The Momentum Operator [Quantum Mechanics I Studied #3]

Picking up right where I left off.

When a particle has wave function (state function) $\psi(x,t)$, the probability — uhh, the expected value — of finding that particle at position $x$ is:

$$\langle j \rangle = \sum j P(j)$$

Same idea, continuous version with $x$:

$$\langle x \rangle = \int x \left| \psi \right|^{2} dx$$

Done.

Now — the thing the professor and the book both hammered on, so we wouldn’t get the wrong idea about $\langle x \rangle$:

It is NOT “measure the same particle $\psi$ over and over, and the measurements settle toward $\langle x \rangle$.” BZZT. Wrong.

It’s “a whole bunch of people each measure this $\psi$ once, then average their results.” That’s $\langle x \rangle$. Ensemble average. Prof kept repeating it, book kept repeating it — and I’m pretty sure the emphasis was specifically on the “ensemble” part.

OK. So now I kinda know what $\langle x \rangle$ is.

Then — if the state function changes with $t$, the probability of the particle being at position $x$ also changes with $t$, right?

So let’s track how $\langle x \rangle$ moves as time ticks. Call it… I dunno, the velocity of the expected value of $x$? $\langle v \rangle$.

Anyway — all I want to do is differentiate $\langle x \rangle$ with respect to $t$.

It’s super similar to the integration gymnastics we did last time. Here we goooo!!! (What this whole calculation actually means will sink in as we go. Let’s just crank through it first.)

$$ \begin{aligned} \frac{d}{dt}\left(\int_{-\infty}^{\infty} x\left|\psi\right|^{2} dx\right) &= \int_{-\infty}^{\infty} x\left(\frac{\partial \psi^{*}}{\partial t}\psi + \psi^{*}\frac{\partial \psi}{\partial t}\right)dx \\ &= \int_{-\infty}^{\infty} x\left(\left(-\frac{i\hbar}{2m}\frac{\partial^{2}\psi^{*}}{\partial x^{2}} + \frac{i}{\hbar}V\psi^{*}\right)\psi + \psi^{*}\left(\frac{i\hbar}{2m}\frac{\partial^{2}\psi}{\partial x^{2}} - \frac{i}{\hbar}V\psi\right)\right)dx \\ &= \int_{-\infty}^{\infty} x\left(-\psi\frac{i\hbar}{2m}\frac{\partial^{2}\psi^{*}}{\partial x^{2}} + \frac{i}{\hbar}V\psi^{*}\psi + \frac{i\hbar}{2m}\psi^{*}\frac{\partial^{2}\psi}{\partial x^{2}} - \frac{i}{\hbar}V\psi\psi^{*}\right)dx \\ &= \int_{-\infty}^{\infty} x\left(-\psi\frac{i\hbar}{2m}\frac{\partial^{2}\psi^{*}}{\partial x^{2}} + \frac{i\hbar}{2m}\psi^{*}\frac{\partial^{2}\psi}{\partial x^{2}}\right)dx \\ &= \frac{i\hbar}{2m}\int_{-\infty}^{\infty} x\left(\frac{\partial^{2}\psi}{\partial x^{2}}\psi^{*} - \psi\frac{\partial^{2}\psi^{*}}{\partial x^{2}}\right)dx \\ &= \frac{i\hbar}{2m}\int_{-\infty}^{\infty} x\frac{d}{dx}\left(\frac{\partial\psi}{\partial x}\psi^{*} - \psi\frac{\partial\psi^{*}}{\partial x}\right)dx \quad \text{integration by parts,} \\ &= \frac{i\hbar}{2m}\left\{\left[x\left(\frac{\partial\psi}{\partial x}\psi^{*} - \psi\frac{\partial\psi^{*}}{\partial x}\right)\right]_{-\infty}^{\infty} - \int_{-\infty}^{\infty}\left(\frac{\partial\psi}{\partial x}\psi^{*} - \psi\frac{\partial\psi^{*}}{\partial x}\right)dx\right\} \\ &= -\frac{i\hbar}{2m}\int_{-\infty}^{\infty}\left(\frac{\partial\psi}{\partial x}\psi^{*} - \psi\frac{\partial\psi^{*}}{\partial x}\right)dx \end{aligned} $$$$ \begin{aligned} \therefore\quad \frac{d}{dt}\langle x \rangle &= -\frac{i\hbar}{2m}\int_{-\infty}^{\infty}\left(\frac{\partial\psi}{\partial x}\psi^{*} - \psi\frac{\partial\psi^{*}}{\partial x}\right)dx \\ &= -\frac{i\hbar}{2m}\left\{\int_{-\infty}^{\infty}\frac{\partial\psi}{\partial x}\psi^{*}\,dx - \int_{-\infty}^{\infty}\psi\frac{\partial\psi^{*}}{\partial x}\,dx\right\} \\ &\quad\quad \left(\text{IBP the red one again!}\right) \\ &= -\frac{i\hbar}{2m}\left\{\int_{-\infty}^{\infty}\frac{\partial\psi}{\partial x}\psi^{*}\,dx - \left(\left[\psi\psi^{*}\right]_{-\infty}^{\infty} - \frac{\partial\psi}{\partial x}\psi^{*}\right)\right\} \\ &\quad\quad \left(\text{the blue term is 0 again}\right) \\ &= --\frac{i\hbar}{2m}\left\{2\int_{-\infty}^{\infty}\frac{\partial\psi}{\partial x}\psi^{*}\,dx\right\} \\ &= -\frac{i\hbar}{m}\int_{-\infty}^{\infty}\frac{\partial\psi}{\partial x}\psi^{*}\,dx \\[6pt] \therefore\quad \frac{d}{dt}\langle x \rangle &= \langle v \rangle \\ &= -\frac{i\hbar}{m}\int_{-\infty}^{\infty}\frac{\partial\psi}{\partial x}\psi^{*}\,dx \end{aligned} $$

