QM II: Momentum and Planck's constant

4.1.Momentum in Classical Mechanics

For motivation, we first recall the interpretation of momentum in classical mechanics as the generator of translations in space.

Recall from last term's lectures that functions of position and momentum generate infinitesimal canonical transformations in Hamiltonian mechanics. The infinitesimal canonical transformation with parameter $\epsilon$ generated by a function $A(x,p)$ is

\begin{align} x & \to x' = x+\epsilon \{ x, A\} \\ p & \to p' = p +\epsilon\{ p,A\} \, . \end{align}

where

\begin{equation} \{ f , g \} := \frac{\partial f}{\partial x}\frac{\partial g}{\partial p} - \frac{\partial f}{\partial p}\frac{\partial g}{\partial x} \end{equation}

is the Poisson bracket. In particular, the canonical transformation generated by the function

\begin{equation} A(x,p) = p \end{equation}

is

\begin{align} x & \to x' = x+\epsilon \{ x,p\} = x + \epsilon \\ p & \to p' = p +\epsilon\{ p,p\} = p \, . \end{align}

Here we have used that

\begin{equation} \{ x,p\} = 1 \, . \end{equation}

The important lesson is that momentum $p$ generates an infinitesimal translation of $x$.

4.2.Momentum in Quantum Mechanics

We now use this idea of momentum to understand momentum in quantum mechanics. Let us consider translating a wave function $\psi(x)$ by a infinitesimal amount $\epsilon$ in the positive $x$ direction - as shown below. As in the previous lecture, we work at some fixed moment in time.

Figure 4.1: Infinitesimal translation of a wave function to the right, $\psi_{\text{trans}}(x) = \psi(x-\epsilon)$ (and note the sign!).

The translated wave function is of course $\psi(x-\epsilon)$. To first order in $\epsilon$, the change in the wave function is found by Taylor expanding,

\begin{align} \delta_\epsilon\psi(x) & = \psi(x-\epsilon) - \psi(x) \\ & = - \epsilon \frac{\partial}{\partial x} \psi(x) + {\cal O}(\epsilon^2) \, . \end{align}

If we want momentum to generate translations, this suggests we should identify momentum $p$ with the derivative with respect to $x$.

Let us therefore define a momentum operator

\begin{equation} \hat p = - i \hbar \frac{\partial}{\partial x} \, , \label{eq:mom-op} \end{equation}

such that

\begin{equation} \delta_\epsilon \psi(x) = - \epsilon\frac{i}{\hbar} \hat p \, \psi(x) \, , \end{equation}

where $\hbar$ is a constant of proportionality. The additional factors of $i$ are introduced for convenience; we will get back to them shortly. Some comments are in order

Note that while the momentum $p$ has units of $MLT^{-1}$, $\partial / \partial x$ has units of $L^{-1}$. This means $\hbar$ must have units $ML^2T^{-1}$, or ‘energy$\cdot$time’.
Since the wave function is complex, we could imagine $\hbar$ is a complex number. In a moment, we will show that $\hbar$ must be real for momentum measurements to yield real results. This is why we introduced the extra factor of $-i$ in the definition.

Planck's constant has to be present in quantum mechanics for dimensional reasons. It has dimension of ‘energy$\cdot$time’ or ‘action’.

The constant $\hbar$ is known as the (reduced) Planck constant, and pronounced ‘h-bar’.

Its value cannot be determined by mathematical arguments. It must be determined by comparing to experimental data, for example atomic spectra. In our universe,

\begin{equation} \hbar \approx 1.05 \times 10^{-34} \, \mathrm{kg} \, \mathrm{m}^2 \, \mathrm{s}^{-1} \, . \end{equation}

The smallness of this number, in units that are natural to humans, is why we do not observe quantum mechanical effects in everyday life. You will in the literature also find $h = 2\pi \hbar$, which is usually called Planck's constant (without the ‘reduced’ prefix).

