19 Momentum-space Wave function

19.1 Motivation

In the Hamiltonian formulation of classical mechanics, position and momentum appear in a symmetrical way as coordinates \((x,p)\) on phase space. Moreover, there is a canonical transformation that exchanges them!

Recall that a canonical transformation is a change of coordinates \((x,p) \to (x',p')\) that leaves Hamilton’s equations invariant. Under the transformation \[(x,p) \to (p,-x) \, ,\] we find \[\begin{aligned} \dot x = + \frac{\partial H}{\partial p} \quad & \to \quad \dot p = -\frac{\partial H}{\partial x} \\ \dot p = - \frac{\partial H}{\partial x} \quad & \to \quad \dot x = + \frac{\partial H}{\partial p} \, . \end{aligned}\] So Hamilton’s equations are indeed unchanged but the role of position and momentum has been reversed.

19.2 The Fourier Transform

In contrast, position and momentum appear quite asymmetrically in our description of quantum mechanics so far. We have introduced a ‘position-space’ wave function \(\psi(x)\) on which position and momentum act as operators \[\begin{aligned} \hat x & = x\,, \\[1ex] \hat p & = - i \hbar \frac{\partial}{\partial x} \, . \end{aligned}\] The canonical transformation \((x,p) \to (p,-x)\) from classical mechanics suggests there should another formulation of quantum mechanics with a ‘momentum-space’ wave function \(\widetilde{\psi}(p)\) on which position and momentum act as operators \[\begin{aligned} \hat x & = i \hbar \frac{\partial}{\partial p}\,, \\[1ex] \hat p & = p \, . \end{aligned}\] We claim that the position and momentum wave functions are in fact related by the following pair of integrals \[\begin{aligned} \psi(x) & = \frac{1}{\sqrt{2\pi \hbar}}\int^\infty_{-\infty} {\rm d}p \, \widetilde{\psi}(p) \, e^{ipx/\hbar} \,, \\[1ex] \widetilde{\psi}(p) & = \frac{1}{\sqrt{2\pi \hbar}}\int^\infty_{-\infty} {\rm d}x \, \psi(x) \, e^{-ipx/\hbar} \, . \end{aligned}\] This is an example of a ‘Fourier transform’.

Let us see that these are consistent with the expected form of the position and momentum operators above.

• First acting with \(i \hbar \displaystyle\frac{\partial}{\partial p}\) on the momentum-space wave function, \[\begin{aligned} i\hbar \frac{\partial}{\partial p} \, \widetilde{\psi}(p) & = \frac{1}{\sqrt{2\pi \hbar}}\int^\infty_{-\infty} {\rm d}x \, \psi(x) \left( i \hbar \frac{\partial}{\partial p} e^{-ipx/\hbar} \right) \\[1ex] & = \frac{1}{\sqrt{2\pi \hbar}}\int^\infty_{-\infty} {\rm d}x \, \psi(x) \, \left( x e^{-ipx/\hbar} \right) \\[1ex] & = \frac{1}{\sqrt{2\pi \hbar}}\int^\infty_{-\infty} {\rm d}x \, \left( x \psi(x) \right) \, e^{-ipx/\hbar} \, , \end{aligned}\] which corresponds to multiplying the position-space wave function by \(x\).

• Second multiplying the momentum-space wave function by \(p\), \[\begin{aligned} p \, \widetilde{\psi}(p) & = \frac{1}{\sqrt{2\pi \hbar}}\int^\infty_{-\infty} {\rm d}x \, \psi(x) \left( p \, e^{-ipx/\hbar} \right) \\[1ex] & = \frac{1}{\sqrt{2\pi \hbar}}\int^\infty_{-\infty} {\rm d}x \, \psi(x) \, \left( i \hbar \frac{\partial}{\partial x} e^{-ipx/\hbar} \right) \\[1ex] & = \frac{1}{\sqrt{2\pi \hbar}}\int^\infty_{-\infty} {\rm d}x \, \left( - i \hbar \frac{\partial}{\partial x} \psi(x) \right) \, e^{-ipx/\hbar} \, , \end{aligned}\] which corresponds to acting with \(\displaystyle- i \hbar \frac{\partial}{\partial x}\) on the position-space wave function.

