frac{\langle p^{2}\rangle}{2m}\ \ \ \ \ (10)
\displaystyle  \displaystyle  = \displaystyle  \frac{\hbar^{2}}{2ma^{2}}\ \ \ \ \ (11)
\displaystyle  \displaystyle  = \displaystyle  \left(\frac{e^{2}}{4\pi\epsilon_{0}}\right)^{2}\frac{m}{2\hbar^{2}}\ \ \ \ \ (12)
\displaystyle  \displaystyle  = \displaystyle  -E_{1} \ \ \ \ \ (13)

where we’ve used the formula for the energy levels in the last line. This agrees with the result of the 3-d virial theorem.

Free particle in momentum space

Required math: calculus

Required physics: Schrödinger equation

References: Griffiths, David J. (2005), Introduction to Quantum Mechanics, 2nd Edition; Pearson Education – Problem 3.40.

Since the Hamiltonian for a free particle is {H=p^{2}/2m}, the Schrodinger equation in momentum space is

\displaystyle  i\hbar\frac{\partial\Phi}{\partial t}=\frac{p^{2}}{2m}\Phi \ \ \ \ \ (1)

so the solution can be found by simply integrating with respect to {t}:

\displaystyle  \Phi(p,t)=e^{-ip^{2}t/2m\hbar}\Phi(p,0) \ \ \ \ \ (2)

We looked at the travelling Gaussian wave packet in free space earlier. Its initial state in position space is

\displaystyle  \Psi(x,0)=\left(\frac{2a}{\pi}\right)^{1/4}e^{-ax^{2}}e^{ilx} \ \ \ \ \ (3)

To find {\Phi(p,0)} we use the conversion to momentum space we found earlier:

\displaystyle  \Phi(p,0)=\frac{1}{\sqrt{2\pi\hbar}}\left(\frac{2a}{\pi}\right)^{1/4}\int_{-\infty}^{\infty}e^{-ax^{2}+ilx-ipx/\hbar}dx \ \ \ \ \ (4)

From the analysis of the travelling Gaussian packet we see that the integral is the same as that done when calculating {\phi(k)} if we replace {k} with {p/\hbar}. Therefore

\displaystyle  \Phi(p,0)=\left(\frac{2}{\pi a}\right)^{1/4}\frac{1}{\sqrt{2\hbar}}e^{-(p/\hbar-l)^{2}/4a} \ \ \ \ \ (5)

Using 2, we have the full solution for {\Phi(p,t)}:

\displaystyle  \Phi(p,t)=\left(\frac{2}{\pi a}\right)^{1/4}\frac{1}{\sqrt{2\hbar}}e^{-ip^{2}t/2m\hbar}e^{-(p/\hbar-l)^{2}/4a} \ \ \ \ \ (6)


\displaystyle  |\Phi(p,t)|^{2}=\frac{1}{\hbar}\frac{1}{\sqrt{2\pi a}}e^{-(p/\hbar-l)^{2}/2a} \ \ \ \ \ (7)

which is independent of time. (As a check, we can integrate this over all {p} and verify that this integral is 1.)

We can calculate the means for momentum in the usual way:

\displaystyle   \langle p\rangle \displaystyle  = \displaystyle  \frac{1}{\hbar}\frac{1}{\sqrt{2\pi a}}\int_{-\infty}^{\infty}pe^{-(p/\hbar-l)^{2}/2a}dp\ \ \ \ \ (8)
\displaystyle  \displaystyle  = \displaystyle  \hbar l\ \ \ \ \ (9)
\displaystyle  \langle p^{2}\rangle \displaystyle  = \displaystyle  \frac{1}{\hbar}\frac{1}{\sqrt{2\pi a}}\int_{-\infty}^{\infty}p^{2}e^{-(p/\hbar-l)^{2}/2a}dp\ \ \ \ \ (10)
\displaystyle  \displaystyle  = \displaystyle  \hbar^{2}(l^{2}+a) \ \ \ \ \ (11)

Both results agree with those in the analysis of the travelling Gaussian packet.