4.3.A Quick Commutator

In order have position and momentum on an equal footing, we can introduce a position operator $\hat x$ that simply multiplies a wave function by $x$. In summary,

\begin{align} \hat x & = x \\ \hat p & = - i \hbar \frac{\partial}{\partial x} \, . \end{align}

The ‘commutator’ of these operators is defined by

\begin{equation} \left[ \hat x , \hat p \right] : = \hat x \, \hat p - \hat p \, \hat x \, . \end{equation}

Here the operators should always be understood to act on everything to the right. Acting with this equation on a wave function $\psi(x)$, we have

\begin{align} \left[ \hat x , \hat p \right] \, \psi(x) & = \hat x (\hat p \, \psi(x) ) - \hat p (\hat x \, \psi(x) ) \\ & = x \left(-i \hbar\frac{\partial}{\partial x}\psi(x) \right) + i \hbar \frac{\partial}{\partial x} \left( x \psi(x) \right) \\ & = i \hbar \, \psi(x) \, . \end{align}

where the final line follows from the product rule. Since this holds for any wave function $\psi(x)$, we can summarise this result by

\begin{equation} \left[ \hat x , \hat p \right] = i \hbar \, . \end{equation}

This is known as the canonical commutation relation.

The commutator is reminiscent of the Poisson bracket formula $\{ x , p\} = 1$ from classical mechanics. In fact, the commutator in quantum mechanics is found by replacing

\begin{equation} \{ \, , \, \} \to -\frac{i}{\hbar}[ \, , \, ] \, . \end{equation}

This replacement rule is known as ‘canonical quantisation’. We study it further in later lectures after introducing some more mathematical machinery.

4.4.Momentum Expectation Values

Just like position, in quantum mechanics we can only compute the probabilities of the outcomes of momentum measurements. For now, we satisfy ourselves with computing expectation values of functions of momentum.

First, recall from the last lecture that the expectation value $\langle x\rangle$ can be written

\begin{align} \langle x\rangle & = \int^\infty_{-\infty} x \, | \psi(x,t) |^2 \, dx \\ & = \int^\infty_{-\infty} \overline{\psi(x,t)} \, \hat x \, \psi(x,t) \, dx \, . \end{align}

We now propose, similarly, that the expectation value of momentum is

\begin{align} \langle p \rangle & = \int^\infty_{-\infty} \overline{\psi(x,t)} \, \hat p \, \psi(x,t) \, dx \\ & = - i \hbar \int^\infty_{-\infty} \overline{\psi(x,t)} \, \frac{\partial}{\partial x} \, \psi(x,t) \, dx \, . \end{align}

As for the position expectation value, we emphasise that $\langle p\rangle$ is interpreted as the average of momentum measurements on an ensemble of particles with the same wave function $\psi(x,t)$.

Let us now return to explain why $\hbar$ must be real. Since the outcomes of momentum measurements are real numbers, we require $\langle p\rangle \in \mathbb{R}$. Let us imagine for a second that $\hbar$ is complex and compute the complex conjugate of $\langle p\rangle$,

\begin{align} \overline{\langle p \rangle} & = i \bar\hbar \int^\infty_{-\infty} dx \, \psi(x,t) \frac{\partial}{\partial x}\overline{\psi(x,t)} \\ & = -i \bar\hbar \int^\infty_{-\infty} dx \, \frac{\partial}{\partial x}\psi(x,t) \, \overline{\psi(x,t)} + i \hbar \left[ |\psi(x,t)|^2 \right]^\infty_{-\infty} \\ & = - i \bar\hbar \int^\infty_{-\infty} dx \, \overline{\psi(x,t)} \frac{\partial}{\partial x}\psi(x,t) \\ & = \frac{\bar\hbar}{\hbar} \langle p\rangle \, . \end{align}

In the passing second line, we have integrated by parts. In passing to the third line, we discarded the boundary term because $|\psi(x,t)|^2$ must vanish as $x \to \pm \infty$ if the wave function is square normalisable. Therefore, the momentum expectation value is real if and only if $\hbar$ is real.

In a similar way, we can compute more general expectation values

\begin{align} \langle f(p) \rangle & = \int^\infty_{-\infty} dx\, \overline{\psi(x,t)} \, f(\hat p) \, \psi(x,t) \\ & = \int^\infty_{-\infty}dx \, \overline{\psi(x,t)} \, f\left(-i\hbar \frac{\partial}{\partial x} \right) \, \psi(x,t) \, . \end{align}

Of particular importance is the momentum uncertainty

\begin{equation} \Delta p = \sqrt{\langle p^2\rangle - \langle p \rangle^2} \, , \end{equation}

which gives a measure of the spread of momentum measurements around $\langle p \rangle$ made on an ensemble of particles with identical wave function $\psi(x,t)$.

