The Uncertainty Principle
We derive the generalized uncertainty principle and chase down exactly when it hits its minimum — turns out the answer is when ψ is Gaussian, and here's why!
So we get
$$\sigma_{A}^{2}\cdot \sigma_{B}^{2} \ge \left\{\frac{1}{2i}\left\langle[\hat{A},\hat{B}]\right\rangle\right\}^{2}$$Yes yes yes yes — so this is the generalized uncertainty principle. This right here is the equation.
And the one we all know and love,
$$\sigma_{x}\cdot \sigma_{p} \ge \frac{\hbar}{2}$$it just pops out if you plug in $A = x$ and $B = p$!!
$$\sigma_{A}^{2}\cdot \sigma_{B}^{2} \ge \left\{\frac{1}{2i}\left\langle[\hat{A},\hat{B}]\right\rangle\right\}^{2} \\ \sigma_{x}^{2}\cdot \sigma_{p}^{2} \ge \left\{\frac{1}{2i}\left\langle\left[x,\frac{\hbar}{i}\frac{d}{dx}\right]\right\rangle\right\}^{2} = \left\{\frac{1}{2i}i\hbar\right\}^{2} = \left(\frac{\hbar}{2}\right)^{2} \\ \therefore \quad \sigma_{x}\cdot \sigma_{p} \ge \left(\frac{\hbar}{2}\right)^{2}$$Like that~~~
Ohhh that was fun, been a while. heh heh heh.
“Hey — so what about when
$$\sigma_{x}\cdot \sigma_{p} = \frac{\hbar}{2}$$like, when it’s at the absolute smallest value it can be!!
When exactly is that??”
We derived uncertainty as an inequality, right?
So “smallest uncertainty” = “the inequality saturates into an equality”!?!!!!
OK let’s hunt that down. When does the inequality become an equality?
Specifically — Heisenberg’s uncertainty principle,
$$\sigma_{x}\cdot \sigma_{p} \ge \frac{\hbar}{2}$$when does this become an equality?
Spoiler: when $\psi$ is Gaussian. That’s the minimum-uncertainty state.
OK so — why on earth is it Gaussian that minimizes uncertainty? Let’s see.
We came here from an inequality of this form:
$$\sigma_{A}^{2}\cdot \sigma_{B}^{2} \ge \left|\langle f|g\rangle\right|^{2} \ge \left\{\text{imaginary part of } \langle f|g\rangle\right\}^{2}$$Stare at it for a sec. You can see it now, right?
“When does equality hold~~~~?”
“When the real part is zero.” Because that’s the little + extra bit on the left that kills equality. Kill the real part, kill the slack.
You following?
OK so
$\left|\langle f|g\rangle\right|$
has real part zero!!!
Which means equality holds exactly when that complex number is purely imaginary!!! And that’s literally the condition that
$$\int f^{*}g\,dx$$has real part zero, i.e.
$g = if$
!!!!
Let me write it out cleanly:
$$f^{*}\cdot f \quad \text{is (purely) real} \\ if^{*}\cdot f \quad \text{is purely imaginary}\;!$$So~~~
Actually, let me loosen it juuust a tiny bit:
$g = iaf$
A constant multiple is fine too~~ I’ve given myself a little more room like that.
Now then,
$g = iaf$
unpacks to
$$g = ia\left(\hat{A} - \langle A\rangle\right)\psi$$Aha haha haha
$$\left(\hat{B} - \langle B\rangle\right)\psi = ia\left(\hat{A} - \langle A\rangle\right)\psi$$This is the condition!!!! When this holds, uncertainty is minimum. That’s what we’re saying.
Let’s plug in $x$ for $A$ and $p$ for $B$ and see what falls out.
$$(\hat{p} - \langle p\rangle)\psi = ia(\hat{x} - \langle x\rangle)\psi \\ \left(\frac{\hbar}{i}\frac{d}{dx} - \langle p\rangle\right)\psi = ia(\hat{x} - \langle x\rangle)\psi \\ \frac{\hbar}{i}\frac{d\psi}{dx} = \left\{\langle p\rangle - iax - ia\langle x\rangle\right\}\psi\; \ldots\text{wait, pull the } ia \text{ out:} \\ = ia\left\{x - \langle x\rangle - \frac{i}{a}\langle p\rangle\right\}\psi \\ \frac{d\psi}{dx} = -\frac{a}{\hbar}\left\{x - \langle x\rangle - \frac{i}{a}\langle p\rangle\right\}\psi$$Is this an easy ODE? Yeah, it’s separable.