OK so — if this is the expected value of the particle’s velocity (or should I say “the velocity of the expected value of the particle’s position”?) —

then multiply by $m$ and we’ve got the expected value of momentum:

$$ \begin{aligned} \langle p \rangle &= m\langle v \rangle \\ &= m\left(-\frac{i\hbar}{m}\int_{-\infty}^{\infty}\frac{\partial\psi}{\partial x}\psi^{*}\,dx\right) \\ &= -i\hbar\int_{-\infty}^{\infty}\frac{\partial\psi}{\partial x}\psi^{*}\,dx \\ &= \int_{-\infty}^{\infty}\frac{\partial\psi}{\partial x}\psi^{*}\,dx \\ &= \int_{-\infty}^{\infty}\frac{\hbar}{i}\psi^{*}\frac{\partial\psi}{\partial x}\,dx \\ &= \int_{-\infty}^{\infty}\psi^{*}\frac{\hbar}{i}\frac{\partial\psi}{\partial x}\,dx \\ &= \int_{-\infty}^{\infty}\psi^{*}\left(\frac{\hbar}{i}\frac{\partial}{\partial x}\right)\psi\,dx \end{aligned} $$

OK OK OK OK, let me yank out just the two headline results from that whole frantic integrating session:

$$ \langle x \rangle = \int \psi^{*}\, x\, \psi\, dx \\ \langle p \rangle = \int \psi^{*}\left(\frac{\hbar}{i}\frac{\partial}{\partial x}\right)\psi\, dx $$

The book’s line is:

‘$x$’ isn’t just “position $x$” — it “represents $x$”

‘$p$’ isn’t just “momentum $p$” — it “represents $p$”

— that’s what it says. And I want to retell that in my own words so it actually clicks.