For the mean energy, we have

\displaystyle   \langle H\rangle \displaystyle  = \displaystyle  \left\langle \frac{p^{2}}{2m}\right\rangle \ \ \ \ \ (12)
\displaystyle  \displaystyle  = \displaystyle  \frac{\hbar^{2}}{2m}(l^{2}+a)\ \ \ \ \ (13)
\displaystyle  \displaystyle  = \displaystyle  \frac{\langle p\rangle^{2}}{2m}+\frac{a\hbar^{2}}{2m} \ \ \ \ \ (14)

Referring back to the stationary Gaussian wave packet in free space, we see that {\langle p^{2}\rangle=a\hbar^{2}}, so the energy is the sum of that for a stationary Gaussian wave packet and the term {\langle p\rangle^{2}/2m}. For the travelling packet, there is a net non-zero average momentum, so {\langle p\rangle} is non-zero. Thus the energy arises from the inherent energy of the wave packet, plus the kinetic energy of motion of the packet.

Momentum space: another example

Required math: calculus

Required physics: Schrödinger equation

References: Griffiths, David J. (2005), Introduction to Quantum Mechanics, 2nd Edition; Pearson Education – Problem 3.30.

Another example of calculations in momentum space. Suppose we have the initial state of a wave function in position space given by

\displaystyle  \Psi(x,0)=\frac{A}{x^{2}+a^{2}} \ \ \ \ \ (1)

where {a} and {A} are constants.

(a) Applying normalization

\displaystyle   1 \displaystyle  = \displaystyle  A^{2}\int_{-\infty}^{\infty}\left(\frac{1}{x^{2}+a^{2}}\right)^{2}dx\ \ \ \ \ (2)
\displaystyle  \displaystyle  = \displaystyle  \frac{\pi A^{2}}{2a^{3}}\ \ \ \ \ (3)
\displaystyle  A \displaystyle  = \displaystyle  \sqrt{\frac{2}{\pi}}a^{3/2} \ \ \ \ \ (4)

(b) We can use the position space wave function to work out {\left\langle x\right\rangle } and {\left\langle x^{2}\right\rangle }. Since {|\Psi(x,0)|^{2}=A^{2}/(x^{2}+a^{2})^{2}} is even, {\langle x\rangle=0}. For {\left\langle x^{2}\right\rangle }, we find

\displaystyle   \left\langle x^{2}\right\rangle \displaystyle  = \displaystyle  \frac{2}{\pi}a^{3}\int_{-\infty}^{\infty}x^{2}\left(\frac{1}{x^{2}+a^{2}}\right)^{2}dx\ \ \ \ \ (5)
\displaystyle  \displaystyle  = \displaystyle  a^{2} \ \ \ \ \ (6)

Thus {\sigma_{x}=\sqrt{\langle x^{2}\rangle-\langle x\rangle^{2}}=a}.

(c) To get the momentum space wave function, we must do the integral

\displaystyle   \Phi(p,0) \displaystyle  = \displaystyle  \frac{1}{\sqrt{2\pi\hbar}}\int_{-\infty}^{\infty}e^{-ipx/\hbar}\sqrt{\frac{2a^{3}}{\pi}}\frac{1}{x^{2}+a^{2}}dx \ \ \ \ \ (7)

Feeding this integral into Maple and telling it to assume {p} is real gives an answer that depends on the sign of p. Simplifying the expression for the two signs gives the single result:

\displaystyle  \Phi(p,0)=\sqrt{\frac{a}{\hbar}}e^{-|p|a/\hbar} \ \ \ \ \ (8)

Note that this is an even function of p, which is correct, since looking at the integral, the complex exponential is the sum of a real, even function (cosine) and an imaginary, odd function (i*sine). The product of this exponential with an even function ({1/(x^{2}+a^{2})}) integrated over a symmetric interval will be non-zero only for the cosine part, and the cosine itself is even in p, so the result of the integral must also be even in p.

To check the normalization of {\Phi(p,0)}, we do the integral:

\displaystyle   \int_{-\infty}^{\infty}\left|\Phi(p,0)\right|^{2}dp \displaystyle  = \displaystyle  2\frac{a}{\hbar}\int_{0}^{\infty}e^{-2pa/\hbar}dp\ \ \ \ \ (9)
\displaystyle  \displaystyle  = \displaystyle  1 \ \ \ \ \ (10)

(d) Since {\Phi(p,0)} is even, {\langle p\rangle=0}. For {\left\langle p^{2}\right\rangle }, we have

\displaystyle   \left\langle p^{2}\right\rangle \displaystyle  = \displaystyle  \int_{-\infty}^{\infty}p^{2}\left|\Phi(p,0)\right|^{2}dp\ \ \ \ \ (11)
\displaystyle  \displaystyle  = \displaystyle  2\frac{a}{\hbar}\int_{0}^{\infty}p^{2}e^{-2pa/\hbar}dp\ \ \ \ \ (12)
\displaystyle  \displaystyle  = \displaystyle  \frac{\hbar^{2}}{2a^{2}} \ \ \ \ \ (13)

Thus {\sigma_{p}=\hbar/\sqrt{2}a} and the uncertainty principle is {\sigma_{p}\sigma_{x}=\hbar/\sqrt{2}}.