The action of position and momentum operators on position and momentum wave functions is summarised in the following table.

\[\begin{aligned} &\quad & \psi(x) &\qquad& \widetilde{\psi}(p) \\[1ex] \hat{x} &\quad & x &\qquad& +i\hbar\frac{\partial}{\partial p} \\[1ex] \hat{p} &\quad& -i\hbar\frac{\partial}{\partial x} &\qquad& p \end{aligned}\]

19.3 The Momentum-space Wave function

The function \(\widetilde{\psi}(x)\) is known as the momentum-space wave function. Everything we have learnt about the position-space wave function has analogues for the momentum-space wave function.

• \(\widetilde P(p) : = |\widetilde{\psi}(p)|^2\) is the momentum probability density. In particular, the probability that a momentum measurement will find \(a < p < b\) is \[\int^{b}_{a} {\rm d}p \, \widetilde P(p) = \int^{b}_{a} {\rm d}p \, |\widetilde{\psi}(p)|^2 \, .\] This is an improvement on previous lectures where we could only compute expectation values of momentum.

• We can compute momentum expectation values using \[\langle f(p) \rangle = \int^\infty_{-\infty} {\rm d}p \, f(p) \, |\widetilde{\psi}(p)|^2\] for any polynomial function \(f(p)\).

• We can compute position expectation values using \[\langle f(x ) \rangle = \int^\infty_{-\infty} {\rm d}p \, \overline{\widetilde{\psi}(p)} \, f\left( i\hbar \frac{\partial}{\partial p} \right) \, \widetilde{\psi}(p)\] for any polynomial function \(f(x)\).

19.4 Examples

Example 1: bound state of the delta-function potential

In general, the momentum-space wave function is different from the position-space wave function (the Fourier transform is non-trivial).

Consider the wave function \[\psi(x) = C e^{-\lambda |x| / \hbar}\] where \(\lambda >0\) is a constant; we have seen this wave function in the discussion of the delta function potential. To find the normalisation \(C\), we require the probability to find the particle anywhere is \(1\), \[\begin{aligned} 1 & = |C|^2 \int^\infty_{-\infty} e^{-2 \lambda |x| / \hbar} \, {\rm d}x \\ & = 2 |C|^2 \int^\infty_0 e^{-2\lambda x / \hbar} \, {\rm d}x \\ & = |C|^2 \frac{\hbar}{\lambda} \, , \end{aligned}\] and therefore \(C = \sqrt{\lambda / \hbar}\) up to a constant phase.

Let us now compute the momentum-space wave function, \[\begin{aligned} \widetilde{\psi}(p) & = \frac{1}{\sqrt{2\pi \hbar}} \int^\infty_{-\infty} {\rm d}p \, e^{-i px / \hbar} \psi(x) \\ & = \sqrt{\frac{\lambda}{2 \pi \hbar^2}} \int^\infty_{-\infty} {\rm d}p \, e^{-i px / \hbar} e^{-\lambda|x| / \hbar} \\ & = \sqrt{\frac{\lambda}{2 \pi \hbar^2}}\left( \int^\infty_0 e^{(-i p-\lambda)x / \hbar} + \int^0_{-\infty} e^{(-i p+\lambda)x / \hbar} \right) \\ & = \sqrt{\frac{\lambda}{2 \pi}}\left( \frac{1}{ip+\lambda} +\frac{1}{-ip+\lambda} \right) \\ & = \sqrt{\frac{2}{\pi}} \frac{\lambda^{3/2}}{p^2+\lambda^2} \, . \end{aligned}\] You may wish to verify that \(\widetilde{\psi}(p)\) is correctly normalised!

Example 2: Gaussian wave function

The momentum wave function corresponding to a Gaussian in position-space is again a Gaussian, but now in momentum-space.

Consider the normalised Gaussian wave function \[\psi(x) = Ce^{-x^2/4\Delta^2} \, .\] where \(C = 1/(2 \pi \Delta^2)^{1/4}\).