4.4.1.Example: Gaussian Wave function

Let us again consider the wave function

\begin{equation} \psi(x) = C e^{-x^2/4\Delta^2} \, , \end{equation}

with normalisation $C = (2\pi\Delta^2)^{-1/4}$. In the last lecture, we showed that the position expectation values are given by $\langle x \rangle = 0$ and $\langle x^2\rangle = \Delta^2$, and therefore the uncertainty in position is $\Delta x = \Delta$.

Figure 4.2: Gaussian wave function with width $\Delta$.

The action of the momentum operator on this wave function is

\begin{align} \hat p \, \psi(x) & = \frac{i\hbar}{2\Delta^2} \, x \, \psi(x) \\ \hat p^2 \, \psi(x) & = \frac{\hbar^2}{2\Delta^2} \, \psi(x) - \frac{\hbar^2}{4\Delta^4} \, x^2 \, \psi(x) \, . \end{align}

Note that the result is always a polynomial in $x$ times the original wave function. This means we can recycle our results for position expectation values to compute momentum expectation values. For example,

\begin{equation} \langle p \rangle = \frac{i \hbar }{2\Delta^2} \int^\infty_{-\infty} x | \psi(x)|^2 = \frac{i \hbar }{2\Delta^2} \langle x \rangle = 0 \end{equation}

and similarly

\begin{align} \langle p^2 \rangle & = \frac{\hbar^2}{2\Delta^2} \int^\infty_{-\infty} |\psi(x)|^2 -\frac{\hbar^2}{4\Delta^4} \int^\infty_{-\infty} x^2 |\psi(x)|^2 \\ & = \frac{\hbar^2}{2\Delta^2} -\frac{\hbar^2}{4\Delta^4} \langle x^2 \rangle \\ & = \frac{\hbar^2}{4\Delta^2} \, . \end{align}

The momentum uncertainty is therefore $\Delta p = \frac{\hbar}{2\Delta}$.

Note that the product of position and momentum uncertainties is independent of $\Delta$,

\begin{equation} \Delta x \, \Delta p = \frac{\hbar}{2} \, . \end{equation}

This means that if we attempt to localise the particle in space by making $\Delta x$ smaller, the the uncertainly in momentum $\Delta p$ necessarily increases, and vice verse.

4.5.Heisenberg's Uncertainty Principle

This example illustrates an important result known as Heisenberg's uncertainty principle. This states that for any normalised wave function,

\begin{equation} \Delta x \Delta p \geq \frac{\hbar}{2} \, . \end{equation}

We will prove this result later in the course. It shows that there is a fundamental limit in quantum mechanics on the degree we can simultaneously reduce the uncertainty in position and momentum. The Gaussian wave function saturates this limit: it is a ‘minimal uncertainty’ wave function.

Remember that $\hbar$ is an extremely small number in human units. So while we cannot arrange for both $\Delta x$ and $\Delta p$ to vanish, both uncertainties can be simultaneously small in human units. This goes some way to explaining why in everyday life, objects appear to have a definite position and momentum.