Momentum space representation of finite wave function

Required math: calculus

Required physics: Schrödinger equation

References: Griffiths, David J. (2005), Introduction to Quantum Mechanics, 2nd Edition; Pearson Education – Problem 3.29.

Suppose we have a wave function that is purely sinusoidal in position space, but is defined only over a finite interval, so we have

\displaystyle  \Psi(x,0)=\begin{cases} \frac{1}{\sqrt{2n\lambda}}e^{i2\pi x/\lambda} & -n\lambda<x<n\lambda\\ 0 & \mathrm{otherwise} \end{cases} \ \ \ \ \ (1)

where {n} is a positive integer and {\lambda} is a positive constant.

We can calculate the momentum space wave function in the usual way:

\displaystyle   \Phi_{n}(p,0) \displaystyle  = \displaystyle  \frac{1}{\sqrt{2\pi h}}\int_{-n\lambda}^{n\lambda}e^{-ipx/\hbar}\Psi_{n}(x,0)dx\ \ \ \ \ (2)
\displaystyle  \displaystyle  = \displaystyle  \frac{1}{\sqrt{2\pi h}}\int_{-n\lambda}^{n\lambda}e^{-ipx/\hbar}\frac{1}{2n\lambda}e^{i2\pi x/\lambda}dx\ \ \ \ \ (3)
\displaystyle  \displaystyle  = \displaystyle  \sqrt{\frac{\lambda\hbar}{\pi n}}\frac{\sin(np\lambda/\hbar)}{(p\lambda-2\pi\hbar)} \ \ \ \ \ (4)

The plot of {|\Psi_{n}(x,0)|^{2}} is constant with value {1/2n\lambda} in the interval {[-n\lambda,n\lambda]}.

To plot {|\Phi_{1}(p,0)|^{2}} we can define the auxiliary variable {\rho\equiv p\lambda/\hbar} so we have

\displaystyle  |\Phi_{1}(p,0)|^{2}=\frac{\lambda}{\pi\hbar}\left[\frac{\sin^{2}\left(\rho\right)}{\left(\rho-2\pi\right)^{2}}\right] \ \ \ \ \ (5)

We get the following plot:

The peak occurs at {\rho=2\pi}, or {p=2\pi\hbar/\lambda}. The first zero on either side of the peak occurs at {\rho=2\pi\pm\pi=\pi,3\pi}, so the width of the central peak is {w_{p}=2\pi\hbar/\lambda}. The width of the position space curve is simply {w_{x}=2\lambda}. If we take these two width as measures of the uncertainties in position and momentum, we get {w_{x}w_{p}=4\pi\hbar}.

For general {n}, the two widths are {w_{x}=2n\lambda} and {w_{p}=2\pi\hbar/n\lambda}. As {n\rightarrow\infty}, the width in {x}-space gets larger, while that in momentum space gets smaller, eventually reaching the point where there is no uncertainty in momentum but total uncertainty in position. The product {w_{x}w_{p}=4\pi\hbar} is independent of {n} and is well above the uncertainty principle limit of {\hbar/2}.

We can attempt to calculate {\left\langle p\right\rangle } and {\left\langle p^{2}\right\rangle } in the usual way using integration. We get

\displaystyle  \left\langle p\right\rangle =\int_{-\infty}^{\infty}p|\Phi_{1}(p,0)|^{2}dp \ \ \ \ \ (6)

Converting to the variable {\rho} we get

\displaystyle  \left\langle p\right\rangle =\frac{\hbar}{\pi\lambda}\int_{-\infty}^{\infty}\rho\frac{\sin^{2}\left(\rho\right)}{\left(\rho-2\pi\right)^{2}}d\rho \ \ \ \ \ (7)

Maple has a problem with the infinite limits, but if we try it for finite limits which are centred on {\rho=2\pi}, we find