$$\frac{1}{\psi}d\psi = -\frac{a}{\hbar}\left\{x - \langle x\rangle - \frac{i}{a}\langle p\rangle\right\}dx = \left\{\frac{i}{\hbar}\langle p\rangle + \frac{a}{\hbar}\langle x\rangle - \frac{a}{\hbar}x\right\}dx$$Integrate both sides, go go go —
$$\ln\psi = \left(\frac{i}{\hbar}\langle p\rangle + \frac{a}{\hbar}\langle x\rangle\right)x - \frac{a}{2\hbar}x^{2} + C \\ \therefore \quad \psi = e^{\left(\frac{i}{\hbar}\langle p\rangle + \frac{a}{\hbar}\langle x\rangle\right)x - \frac{a}{2\hbar}x^{2} + C} = e^{\frac{i}{\hbar}\langle p\rangle x}\cdot e^{\frac{2a}{2\hbar}\langle x\rangle}\cdot e^{-\frac{a}{2\hbar}x^{2}}\cdot e^{C} \\ = Ae^{-\frac{a}{2\hbar}(x-\langle x\rangle)^{2}}\cdot e^{\frac{i\langle p\rangle x}{\hbar}}$$It really is Gaussian!? heh heh heh. Wild. Truly wild. heh heh heh heh.
Another flavor of uncertainty — energy and time
We derived
$$\sigma_{x}\cdot \sigma_{p} \ge \frac{\hbar}{2}$$and apparently there’s one more famous uncertainty relation besides the $x$–$p$ one.
“What other pair can’t be measured precisely, besides $x$ and $p$?”
Energy and time!!!
Now, this got derived very handwavily back in modern physics class. Let me quickly recap how that went.
We want to measure the energy $E$ emitted by some atom during a time interval $\Delta t$. That means measuring the $E$ of the emitted EM wave…
The energy of the EM wave is
$$E = h\nu$$so measuring $E$ means we have to count the number of oscillations of the frequency $\nu$!?
And we have to count oscillations during $\Delta t$ — and obviously you’re not going to count oscillations exactly right.
“Let’s say the uncertainty is 1.”
Meaning: we might miscount by at least 1 oscillation. Over a window of $\Delta t$ seconds, miscounting by 1 means the measured frequency $\nu$ can be off by as much as $1/\Delta t$.
So the uncertainty $\Delta\nu$ in frequency satisfies
$$\Delta\nu \ge \frac{1}{\Delta t}$$And the uncertainty in energy is easy!!!
$$\Delta E = h\cdot\Delta\nu$$Combine them. Take
$$\Delta\nu \ge \frac{1}{\Delta t}$$and sub in
$$\Delta E = h\cdot\Delta\nu$$to get
$$\frac{\Delta E}{h} \ge \frac{1}{\Delta t} \\ \therefore \quad \Delta E\cdot\Delta t \ge h \\ \text{and if you do it more carefully,} \\ \Delta E\cdot\Delta t \ge \frac{\hbar}{2}$$That’s how we did it in modern physics.
And there’s this painful memory of just muttering “$\Delta t$ means ’lifetime’” and speedrunning past it without really unpacking anything.
This time, since we’re in quantum mechanics class, let’s do it properly.
Instead of pinning $\Delta t$ to “lifetime”, the book introduces it more broadly and then does the derivation:
$\Delta t$: “the time it takes for the given physical system to change.”
Hmm~~ like that.
“How fast or slow does $E$ change, how fast or slow does $p$ change~~ how fast or slow does this-and-that change.”
We want to look at
$$\frac{d}{dt}\langle Q\rangle$$OK let’s do it.
$$\frac{d}{dt}\langle\hat{Q}\rangle = \frac{d}{dt}\langle\psi|\hat{Q}\psi\rangle \\ = \left\langle\frac{d\psi}{dt}\middle|\hat{Q}\psi\right\rangle + \left\langle\psi\middle|\frac{d\hat{Q}}{dt}\psi\right\rangle + \left\langle\psi\middle|\hat{Q}\frac{d\psi}{dt}\right\rangle \\ \left(\text{since}\; i\hbar\frac{d\psi}{dt} = \hat{H}\psi\right) \\ = \left\langle\frac{1}{i\hbar}\hat{H}\psi\middle|\hat{Q}\psi\right\rangle + \left\langle\psi\middle|\frac{d\hat{Q}}{dt}\psi\right\rangle + \left\langle\psi\middle|\hat{Q}\frac{1}{i\hbar}\hat{H}\psi\right\rangle \\ = -\frac{1}{i\hbar}\langle\hat{H}\psi|\hat{Q}\psi\rangle + \frac{1}{i\hbar}\langle\psi|\hat{Q}\hat{H}\psi\rangle + \left\langle\frac{d\hat{Q}}{dt}\right\rangle \\ = \frac{i}{\hbar}\langle\psi|\hat{H}\hat{Q}\psi\rangle - \frac{i}{\hbar}\langle\psi|\hat{Q}\hat{H}\psi\rangle + \left\langle\frac{d\hat{Q}}{dt}\right\rangle \\ = \frac{i}{\hbar}\left\langle[\hat{H},\hat{Q}]\right\rangle + \left\langle\frac{d\hat{Q}}{dt}\right\rangle \\ \therefore \quad \frac{d}{dt}\langle\hat{Q}\rangle = \frac{i}{\hbar}\left\langle[\hat{H},\hat{Q}]\right\rangle + \left\langle\frac{d\hat{Q}}{dt}\right\rangle$$That thing we just derived… the rate of change of the expectation value with respect to time… isn’t that kinda amazing??
