my biggest goal this semester is to know the physics I’m learning well, and one of the steps is to cement my understanding of important tools, procedures, and results. i’m also working on an 8.20 (sr) post but that’s second priority to 8.04/5 review for 8.06 this semester.

general uncertainty principle

schwarz inequality

An important inequality you must know is the Schwarz inequality–$\langle a, a \rangle \langle b, b \rangle \geq |\langle a, b \rangle|^2$. Here we are using our complex inner product, which can yield complex values. In order to derive the inequality, first write $a = cb + a_\perp$, splitting $a$ into a part parallel to $b$ and a part perpendicular to $b$. In order for this to work, $c = \frac{\langle a, b\rangle}{\langle b,b\rangle}$. Think of it as taking the dot product of $a$ with the $\hat b$ direction. We can verify that $\langle a_\perp, b\rangle = \langle a - cb, b \rangle = \langle a,b \rangle - c \langle b,b \rangle = \langle a, b \rangle - \langle a, b \rangle = 0$ as desired. Then $$\begin{align*} \langle a,a\rangle \langle b,b \rangle &= \langle cb + a_\perp, cb + a_\perp \rangle \langle b,b \rangle \\ &= (|c|^2 \langle b,b \rangle + |a_\perp|^2)\langle b,b \rangle \\ &= |c|^2|b|^2 + |a_\perp|^2|b|^2 \\ &= |\langle a,b, \rangle|^2 + |a_\perp|^2|b|^2 \\ &\geq |\langle a,b, \rangle|^2 \end{align*}$$ as desired.

application to wavefunctions

The uncertainty of a Hermitian operator $\Delta A$ is $\langle A - \langle A \rangle \rangle$, or average deviation from the mean. All expectations of Hermitian operators must be real, so we can write the Schwarz inequality for $a = (A - \langle A \rangle) \psi$ and $b = (B - \langle B \rangle) \psi$: $(\Delta A)^2 (\Delta B)^2 \geq |\langle \psi |(A - \langle A \rangle)(B - \langle B \rangle| \psi \rangle|^2$, as $A - \langle A \rangle$ is Hermitian. Then multiplying things out, we see that $\langle \psi |(A - \langle A \rangle)(B - \langle B \rangle| \psi \rangle = \langle AB \rangle - \langle A \rangle \langle B \rangle$. Note that $AB$ is not necessarily Hermitian, as $(AB)^\dagger = B^\dagger A^\dagger = BA$ for Hermitian operators $A$,$B$. Therefore $\langle AB \rangle$ may be complex. The norm squared of a complex number $\langle AB \rangle - \langle A \rangle \langle B \rangle$ is the sum of the squares of its real and complex parts, so the norm squared must be at least the square of the imaginary part. The imaginary part is the complex number minus its conjugate divided by $2i$. The conjugate of $\langle (A - \langle A \rangle)\psi, (B - \langle B \rangle) \psi \rangle = \langle AB \rangle - \langle A \rangle \langle B \rangle$ is $\langle (B - \langle B \rangle) \psi, (A - \langle A \rangle) \psi \rangle = \langle BA \rangle - \langle B \rangle \langle A \rangle$ by symmetry, where I’ve rewritten the bra-ket notation as inner products in order to emphasize $\langle a,b \rangle^* = \langle b,a \rangle$. Then the imaginary part of $\langle AB \rangle - \langle A \rangle \langle B \rangle$ is $\frac{\langle AB \rangle - \langle A \rangle \langle B \rangle - \langle BA \rangle + \langle A \rangle \langle B \rangle}{2i} = \frac{1}{2i} \langle [A,B] \rangle$, so plugging it back into our result from the Schwarz inequality, we get $$\begin{align*}(\Delta A)^2 (\Delta B)^2 &\geq |\langle \psi |(A - \langle A \rangle)(B - \langle B \rangle| \psi \rangle|^2\\ &\geq (\text{Im}(\langle AB \rangle - \langle A \rangle \langle B \rangle))^2 \\ &= (\frac{1}{2i} \langle [A,B] \rangle)^2\end{align*}$$Don’t let this notation fool you; $\langle [A,B] \rangle$ is imaginary because $[A,B]$ is not generally Hermitian! $\frac{1}{2i}[A,B]$ is Hermitian though: $(\frac{1}{2i}[A,B])^\dagger = -\frac{1}{2i} ((AB)^\dagger - (BA)^\dagger) = -\frac{1}{2i} (BA - AB) = \frac{1}{2i} [A,B]$. Therefore we would rather write the RHS as the expectation of a Hermitian operator: $(\Delta A)^2 (\Delta B)^2 \geq \langle \frac{1}{2i} [A,B] \rangle^2$. This is the uncertainty principle.

energy-time uncertainty

Motivation: if you have a light ray that’s localized to be nonzero over some distance, it takes $T$ time for the light ray for any point in its trajectory to experience the nonzero part of the wave from start to finish. Then there are $Tv = \frac{T\omega}{2\pi} \pm 1$ wavelengths within this time, the uncertainty being because it’s hard to tell where the first and last wavelengths would have started or ended if they hadn’t gone to 0. Then if $E = \hbar w$, and $\Delta Tv = 1 \implies \Delta w = 2 \pi/T$, then $\Delta E = 2 \pi \hbar / T$, so $\Delta E \cdot T = 2 \pi \hbar$. So you see in some sense if the light wave is very localized in time, its energy uncertainty must be large as it is hard to tell its wavelength.

From the general uncertainty principle $(\Delta A)^2(\Delta B)^2 \geq (\langle \psi | \frac{1}{2 i}[A,B] \psi\rangle)^2$ and plugging in $A=H$ and $B=Q$ (both time-independent), we get $(\Delta H)^2(\Delta Q)^2 \geq (\langle \psi | \frac{1}{2 i}[H,Q] \psi\rangle)^2$. Now we take $\frac{d \langle Q \rangle}{d t} = \frac{d}{dt}\langle \psi | Q | \psi \rangle = \frac{d}{dt} \int \psi^* Q \psi dx = \int \frac{d}{dt} \psi^* Q \psi dx + \int \psi^* Q \frac{d}{dt} \psi dx$. Now by the Schrodinger equation, $i \hbar \frac{d}{dt} \psi = H \psi$, so this becomes $$\begin{align*}\int (-\frac{H}{i \hbar}) \psi^* Q \psi dx + \int \psi^* Q (\frac{H}{i \hbar}) \psi dx &= \frac{i}{\hbar} \langle \psi | HQ - QH | \psi \rangle \\ &= \frac{i}{\hbar} \langle \psi | [H,Q] | \psi \rangle\end{align*}$$ Bringing back the uncertainty inequality, we get $(\Delta H)^2(\Delta Q)^2 \geq (\langle \psi | \frac{1}{2 i}[H,Q] \psi\rangle)^2 = \big( \frac{\hbar}{2} \frac{d\langle Q \rangle}{dt} \big)^2$, so $\Delta H\Delta Q \geq \frac{\hbar}{2} \frac{d\langle Q \rangle}{dt}$. If we say $\Delta T = \frac{\Delta Q}{\frac{d\langle Q \rangle}{dt}}$ is the time it takes for $\langle Q \rangle$ to change by $\Delta Q$ at some time, then we can rewrite this as $\Delta H \Delta T \geq \frac{\hbar}{2}$. This is the energy-time uncertainty.