\displaystyle  \frac{\hbar}{\pi\lambda}\int_{2\pi-a}^{2\pi+a}\rho\frac{\sin^{2}\left(\rho\right)}{\left(\rho-2\pi\right)^{2}}d\rho=\frac{4\hbar}{\lambda a}\left(\cos^{2}a+a\mathrm{Si}\left(2a\right)-1\right) \ \ \ \ \ (8)

Here {\mathrm{Si}\left(x\right)} is the sine integral, defined as

\displaystyle  \mathrm{Si}\left(x\right)\equiv\int_{0}^{x}\frac{\sin t}{t}dt \ \ \ \ \ (9)

The important property of the sine integral is

\displaystyle  \lim_{x\rightarrow\infty}\mathrm{Si}\left(x\right)=\frac{\pi}{2} \ \ \ \ \ (10)

so we can now take the limit of the above integral when {a\rightarrow\infty} and get

\displaystyle   \left\langle p\right\rangle \displaystyle  = \displaystyle  \lim_{a\rightarrow\infty}\frac{4\hbar}{\lambda a}\left(\cos^{2}a+a\mathrm{Si}\left(2a\right)-1\right)\ \ \ \ \ (11)
\displaystyle  \displaystyle  = \displaystyle  \frac{2\pi\hbar}{\lambda} \ \ \ \ \ (12)

This result isn’t much of a surprise, since it says the mean momentum is just that at the peak in the graph above.

Attempting to calculate {\langle p^{2}\rangle} requires integrating the function {p^{2}|\Phi_{n}(p,0)|^{2}} over an infinite range. This involves the product of a sine-squared function with the term {p^{2}/(p\lambda-2\pi\hbar)^{2}=1/(\lambda-2\pi\hbar/p)^{2}}. As {p\rightarrow\pm\infty}, this second term tends to a constant ({1/\lambda}), so the integrand tends to the sine-squared term (which is always non-negative), so the integral diverges to infinity. This probably results from the fact that {\Psi(x,0)} has two step functions, and since the momentum operator involves the derivative of position, this will produce a delta function at the two end points. Finding {\left\langle p^{2}\right\rangle } thus involves integrating the square of a delta function and I wouldn’t want to speculate on what that would produce.

Infinite square well: momentum space wave functions

Required math: calculus

Required physics: Schrödinger equation

References: Griffiths, David J. (2005), Introduction to Quantum Mechanics, 2nd Edition; Pearson Education – Problem 3.28.

We’ve calculated the momentum space wave function for the ground state of the harmonic oscillator, and we can use the same technique to investigate the infinite square well. We have:

\displaystyle   \Phi_{n}(p,t) \displaystyle  = \displaystyle  \frac{1}{\sqrt{2\pi h}}\int_{0}^{a}e^{-ipx/\hbar}\Psi_{n}(x,t)dx\ \ \ \ \ (1)
\displaystyle  \displaystyle  = \displaystyle  \frac{1}{\sqrt{2\pi h}}\int_{0}^{a}e^{-ipx/\hbar}\sin\left(\frac{n\pi x}{a}\right)e^{-i(n^{2}\pi^{2}\hbar/2ma^{2})t}dx\ \ \ \ \ (2)
\displaystyle  \displaystyle  = \displaystyle  \frac{n\hbar^{3/2}\sqrt{\pi a}}{p^{2}a^{2}-n^{2}\pi^{2}\hbar^{2}}((-1)^{n}e^{ipa/\hbar}-1)e^{-i(n^{2}\pi^{2}\hbar/2ma^{2})t} \ \ \ \ \ (3)

The square modulus of this function is

\displaystyle  |\Phi_{n}(p,t)|^{2}=2\pi n^{2}\hbar^{3}a\frac{1+(-1)^{n+1}\cos(ap/\hbar)}{(p^{2}a^{2}-n^{2}\pi^{2}\hbar^{2})^{2}} \ \ \ \ \ (4)

We can write this in terms of the auxiliary variable {\rho\equiv pa/\hbar}:

\displaystyle  |\Phi_{n}(p,t)|^{2}=\frac{2\pi n^{2}a}{\hbar}\left[\frac{1+\left(-1\right)^{n+1}\cos\rho}{\left(\rho^{2}-n^{2}\pi^{2}\right)^{2}}\right] \ \ \ \ \ (5)