4.6.Problems

Momentum expectation value
The expectation value of momentum measurements on a particle with wave function $\psi(x)$ is
\[ \langle p \rangle = - i \hbar \int_{-\infty}^\infty dx \, \overline{\psi(x)} \, \partial_x \psi(x) \, . \]
You may assume the wave function is normalised, $\displaystyle\int^\infty_{-\infty} dx | \psi(x)|^2 =1$.
1. Show that $\langle p\rangle$ is real.
2. Show that if $\psi(x)$ is real then $\langle p \rangle = 0$.
3. Suppose a wave function $\phi(x)$ has momentum expectation value $\langle p \rangle_\phi = P$.
  Compute the expectation value $\langle p \rangle_\psi$ for
  \[ \psi(x) = e^{i x p_0/\hbar} \phi(x) \, . \]
Solution:
▶
1. We compute the conjugate
  \begin{align}\nonumber \overline{ \langle p \rangle} & = + i \hbar \int_{-\infty}^\infty dx \, \psi(x) \partial_x \overline{\psi(x)} \\ & = - i \hbar \int_{-\infty}^\infty dx \, \partial_x \psi(x) \overline{\psi(x)} +i\hbar \left[ |\psi(x)|^2 \right]^\infty_{-\infty} \\ & = - i \hbar \int_{-\infty}^\infty dx \, \overline{\psi(x)} \partial_x \psi(x) \\ & = \langle p\rangle \, . \end{align}
  In passing to the second line we have integrated by parts. The boundary term vanishes provided the probability density $|\psi(x)|^2$ vanishes at $x\to\pm\infty$. This is a necessary condition for the wave function to be normalizable, $\int^\infty_{-\infty} |\psi(x)|^2 \lt{} \infty$.
2. Assuming the wave function is real $\overline{\psi(x)} = \psi(x)$, then
  \begin{align} \langle p \rangle & = - i \hbar \int^\infty_{-\infty}dx \, \psi(x) \partial_x \psi(x) \\ \nonumber & = -\frac{i\hbar}{2}\int^\infty_{-\infty} dx \, \partial_x(\psi(x)^2) \\ & = - \frac{i\hbar}{2} \left[ \psi(x)^2 \right]^\infty_{-\infty} \, , \end{align}
  which vanishes for the same reason as the boundary term in part (a).
3. By direct computation
  \begin{align} \langle p\rangle_{\psi} & = - i \hbar \int^\infty_{-\infty} dx\, \overline{\psi(x)} \partial_x \psi(x) \\ \nonumber & = - i \hbar \int^\infty_{-\infty} dx \, e^{-ixp_0/\hbar} \overline{\phi(x)} \, \partial_x \left( e^{ixp_0/\hbar}\phi(x) \right)\\ & = - i \hbar \int^\infty_{-\infty} dx \, \overline{\phi(x)} \left( \partial_x \phi(x) + \frac{ip_0}{\hbar} \phi(x) \right) \\ & = - i \hbar \int^\infty_{-\infty} dx \, \overline{\phi(x)} \partial_x \phi(x) + p_0 \int^\infty_{-\infty} dx |\phi(x)|^2 \\ & = P+p_0 \end{align}
  where in the last step we used that the wave function is normalized.
Gaussian wave function
Reconsider the Gaussian wave function which you have seen in one of the problems in the previous chapter.
1. Explain why the momentum expectation value is zero, $\langle p \rangle = 0$.
2. Compute the momentum uncertainty $\Delta p$.
3. Show that the wave function saturates Heisenberg's uncertainty principle,
  \[ \Delta x \, \Delta p \geq \frac{\hbar}{2} \, . \]
4. What happens to $\Delta p$ when the particle is localised in space?
5. What changes if you multiply the wave function by $e^{i x p_0/\hbar}$?
Solution:
▶
1. The wave function is real.
2. First compute the action of the momentum operator $p = - i \hbar \frac{\partial}{\partial x}$ on the wave function,
  \begin{align} \nonumber p \, \psi(x) & = - i\hbar \frac{\partial}{\partial x} \, \left( C e^{-(x-x_0)^2/4\Delta^2} \right) \\ & = \frac{i \hbar}{2\Delta^2} (x-x_0) \, C e^{-(x-x_0)^2/4\Delta^2} \\ & = \frac{i \hbar}{2\Delta^2} (x-x_0) \, \psi(x) \, . \end{align}
  Acting again, we find
  \begin{align} \nonumber p^2 \, \psi(x) & = -\frac{\hbar^2}{4\Delta^4} \left( (x-x_0)^2-2\Delta^2 \right) \psi(x) \, . \end{align}
  We can now re-use the position expectation values $\langle x \rangle = x_0$ and $\langle x^2 \rangle = x_0^2+\Delta^2$ computed in question 3 (together with the normalisation condition $\langle 1\rangle = 1$) to compute the momentum expectation values,
  \begin{align} \langle p \rangle & = \frac{i\hbar}{2\Delta^2} \langle x -x_0 \rangle =0 \\ \nonumber \langle p^2 \rangle & = -\frac{\hbar^2}{4\Delta^2} \langle (x-x_0)^2-2\Delta^2 \rangle \\ & = -\frac{\hbar^2}{4\Delta^2} \left( \langle x^2\rangle - 2 x_0\langle x\rangle + x_0^2-2\Delta^2\right) \\ & = \frac{\hbar^2}{4\Delta^2} \, . \end{align}
  If you are uncomfortable with the above manipulations, you can first compute the momentum expectation values using the integral definitions
  \begin{align} \nonumber \langle p \rangle & = - i \hbar \int_{-\infty}^\infty dx\, \overline{\psi(x)} \partial_x \psi(x) \\ \langle p^2 \rangle & = - \hbar^2 \int_{-\infty}^\infty dx\, \overline{\psi(x)} \partial^2_x \psi(x) \, . \end{align}
  You will find that the computation is equivalent to the one above! Finally, $\Delta p = \sqrt{\langle p^2\rangle - \langle p\rangle^2} = \hbar / 2\Delta $.
3. Recalling that $\Delta x = \Delta$, we have $\Delta x \Delta p = \hbar / 2$.
4. The uncertainty in momentum increases.
5. If you multiply the wave function by $e^{i x p_0/\hbar}$, the momentum expectation value changes to $\langle p\rangle = p_0$.
1
Non-normalisable wave functions
Consider the wave function
\begin{equation} \psi_1(x) = C e^{i k x}\,, \end{equation}
for some constant $C$ and $k$. The total integrated probability is not defined, because the integral
\begin{equation} \int_{-\infty}^{\infty} |\psi(x)|^2 {\rm d}x \end{equation}
does not exist for any non-zero value of $C$. However, we can make sense of it by ‘cutting off’ the integral so that $-L \lt{} x \lt{} L$ (‘putting the particle in a box’), doing all computations at finite $L$, and then taking the limit $L\rightarrow\infty$. If you use this ‘regularisation’, what is the expectation value $\langle \hat{p}\rangle$ when the system is described by the wave function above? Also compute $\langle p^2 \rangle$.