The momentum-space wave function is computed by completing the square in the exponential, \[\begin{aligned} \widetilde\psi(p) & = \frac{C}{\sqrt{2\pi \hbar}} \int^\infty_{-\infty} {\rm d}x \, e^{-x^2/4\Delta^2} e^{-ipx/\hbar} \\ & = \frac{C}{\sqrt{2\pi \hbar}} e^{-p^2 \Delta^2 / \hbar^2} \int^\infty_{-\infty} {\rm d}x \, e^{-(x+2ip\Delta/\hbar)^2/4\Delta^2} \\ \label{e:ycontourint} & =\frac{C}{\sqrt{2\pi \hbar}} e^{-p^2 \Delta^2 / \hbar^2} \int_\gamma \, {\rm d}y \, e^{-y^2/4\Delta^2} \end{aligned}\] where \[y = x + 2ip\Delta^2/\hbar \, .\]

In performing the substitution, we are now integrating over a ‘contour’ \(\gamma\) in the complex \(y\)-plane that is shifted by an amount \(2ip\Delta^2/\hbar\) in the imaginary direction. However, as there are no poles in the intermediate region we can deform the contour back to the real axis. We then have a standard Gaussian integral, \[\begin{aligned} \widetilde\psi(p) & = \frac{C}{\sqrt{2\pi \hbar}} e^{-p^2 \Delta^2 / \hbar^2} \int^\infty_{-\infty} {\rm d}y e^{-y^2/4\Delta^2} \\ & =\frac{C\sqrt{4\pi \Delta^2}}{\sqrt{2\pi \hbar}} e^{-p^2 \Delta^2 / \hbar^2} \, . \end{aligned}\] Now defining \[\widetilde \Delta := \hbar / 2\Delta \, ,\] and substituting in the normalisation factor \(C\), this becomes a normalised Gaussian wave function in momentum-space \[\widetilde{\psi}(p) = \frac{1}{(2\pi \widetilde\Delta^2)^{1/4}}e^{-p^2/4\widetilde\Delta^2} \, .\] We can therefore immediately determine that \(\langle p \rangle = 0\) and \(\Delta p = \widetilde \Delta\) in complete agreement with what we computed earlier explicitly in position space using \(\Delta p = \sqrt{\langle p^2\rangle - (\langle p\rangle)^2}\).

• This is a very important result: the Fourier transformation of a Gaussian wave function is a Gaussian wave function with uncertainties related by \[\Delta x \, \Delta p = \frac{\hbar}{2} \, .\]

19.5 Properties

We conclude by listing a couple of important properties of the Fourier transforms relating position and momentum wave functions.

First, you may have noticed from the Gaussian example that the momentum wave function was automatically normalised. This is generally true: \(\psi(x)\) is normalised if and only if \(\widetilde{\psi}(p)\) is normalised.

Second, the Fourier transformations interchange ‘translations’ and ‘phases’. To be concrete, it is straightforward to check from the definitions that if \[\psi(x) \quad \longleftrightarrow \quad \widetilde\psi(p)\] are related by Fourier transform then so are \[\begin{aligned} & \psi(x-x_0) \quad && \longleftrightarrow \quad \widetilde\psi(p)e^{-ipx_0/\hbar} \\ & \psi(x)e^{ip_0x/\hbar} \quad && \longleftrightarrow \quad \widetilde\psi(p-p_0) \end{aligned}\] In words:

• If I translate the position wave function \(\psi(x)\) by an amount \(x_0\), this is equivalent to multiplying the momentum wave function \(\widetilde{\psi}(p)\) by the phase \(e^{-ip x_0/\hbar}\).

• If I translate the momentum wave function \(\widetilde{\psi}(p)\) by an amount \(p_0\), this is equivalent to multiplying the position wave function \(\psi(x)\) by the phase \(e^{ip_0 x/\hbar}\).

This has some important consequences. For example, suppose \(\psi(x) = \phi(x) e^{ip_0x/\hbar}\). Then we should expect the momentum expectation values obey \(\langle p \rangle_\psi = \langle \phi\rangle + p_0\). To see this explicitly using the momentum wave function \[\begin{aligned} \langle p\rangle_\psi & = \int {\rm d}p \, p \, | \widetilde\phi(p-p_0)|^2 \\ & = \int {\rm d}p' \, (p'+p_0) \, | \widetilde\phi(p')|^2 \\ & = \langle p \rangle_{\phi} + p_0 \, , \end{aligned}\] assuming \(\widetilde{\phi}(p)\) is normalised.

19.6 Problems

1. Gaussian in momentum space:

   Consider the normalized Gaussian wave function \[\psi(x) = \frac{1}{(2\pi\Delta^2)^{1/4}} e^{-x^2/4\Delta^2} e^{i p_0 x/\hbar}\]

   1. Compute the momentum expectation values \(\langle p\rangle\), \(\langle p^2\rangle\) and uncertainty \(\Delta p\) using the momentum operator \(\hat p = - i \hbar \partial_x\).