To investigate the behaviour of this formula around the points {\rho=n\pi}, we can write it as

\displaystyle  |\Phi_{n}(p,t)|^{2}=\frac{2\pi n^{2}a}{\hbar}\left[\frac{1+\left(-1\right)^{n+1}\cos\rho}{\left(\rho+n\pi\right)^{2}\left(\rho-n\pi\right)^{2}}\right] \ \ \ \ \ (6)

We can now expand the numerator in a Taylor series about the point {\rho=n\pi}:

\displaystyle   1+\left(-1\right)^{n+1}\cos\rho \displaystyle  = \displaystyle  1+\left(-1\right)^{2n+1}-\frac{\left(-1\right)^{2n+1}}{2}\left(\rho-n\pi\right)^{2}+\frac{\left(-1\right)^{2n+1}}{24}\left(\rho-n\pi\right)^{4}+\ldots\ \ \ \ \ (7)
\displaystyle  \displaystyle  = \displaystyle  \frac{1}{2}\left(\rho-n\pi\right)^{2}-\frac{1}{24}\left(\rho-n\pi\right)^{4}+\ldots \ \ \ \ \ (8)

where we’ve used the fact that {\left(-1\right)^{2n+1}=-1} for all {n}, since the exponent is always odd.

Dividing by the denominator, we get

\displaystyle  |\Phi_{n}(p,t)|^{2}=\frac{2\pi n^{2}a}{\hbar\left(\rho+n\pi\right)^{2}}\left(\frac{1}{2}-\frac{1}{24}\left(\rho-n\pi\right)^{2}+\ldots\right) \ \ \ \ \ (9)

If we now take the limit, we get

\displaystyle  \lim_{\rho\rightarrow n\pi}|\Phi_{n}(p,t)|^{2}=\frac{a}{4\pi\hbar} \ \ \ \ \ (10)

The limit is independent of {n}.

The plots for {|\Phi_{1}|^{2}} and {|\Phi_{2}|^{2}} are shown below.

Here is {|\Phi_{1}|^{2}} :

Here is {|\Phi_{2}|^{2}}:

By direct calculation using 4 (and Maple), we can get

\displaystyle   \langle p^{2}\rangle \displaystyle  = \displaystyle  \int_{-\infty}^{\infty}p^{2}|\Phi_{n}(p,t)|^{2}dp\ \ \ \ \ (11)
\displaystyle  \displaystyle  = \displaystyle  (n\pi\hbar/a)^{2} \ \ \ \ \ (12)

which is the same as that obtained by doing the calculation in position space. For good measure, we can also calculate {\langle p\rangle=0}, either by direct calculation or by observing that since {|\Phi_{n}(p,t)|^{2}} is even for all {n}, {\langle p\rangle} is always the integral of an odd function over a symmetric interval, so is always zero.

Hamiltonian matrix elements

Required math: calculus

Required physics: Schrödinger equation

References: Griffiths, David J. (2005), Introduction to Quantum Mechanics, 2nd Edition; Pearson Education, sec 3.6.

A wave function can be expressed in its more usual form as a function of {x} and {t}: {\Psi(x,t)}. We can also express the same function as the Fourier transform of its momentum space form:

\displaystyle  \Psi(x,t)=\frac{1}{\sqrt{2\pi\hbar}}\int\Phi(p,t)e^{ipx/\hbar}dp \ \ \ \ \ (1)

Here, the momentum space wave function is the inverse Fourier transform of the position space version:

\displaystyle  \Phi(p,t)=\frac{1}{\sqrt{2\pi\hbar}}\int\Psi(x,t)e^{-ipx/\hbar}dx \ \ \ \ \ (2)

Since either function can be obtained from the other with no loss of information, they are equivalent ways of expressing the wave function. In the language of linear algebra, the vector representing the wave function can be expressed in two different bases (plural of ‘basis’). The position space wave function is given in the first equation by a vector in the momentum basis, where {\Phi(p,t)} is the coordinate of the wave function for that particular value of {p}.

If you like thinking of vectors in 3 dimensions (rather than the infinite number of dimensions we’re using here), this is analogous to saying that {\Phi(p,t)} is the coordinate of {\Psi(x,t)} ‘along the {p} direction’.

Similarly, the second equation expressed the momentum space wave function as a vector in position space, with {\Psi(x,t)} the coordinate ‘along the {x} direction’.