Look at the result for $p$. The thing sitting in the middle,

$$\left(\frac{\hbar}{i}\frac{\partial}{\partial x}\right)$$

this is what you apply to something to get a momentum value. And here’s the crucial bit: in QM, $p$ — momentum — is NOT a value.

It’s the thing you have to apply to something in order to get the expected value $\langle p \rangle$. (From here on: $p$ is not a value. It’s an operator.)

That’s how I parsed it.

And once you think of it that way —

(warning: full nonsense from here on)

— it kind of feels like something like this is going on.

So when you want to know the $p$ of some particle…

now…

you go talk to this guy.

Me: “Hey! Mr. p-measurer! Measure my $p$ for me!!”

Him: “Sure thing! Bring me the particle’s state function and I’ll measure it for ya!!”

Me: “Yeahhhh here —

$$\psi$$

this guy~~~

$$\psi^{*}$$

— and this is the complex conjugate!!!!”

Him: “Yes!!! Put those two on either side of me and SLAM an integral on top!!!”

Him: “Then I’ll disappear (T_T)(T_T)(T_T)(T_T). Once I’m gone, just look at what’s left. That’s your answer, got it?!!”

$$\int{\quad}$$$$\psi$$

$$\psi^{*}dx$$

Bye bye…

$$!!!!!\ \langle p \rangle\ !!!!!$$

(I was trying to say “$p$ is an operator,” but I don’t know if it’s OK to say it in such a drug-fueled way (T_T)(T_T)(T_T).)

But — ranting nonsense like this actually made me feel, a little, why the Copenhagen crowd went so hard on the interaction between the thing being measured and the measurer. heh. heh.

cf.) Writing $p$ in 1D as

$$\left(\frac{\hbar}{i}\frac{\partial}{\partial x}\right)$$

— and extending to 3D~

$$p = \left(\frac{\hbar}{i}\nabla\right) = -i\hbar\nabla$$

Clean, right?

One more thing. Energy! — which we originally thought of as just a “value”…

after the whole momentum-is-an-operator makeover…

let’s see what energy turns into in QM.

Kinetic energy:

$$T = \frac{p^{2}}{2m}$$

so~

$$T = \frac{p^{2}}{2m} = \frac{1}{2m}\left(\frac{\hbar}{i}\frac{\partial}{\partial x}\right)\left(\frac{\hbar}{i}\frac{\partial}{\partial x}\right) = -\frac{\hbar^{2}}{2m}\frac{\partial^{2}}{\partial x^{2}}$$

Then $\langle T \rangle$ is… heh

$$\langle T \rangle = \int \psi\left(-\frac{\hbar^{2}}{2m}\frac{\partial^{2}}{\partial x^{2}}\right)\psi^{*}\,dx$$

I keep sandwiching the operator right in the middle between $\psi^{*}$ and $\psi$ whenever I compute an expected value. The actual reason for that — we learn it in chapter 3.

…yeah, at the time I was like “fine, that’s just how it is for now~~,” but honestly I think it’s totally fine to take it as “that’s the definition~” and move on.

P.S.

This whole post leaned hard on the idea of a “measurer” as an operator.

What do we call this act-of-measuring, mathematically?

…couldn’t we call it a matrix?

And that means — momentum is a matrix, and by the same logic —

differentiation itself is a matrix.

“Differentiation is a matrix.”

There’s a YouTube video I recently watched and got really into, and that’s exactly what it’s about. I think it’ll help — a lot — with grokking this post. Adding it way after the original writing date, but here’s the link:

(Gave me goosebumps. Serious goosebumps…)

https://youtu.be/BaEillNU3Nk

Video

The Simulation Universe Discovered in Quantum Mechanics

【Related videos and blog posts】— Differentiation is a matrix (https://youtu.be/RZkTxmUWcns)— Differentiation is a matrix (https://moe34.tistory.com/18)— dif…

youtu.be

Originally written in Korean on my Naver blog (2015-08). Translated to English for gdpark.blog.