Solution:
▶
If we cut off the integrals,
\begin{equation} C^{-2} = \int_{-L}^{L} {\rm d}x = 2L\,,\quad\rightarrow\quad C = 1/\sqrt{2L}\,, \end{equation}
and the expectation value of the momentum becomes
\begin{equation} \langle p \rangle = C^2 \int_{-L}^{L} (-i\hbar)(i k) {\rm d}x = \hbar k\,. \end{equation}
This is independent of $L$. However, manipulations like these are on a shoddy mathematical ground (we e.g. did not discuss the boundary conditions on the wave function), and we will see some examples where naive reasoning like this goes flat on its face.
Momentum in a box
Consider a particle in an infinite potential well, so that $-L/2 \lt{} x \lt{} L/2$. If you want to show that $\langle p\rangle$ is real (see above), you will encounter boundary terms which need to vanish. Do they? Why?

Solution:
▶
The boundary term you encounter when computing $\overline{\langle p\rangle} - \langle p \rangle$ reads
\begin{equation} i\hbar \Big[ \psi(x)\bar\psi(x) \Big]^{L/2}_{-L/2}\,. \end{equation}
For generic wave functions $\psi(x)$ this would not vanish. It will, however, vanish if the wave function vanishes at the edge of the box, which is precisely the condition we have used in the chapter where we first introduced the wave function.

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1.	Why Quantum Mechanics?
2.	The Double Slit Experiment
3.	Wave function and Probabilities
4.	Momentum and Planck's constant
4.1.	Momentum in Classical Mechanics
4.2.	Momentum in Quantum Mechanics
4.3.	A Quick Commutator
4.4.	Momentum Expectation Values
4.5.	Heisenberg's Uncertainty Principle
4.6.	Problems
5.	Schrödinger's Equation
6.	The Hilbert Space
7.	Hermitian Operators
8.	The Spectrum of a Hermitian Operator
9.	Postulates of Quantum Mechanics
10.	Commutators and Uncertainty Principle
11.	Energy Revisited
12.	Stationary states
13.	Case Study: The Free Particle
14.	Two-particle systems
15.	Simple Harmonic Oscillator
16.	The Continuity Equation
17.	Scattering Problems
18.	Tunnelling
19.	Momentum-space Wave function
20.	Ehrenfest's Theorem
21.	Bibliography