For a Hamiltonian with discrete energies, such as the harmonic oscillator, we know that the wave function can be expressed as a linear combination of the stationary states {\psi_{n}(x)}, as in

\displaystyle  \Psi(x,t)=\sum_{n}c_{n}e^{-iE_{n}t/\hbar}\psi_{n}(x) \ \ \ \ \ (3)

In this case, the basis consists of the stationary state functions multiplied by the exponential, and the ‘coordinate along the {\psi_{n}} direction’ is the coefficient {c_{n}}.

A hermitian operator (which represents an observable) transforms one vector into another. For example, the hamiltonian operator, when operating on one of its eigenvectors, multiplies that vector by a constant, which is the energy:

\displaystyle  H\psi_{n}(x)=E_{n}\psi_{n}(x) \ \ \ \ \ (4)

With respect to a given basis, each operator can be represented as a matrix, with one dimension of the matrix for each dimension of the vector space (which is infinite in the examples so far). The matrix elements can be represented in bra-ket notation as

\displaystyle  H_{mn}=\left\langle e_{m}\left|H\right|e_{n}\right\rangle \ \ \ \ \ (5)

where {e_{n}} is the {n}th basis vector (or function).

If the basis consists of the eigenfunctions of the hamiltonian, then the matrix is diagonal, since the eigenfunctions are orthogonal. Things are trickier if we want to find the matrix elements of the hamiltonian with respect to a continuous basis, like momentum. The (non-normalizable, except in the delta function sense) eigenfunctions of momentum are

\displaystyle  f_{p}(x)=\frac{1}{\sqrt{2\pi\hbar}}e^{ipx/\hbar} \ \ \ \ \ (6)

In the case of the harmonic oscillator, the hamiltonian is

\displaystyle  H=-\frac{\hbar^{2}}{2m}\frac{d^{2}}{dx^{2}}+\frac{1}{2}m\omega^{2}x^{2} \ \ \ \ \ (7)

so if we apply this to the momentum eigenfunctions, we get

\displaystyle  Hf_{p}(x)=\frac{1}{2\sqrt{2\pi\hbar}}e^{ipx/\hbar}\left(\frac{p^{2}}{m}+m\omega^{2}x^{2}\right) \ \ \ \ \ (8)

(By the way, although this might look like {f_{p}} is an eigenfunction of {H} since the RHS has the form {(factor)\times f_{p}}, it’s not, since the ‘factor’ is not a constant.)

To get the matrix elements of {H} we take the inner product with another momentum eigenfunction for momentum:

\displaystyle   H_{mn} \displaystyle  = \displaystyle  \left\langle p_{m}\left|H\right|p_{n}\right\rangle \ \ \ \ \ (9)
\displaystyle  \displaystyle  = \displaystyle  \frac{1}{8\pi\hbar}\int e^{i\left(p_{n}-p_{m}\right)x/\hbar}\left(\frac{p_{n}^{2}}{m}+m\omega^{2}x^{2}\right)dx \ \ \ \ \ (10)

The first term of this integral evaluates to a delta function of form {A\delta(p_{n}-p_{m})} for a constant {A}, but the second term involves a more problematic integral containing the product {x^{2}e^{i\left(p_{n}-p_{m}\right)x/\hbar}}. This integral gives a real result (since {x^{2}} is even and the imaginary part of a complex exponential is odd, being a sine), and if we take the limits of the integral to be symmetric about {x=0}, the integral oscillates about zero with an increasing amplitude, the wider we take the limits. The integral is also clearly infinite if {p_{n}=p_{m}}.

Working out the integral in Maple, we get

\displaystyle  \int_{-a}^{a}x^{2}e^{i\left(p_{n}-p_{m}\right)x/\hbar}dx=\frac{2\hbar}{\left(p_{n}-p_{m}\right)^{3}}\left[\sin\left(\frac{\left(p_{n}-p_{m}\right)a}{\hbar}\right)\left(a^{2}\left(p_{n}-p_{m}\right)^{2}-2\hbar^{2}\right)+\cos\left(\frac{\left(p_{n}-p_{m}\right)a}{\hbar}\right)2a\hbar\left(p_{n}-p_{m}\right)\right] \ \ \ \ \ (11)