   2. Show that the momentum space wave function has the form \[\widetilde\psi(p) = \frac{1}{(2\pi\widetilde\Delta^2)^{1/4}} e^{-(p-p_0)^2/4\widetilde\Delta^2}\] up to a constant phase factor and determine \(\widetilde \Delta\).

   3. Repeat part (a) using the momentum probability density.

2. Momentum-space wavefunction for a particle in a box:

   A particle confined to the region \(-a < x < a\) has wave function \[\psi(x) = \begin{cases} \displaystyle C \sin\left(\frac{\pi x}{a}\right) & \text{if} \; -a < x < a\\ 0 & \text{otherwise} \end{cases} \, .\]

   1. Find the normalisation \(C\).

   2. Using the momentum operator \(\hat p = - \hbar \partial_x\) show that \(\langle p\rangle = 0\).

   3. Show that the momentum space wave function is \[\widetilde{\psi}(p) = i \sqrt{\frac{2\pi \hbar^3}{a^3} } \, \frac{\sin(pa/\hbar)}{p^2-(\hbar\pi / a)^2} \, .\]

   4. Sketch the momentum probability density \(|\widetilde{\psi}(p)|^2\) and hence explain why

      1. \(\langle p\rangle = 0\), compatible with part (b).

      2. The mostly likely outcomes of a momentum measurement are

      \[p = \pm \frac{\pi \hbar}{a} \, .\]

   Hints:

   1. Integrate by parts twice or convert the sine to complex exponentials.

   2. If you are having difficulty with the sketch, try Wolfram Alpha!

   Solution ▶

   1. To determine the normalisation, \[\begin{aligned} \nonumber 1 & = |C|^2 \int^a_{-a} \sin^2\left( \frac{\pi x}{a}\right) \, dx \\ & = \frac{|C|^2}{2} \int^a_{-a} \left( 1- \cos\left( \frac{2\pi x}{a}\right) \right) \, dx \\ & = |C|^2 a\, , \end{aligned}\] so we can choose \(C = 1 / \sqrt{a}\).

   2. The momentum expectation value is \[\begin{aligned} \nonumber \langle p \rangle & = - i \hbar \int^\infty_{-\infty} \overline{\psi(x)} \partial_x \psi(x) \\ & = - i \hbar \frac{\pi}{a^2} \int^a_{-a} \sin\left( \frac{\pi x}{a}\right) \cos\left( \frac{\pi x}{a}\right) \\ & = 0 \end{aligned}\] because the integrand is odd. Alternatively, the wave function is normalised and real, which implies that \(\langle p\rangle = 0\) by problem 1.3.

   3. The momentum space wave function is \[\begin{aligned} \nonumber \widetilde{\psi}(p) & = \frac{1}{\sqrt{2\pi \hbar}} \int^\infty_{-\infty} e^{-ipx/\hbar} \psi(x) dx \\ & = \frac{1}{\sqrt{2\pi \hbar a}} \int^a_{-a} e^{-ipx/\hbar} \sin\left(\frac{\pi x}{a}\right) \\ & = \frac{1}{\sqrt{2\pi \hbar a}} \, \frac{1}{2i} \int^a_{-a} \left( e^{-i(p/\hbar-\pi/a)x} - e^{-i(p/\hbar+\pi/a)x}\right) \\ & = \frac{1}{\sqrt{2\pi \hbar a}} \, \frac{1}{2} \left( \frac{e^{ipa/\hbar}-e^{-ipa/\hbar}}{p/\hbar-\pi/a} - \frac{e^{ipa/\hbar}-e^{-ipa/\hbar}}{p/\hbar+\pi/a}\right) \\ & = i \sqrt{\frac{2\pi \hbar^3}{a^3} } \frac{\sin(pa/\hbar)}{p^2-(\hbar\pi / a)^2} \, . \end{aligned}\] You may also integrate by parts twice to find the same result!

   4. The probability density is \[\widetilde{P}(p) = \frac{2\pi \hbar^3}{a^3} \left( \frac{\sin(pa/\hbar)}{p^2-(\hbar\pi /a)^2 }\right)^2 \, .\]

      

      1. The momentum probability density is an even function so \(\langle p \rangle = 0\).

      2. The momentum probability density has global maxima at \(p = \pm \hbar\pi / a\).