We just mashed the Schrödinger equation in mindlessly, and out pops the Hamiltonian, hahahahaha. Wild.
I mean, it’s the Schrödinger equation, so sure, fine, it tracks. Anyway!!!
Moving on.
There’s a footnote at the bottom of the page that says something like:
“It is very rare for an operator to explicitly depend on time. So in almost all cases you can treat
$$\frac{d\hat{Q}}{dt}=0$$as holding.”
Oh god — meaning in almost all cases,
$$\left\langle\frac{d\hat{Q}}{dt}\right\rangle = 0$$So we’re left with
$$\therefore \quad \frac{d}{dt}\langle\hat{Q}\rangle = \frac{i}{\hbar}\left\langle[\hat{H},\hat{Q}]\right\rangle$$Oh my god!!! The commutator between operators?!?!?!
Where have I seen that a lot?! Let me massage this equation a bit.
$$\therefore \quad \frac{d}{dt}\langle\hat{Q}\rangle = \frac{i}{\hbar}\left\langle[\hat{H},\hat{Q}]\right\rangle \\ \frac{i}{\hbar}\left\langle[\hat{H},\hat{Q}]\right\rangle = -\frac{1}{i\hbar}\left\langle[\hat{H},\hat{Q}]\right\rangle \\ = -\frac{2}{2i\hbar}\left\langle[\hat{H},\hat{Q}]\right\rangle \\ = -\frac{2}{\hbar}\cdot\left(\frac{\left\langle[\hat{H},\hat{Q}]\right\rangle}{2i}\right) \\ \text{i.e.,}\quad -\frac{2}{\hbar}\cdot\left(\frac{\left\langle[\hat{H},\hat{Q}]\right\rangle}{2i}\right) = \frac{d}{dt}\langle\hat{Q}\rangle \\ \left(\frac{\left\langle[\hat{H},\hat{Q}]\right\rangle}{2i}\right) = -\frac{\hbar}{2}\frac{d}{dt}\langle\hat{Q}\rangle$$That highlighted chunk — we have absolutely seen that before!!
$$\sigma_{A}^{2}\cdot \sigma_{B}^{2} \ge \left\{\frac{1}{2i}\left\langle[\hat{A},\hat{B}]\right\rangle\right\}^{2}$$Where have I seen it a lot? Right here!!~!!
$$\sigma_{H}^{2}\cdot \sigma_{Q}^{2} \ge \left\{\frac{1}{2i}\left\langle[\hat{H},\hat{Q}]\right\rangle\right\}^{2} = \left(-\frac{\hbar}{2}\frac{d}{dt}\langle\hat{Q}\rangle\right)^{2} \\ \therefore \quad \sigma_{H}\cdot \sigma_{Q} \ge \left|\frac{\hbar}{2}\frac{d\langle\hat{Q}\rangle}{dt}\right|$$Now the sleight of hand.
$\sigma_{H}$
— can we call this $\Delta E$? “the uncertainty of energy,” sure.
And now~~
$\dfrac{d\langle\hat{Q}\rangle}{dt}$
— I want to massage this with $\sigma_{Q}$ into the form
$$\left|\frac{d\langle\hat{Q}\rangle}{dt}\right|\cdot\Delta t = \sigma_{Q}$$“the uncertainty of $\langle Q\rangle$.”
(cf. rate of change of $\langle Q\rangle$ × uncertainty of time = uncertainty of $\langle Q\rangle$.)
$$\therefore \quad \Delta t = \frac{\sigma_{Q}}{\left|\dfrac{d\langle\hat{Q}\rangle}{dt}\right|}$$OK so~
$$\sigma_{H}\cdot \sigma_{Q} = \Delta E\cdot\left(\left|\frac{d\langle\hat{Q}\rangle}{dt}\cdot\Delta t\right|\right) \ge \left|\frac{\hbar}{2}\frac{d\langle\hat{Q}\rangle}{dt}\right| \\ \therefore \quad \Delta E\cdot\Delta t \ge \frac{\hbar}{2}$$Hmmmm~~~~ So in modern physics class we learned $\Delta t$ as lifetime, but in QM class $\Delta t$ is
Ahhh~~ so I think you can actually interpret $\Delta t$ — “the time it takes for some physical quantity to change” — as lifetime! That’s my read!~
“If some physical quantity changes quickly ($\Delta t \downarrow$), the uncertainty in energy gets bigger ($\Delta E \uparrow$).”
(Think about the quantum jump in a hydrogen atom~~!!~ Apparently, in a state with a high principal quantum number, the electron can only hang around for a short time~!?)
Originally written in Korean on my Naver blog (2015-08). Translated to English for gdpark.blog.