This integral is zero whenever

\displaystyle  \tan\left(\frac{\left(p_{n}-p_{m}\right)a}{\hbar}\right)=\frac{2\hbar^{2}-a^{2}\left(p_{n}-p_{m}\right)^{2}}{2a\hbar\left(p_{n}-p_{m}\right)} \ \ \ \ \ (12)

This is another of those transcendental equations (assuming we’re solving for {a} to find out what limits make the integral zero). If we define {q\equiv\left(p_{n}-p_{m}\right)a/\hbar} we can write this as

\displaystyle  \tan q=\frac{2-q^{2}}{2q} \ \ \ \ \ (13)

By plotting the two sides on the same graph, we see there are an infinite number of intersections, so we could make a similar argument as in the delta function case that this integral’s average value as {a\rightarrow\infty} is zero, although I wouldn’t want to bet anything significant on it.

In the plot {\tan q} is in red, and {\frac{2-q^{2}}{2q}} is in blue.

Momentum space: mean position

Required math: calculus

Required physics: Schrödinger equation

References: Griffiths, David J. (2005), Introduction to Quantum Mechanics, 2nd Edition; Pearson Education – Problem 3.12.

The momentum space wave function is the Fourier transform of the regular position space wave function:

\displaystyle  \Phi(p,t)=\frac{1}{\sqrt{2\pi\hbar}}\int_{-\infty}^{\infty}\Psi(x,t)e^{-ipx/\hbar}dx \ \ \ \ \ (1)

In momentum space, the mean value of the momentum is simply

\displaystyle   \left\langle p\right\rangle \displaystyle  = \displaystyle  \left\langle \left.\Phi\right|p\Phi\right\rangle \ \ \ \ \ (2)
\displaystyle  \displaystyle  = \displaystyle  \int_{-\infty}^{\infty}\Phi^*p\Phi dp \ \ \ \ \ (3)

The mean position turns out to be

\displaystyle  \left\langle x\right\rangle =\int\Phi^*\left(-\frac{\hbar}{i}\frac{\partial}{\partial p}\right)\Phi dp \ \ \ \ \ (4)

We can show this as follows. Starting with the definition above, we get

\displaystyle  -\frac{\hbar}{i}\frac{\partial}{\partial p}\Phi=\frac{1}{\sqrt{2\pi\hbar}}\int xe^{-ipx/\hbar}\Psi dx \ \ \ \ \ (5)

Substituting this together with the expression for {\Phi^*} from above into the RHS of 4 we get:

\displaystyle   \int\Phi^*\left(-\frac{\hbar}{i}\frac{\partial}{\partial p}\right)\Phi dp \displaystyle  = \displaystyle  \frac{1}{2\pi\hbar}\int\left(\int e^{ipx'/\hbar}\Psi^*(x')dx'\right)\left(\int xe^{-ipx/\hbar}\Psi(x)dx\right)dp\ \ \ \ \ (6)
\displaystyle  \displaystyle  = \displaystyle  \frac{1}{2\pi\hbar}\int\left(\int\int e^{ip(x'-x)/\hbar}\Psi^*(x')x\Psi(x)dx'dx\right)dp \ \ \ \ \ (7)

We can now do the integral over {p} first, and use the dodgy formula we obtained earlier, which expressed a delta function as an integral over a complex exponential:

\displaystyle  \frac{1}{2\pi\hbar}\int e^{ip(x'-x)/\hbar}dp=\delta(x'-x) \ \ \ \ \ (8)


\displaystyle   \int\Phi^*\left(-\frac{\hbar}{i}\frac{\partial}{\partial p}\right)\Phi dp \displaystyle  = \displaystyle  \int\int\delta(x'-x)\Psi^*(x')x\Psi(x)dx'dx\ \ \ \ \ (9)
\displaystyle  \displaystyle  = \displaystyle  \int\Psi^*(x)x\Psi(x)dx\ \ \ \ \ (10)
\displaystyle  \displaystyle  = \displaystyle  \langle x\rangle \ \ \ \ \ (11)


Momentum space: harmonic oscillator

Required math: calculus

Required physics: Schrödinger equation

References: Griffiths, David J. (2005), Introduction to Quantum Mechanics, 2nd Edition; Pearson Education – Problem 3.11.

