Lectures on Quantum Mechanics 5: The Heisenberg Uncertainty Principle

Conjugate Variables

Before I derive the uncertainty principle, I would like to introduce the notion of conjugate variables. The uncertainty principle relates the standard deviation two conjugate variables, which was briefly mentioned in Lecture 3. In Lecture 1 we described the wave-packet of a free particle being

$\Psi(x,t) = \frac{1}{\sqrt{2\pi \hbar}}\int_{-\infty}^{\infty}{\phi(p)}{e}^{i(px-Et)/ \hbar}dp.$

If we let $\Phi(p,t)={\phi(p)}{e}^{iEt/ \hbar}$, the integral becomes

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

Here we find that the position-space wavefunction $\Psi(x,t)$ is indeed the Fourier transform of another wavefunction that is dependent on momentum instead. We will call $\Phi(p,t)$ the momentum-space wavefunction.

Following this logic, we can recover the momentum-space wavefunction given the position-space wavefunction by taking the inverse Fourier transform:

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

So what? What did we learn by doing this? Well, what this bit of Fourier analysis has achieved is very interesting, because we see a direct relationship between two variables $x$ and $p$, that describe the same wave-packet in different spaces. In this example, a free particle is measured in the position space by $\Psi(x,t)$, but measured in the momentum space by $\Phi(p,t)$. Yet regardless which space we choose to measure the particle, the physics doesn't change. Thus by discovering this relationship of conjugate variables, we can make measurements on particles in physical experiments in two ways. In fact, this is how crystallography and much of solid-state physics is done (see Reciprocal Lattice).

Heisenberg's Uncertainty Principle (HUP)

From Lecture 3 we discussed the definition of standard deviation for some variable, say $x$, ${\sigma}_{x}^{2} = \left<{(x-\left<x \right>)}^{2}\right>$. Back then we were concerned with numbers, but to derive the uncertainty principle, we need to write the standard deviation in terms of operators acting on wavefunctions.

Thus we consider the inner product

${\sigma}_{x}^{2}=\left<(\hat{x}-\left<\hat{x}\right>)\Psi|(\hat{x}-\left<\hat{x}\right>)\Psi\right>.$

Let $f=(\hat{x}-\left<\hat{x}\right>)\Psi$, thus ${\sigma}_{x}^{2}=\left<f|f\right>$.

Similarly we have ${\sigma}_{p}^{2}=\left<(\hat{p}-\left<\hat{p}\right>)\Psi|(\hat{p}-\left<\hat{p}\right>)\Psi\right>=\left<g|g\right>$ where $g=(\hat{p}-\left<\hat{p}\right>)\Psi$.

By the Cauchy-Schwarz inequality, we get ${\sigma}_{x}^{2}{\sigma}_{p}^{2} = \left<f|f\right>\left<g|g\right> \ge {|\left<f|g\right>|}^{2}.$

Now let's simplify $\left<f|g\right>$:

$\begin{aligned} &\left<f|g\right>=\left<(\hat{x}-\left<\hat{x}\right>)\Psi|(\hat{p}-\left<\hat{p}\right>)\Psi\right> \\ &=\left<\Psi|(\hat{x}-\left<\hat{x}\right>)(\hat{p}-\left<\hat{p}\right>)\Psi\right> \\ &=\left<\Psi|(\hat{x}\hat{p}-\hat{x}\left<\hat{p}\right>)-\hat{p}\left<\hat{x}\right>+\left<\hat{x}\right>\left<\hat{p}\right>)\Psi\right> \\ &=\left<\Psi|\hat{x}\hat{p}\Psi\right>-\left<p\right>\left<\Psi|\hat{x}\Psi\right>-\left<x\right>\left<\Psi|\hat{p}\Psi\right>+\left<x\right>\left<p\right>\left<\Psi|\Psi\right>\\ &=\left<\hat{x}\hat{p}\right>-\left<\hat{p}\right>\left<\hat{x}\right>-\left<\hat{x}\right>\left<\hat{p}\right>+\left<\hat{x}\right>\left<\hat{p}\right>\\ &=\left<\hat{x}\hat{p}\right>-\left<\hat{p}\right>\left<\hat{x}\right> \end{aligned}$

Similarly, $\left<g|f\right>=\left<\hat{p}\hat{x}\right>-\left<\hat{x}\right>\left<\hat{p}\right>.$

I know the bra-ket notation and expectation of an operator both use angle brackets, which makes the above algebra pretty confusing. Just remember that $\left<\hat{x}\right>$ is the expectation of $\hat{x}$, which is a number.

There is a useful property about complex numbers: ${z}^{*}z \ge {\left[\frac{1}{2i}(z-{z}^{*})\right]}^{2}.$

Therefore if we replace $z$ with $\left<f|g\right>$ and ${z}^{*}$ with $\left<g|f\right>$, we get

$\left<f|f\right>\left<g|g\right>\ge {\left[\frac{1}{2i}(\left<f|g\right>-\left<g|f\right>)\right]}^{2}.$

This translates to ${\sigma}_{x}^{2}{\sigma}_{p}^{2} \ge {\left(\frac{1}{2i}[\hat{x},\hat{p}]\right)}^{2}.$

Remember from Lecture 4 we found out that $[\hat{x},\hat{p}]=i\hbar$. Hence, by substituting the above commutator relation and taking the square root, we arrive at the HUP ${\sigma}_{x}{\sigma}_{p} \ge \frac{\hbar}{2}.$

Visit my set Lectures on Quantum Mechanics for more notes.

Problems

1. Prove that $\int_{-\infty}^{\infty} {\Psi}^{*}\Psi dx = \int_{-\infty}^{\infty} {\Phi}^{*}\Phi dp$ given the position-space and momentum-space wavefunctions.

2. Operators in momentum-space The position operator in momentum space is $\hat{x} = i\hbar\frac{\partial}{\partial p}$ and the momentum operator is $\hat{p} = p.$ Verify that $\left[\hat{x},\hat{p}\right]=i\hbar$ is valid in momentum space.

3. The position-momentum uncertainty principle is often written loosely as $(\Delta x)(\Delta p) \ge \frac{\hbar}{2}.$ Show that the energy-time uncertainty holds $(\Delta E)(\Delta t) \ge \frac{\hbar}{2}$ hence, proving ${\sigma}_{E} {\sigma}_{t} \ge \frac{\hbar}{2}.$