We’ve seen that any function {g(x)} can be expressed in terms of the eigenfunctions {f_{p}(x)} of the momentum operator, in the sense that, for a function {g(x)} we can write

\displaystyle   g(x) \displaystyle  = \displaystyle  \int_{-\infty}^{\infty}c\left(p\right)f_{p}\left(x\right)dp\ \ \ \ \ (1)
\displaystyle  \displaystyle  = \displaystyle  \frac{1}{\sqrt{2\pi\hbar}}\int_{-\infty}^{\infty}c\left(p\right)e^{ipx/\hbar}dp \ \ \ \ \ (2)

where {c(p)} is the inverse Fourier transform:

\displaystyle   c(p) \displaystyle  = \displaystyle  \frac{1}{\sqrt{2\pi\hbar}}\int_{-\infty}^{\infty}g\left(x\right)e^{-ipx/\hbar}dx\ \ \ \ \ (3)
\displaystyle  \displaystyle  = \displaystyle  \left\langle \left.f_{p}\right|g\right\rangle \ \ \ \ \ (4)

If {g} is a wave function {\Psi(x,t)}, we can view {c(p,t)} as the momentum-space wave function, often given the symbol {\Phi(p,t)}. That is

\displaystyle  \Phi(p,t)=\frac{1}{\sqrt{2\pi\hbar}}\int_{-\infty}^{\infty}\Psi(x,t)e^{-ipx/\hbar}dx \ \ \ \ \ (5)

An example is the ground state wave function for the harmonic oscillator:

\displaystyle  \psi_{0}(x,t)=\left(\frac{m\omega}{\pi\hbar}\right)^{1/4}e^{-m\omega x^{2}/2\hbar}e^{-i\omega t/2} \ \ \ \ \ (6)

Using a couple of shorthand parameters for simplicity: {\alpha\equiv(m\omega/\pi\hbar)^{1/4}}; {\beta\equiv m\omega/2\hbar}, we can get the momentum space function:

\displaystyle   \Phi_{0}(p,t) \displaystyle  = \displaystyle  \frac{\alpha e^{-i\omega t/2}}{\sqrt{2\pi\hbar}}\int_{-\infty}^{\infty}e^{-ipx/\hbar}e^{-\beta x^{2}}dx\ \ \ \ \ (7)
\displaystyle  \displaystyle  = \displaystyle  \frac{e^{-i\omega t/2}}{(\pi m\omega\hbar)^{1/4}}e^{-p^{2}/2\hbar m\omega} \ \ \ \ \ (8)

Here we have used Maple to do the integral, and simplified the result by expanding {\alpha} and {\beta}. Note that although the integrand contains a complex exponential, the result is real. This is because the imaginary part of the integrand is the product of an odd function ({\sin(px/\hbar)}) and an even function ({e^{-\beta x^{2}}}) so it will integrate to zero over any interval symmetric about the origin.

To find the probability that the momentum is outside the classical range, we note that the ground state energy is {E_{0}=\hbar\omega/2}, so the maximum ground state momentum is {p_{0}=\sqrt{2mE_{0}}=\sqrt{m\hbar\omega}.} The classical momentum thus ranges from {-\sqrt{m\hbar\omega}} to {\sqrt{m\hbar\omega}}. The probability that the quantum momentum is outside this range is therefore

\displaystyle   Prob \displaystyle  = \displaystyle  2\int_{\sqrt{m\hbar\omega}}^{\infty}|\Phi_{0}(x,t)|^{2}dp\ \ \ \ \ (9)
\displaystyle  \displaystyle  = \displaystyle  \frac{2}{\sqrt{\pi m\hbar\omega}}\int_{\sqrt{m\hbar\omega}}^{\infty}e^{-p^{2}/m\hbar\omega}dp\ \ \ \ \ (10)
\displaystyle  \displaystyle  = \displaystyle  1-\text{erf }(1)\ \ \ \ \ (11)
\displaystyle  \displaystyle  = \displaystyle  0.1573 \ \ \ \ \ (12)

The integral and evalution were done using Maple. Here, erf is the error function, defined as:

\displaystyle  \text{erf}(x)\equiv\frac{2}{\sqrt{\pi}}\int_{0}^{x}e^{-t^{2}}dt \ \ \ \ \ (13)

You might wonder why the momentum doesn’t have a fixed value, since in the ground state, the energy is fixed. The energy, however, is composed of kinetic and potential parts, and the momentum is derived from the kinetic part, which varies as the position of the particle varies. Still, it’s quite odd that the quantum momentum has a non-zero probability of being greater than any finite value, no matter how large.