How to interpret integration by parts

[Tony Hau]

TL;DR

I don't know where this question suits; does it belong to calculus or quantum mechanics? But anyway, I would like to ask a question about deriving the expected value of velocity of wave function.

So I am confused about a proof in which the formula for expected value of velocity, $\frac{d\langle x \rangle}{dt}$, is derived.

Firstly, because the expected value of the position of wave function is $$\langle x \rangle =\int_{-\infty}^{+\infty} x|\Psi(x,t)|^2 dx$$Therefore, $$\frac{d\langle x \rangle}{dt} = \int x\frac {\partial } {\partial t} |\Psi|^2 dx = \frac{i \hbar}{2m}\int x\frac {\partial } {\partial x} (\Psi^{*}\frac{\partial \Psi} {\partial x}-\frac{\partial \Psi^{*}} {\partial x}\Psi)dx -(1)$$

The author then writes:
This expression can be simplified using integration-by-parts, which gives: $$\frac{d\langle x \rangle}{dt} =-\frac{i\hbar}{2m}\int (\Psi^{*}\frac{\partial \Psi} {\partial x}-\frac{\partial \Psi^{*}} {\partial x}\Psi)dx -(2)$$

(By the way, he has also written this line of notes: I used the fact that $\frac{\partial x} {\partial x} =1$, and threw away the boundary term, on the grounds that $\Psi$ goes to zero at $(\pm)$ infinity.)

At the bottom of the page, he makes a note as follows:
For $$ \int_a^b f\frac{dg}{dx}dx = -\int_a^b\frac{df}{dx}g dx +fg |_a^b$$
Under the integral sign, then, you can peel a derivative off one factor in a product, and slap it onto the other one - it'll cost you a minus sign, and you'll pick up a boundary condition.

To be honest with, I have no idea of what it means by "you can peel a derivative off one factor in a product, and slap it onto the other one". Plus, I don't know how the author transits from eq(1) to eq(2). According to my previous calculus II knowledge, whenever we see an integrand composed of a product of two functions, we should try to evaluate the antiderivative of one of the functions and expand it into the integration by parts. However, it seems that there is a more sophisticated way of understanding it in physics.

 

[PeroK]

I think by this stage in your physics studies, you are expected to have seen this sort of thing before. If you haven't there are quite a few steps here.

Note that you've dropped the $x$ from the integrand. What you have doesn't make sense, therefore. In terms of integration by parts, the first function is $x$ and the second function is the rest of the integrand, which is a derivative wrt the integration variable. Using the parts rule (integrate the second function) this cancels the derivative.

It's just the usual integration by parts here, with a vanishing boundary term.

 

[Tony Hau]

I think by this stage in your physics studies, you are expected to have seen this sort of thing before. If you haven't there are quite a few steps here.

Note that you've dropped the $x$ from the integrand. What you have doesn't make sense, therefore. In terms of integration by parts, the first function is $x$ and the second function is the rest of the integrand, which is a derivative wrt the integration variable. Using the parts rule (integrate the second function) this cancels the derivative.

It's just the usual integration by parts here, with a vanishing boundary term.

Yes, I have seen a few times, for example in electrodynamics. But I chose to ignore the proofs at that time. Now I feel obliged to understand it...

 

[PeroK]

Yes, I have seen a few times, for example in electrodynamics. But I chose to ignore the proofs at that time. Now I feel obliged to understand it...

Well, you can hardly complain now!

 

[Tony Hau]

I think by this stage in your physics studies, you are expected to have seen this sort of thing before. If you haven't there are quite a few steps here.

Note that you've dropped the $x$ from the integrand. What you have doesn't make sense, therefore. In terms of integration by parts, the first function is $x$ and the second function is the rest of the integrand, which is a derivative wrt the integration variable. Using the parts rule (integrate the second function) this cancels the derivative.

It's just the usual integration by parts here, with a vanishing boundary term.

I have added back that $x$. So here is what I think: $$\frac{d\langle x \rangle}{dt} = \frac{i \hbar}{2m}\int_{-\infty}^{+\infty} x\frac {\partial } {\partial x} (\Psi^{*}\frac{\partial \Psi} {\partial x}-\frac{\partial \Psi^{*}} {\partial x}\Psi)dx $$
$$= -\int_{-\infty}^{+\infty} \frac{\partial x} {\partial x} (\Psi^{*}\frac{\partial \Psi} {\partial x}-\frac{\partial \Psi^{*}} {\partial x}\Psi)dx + \left. (\Psi^{*}\frac{\partial \Psi} {\partial x}-\frac{\partial \Psi^{*}} {\partial x}\Psi) x \right| _{-\infty}^{+\infty}$$
Because wave function is $0$ at $\pm \infty$, the second term on the right hand side is $0$.
$$\frac{d\langle x \rangle}{dt} =-\frac{i \hbar}{2m} \int_{-\infty}^{+\infty} (\Psi^{*}\frac{\partial \Psi} {\partial x}-\frac{\partial \Psi^{*}} {\partial x}\Psi)dx$$
For $$ \int_{-\infty}^{+\infty} (\Psi^{*}\frac{\partial \Psi} {\partial x})dx-\int_{-\infty}^{+\infty} (\frac{\partial \Psi^{*}} {\partial x}\Psi)dx$$
$$\int_{-\infty}^{+\infty} (\Psi^{*}\frac{\partial \Psi} {\partial x})dx = -\int_{-\infty}^{+\infty} \frac{\partial \Psi^*} {\partial x}\Psi dx + \left. \Psi \Psi^* \right|_{-\infty}^{\infty}$$
Again, the second term is $0$. Substituting the result back to the equation, we get: $$\int_{-\infty}^{+\infty} (\Psi^{*}\frac{\partial \Psi} {\partial x})dx-\int_{-\infty}^{+\infty} (\frac{\partial \Psi^{*}} {\partial x}\Psi)dx = 2\int_{-\infty}^{+\infty} (\Psi^{*}\frac{\partial \Psi} {\partial x})dx$$
Therefore, $$\frac{d\langle x \rangle}{dt} =-\frac{i\hbar}{m}\int_{-\infty}^{+\infty} (\Psi^{*}\frac{\partial \Psi} {\partial x})dx$$
$$\langle p \rangle=m\frac{d\langle x \rangle}{dt}=-i\hbar\int_{-\infty}^{+\infty} (\Psi^{*}\frac{\partial \Psi} {\partial x})dx$$

 

[Tony Hau]

Well, you can hardly complain now!

A very silly question: when performing the integration wrt to dx, we keep the variable in $\Psi$ constant, right? Sorry but I am not very familiar with such type of multivariable integration.

 

[PeroK]

A very silly question: when performing the integration wrt to dx, we keep the varaible in $\Psi$ constant, right? Sorry but I am not very familiar with such type of multivariable integration.

Actually, $\Psi(x, t)$ is a function of two variables. If you integrate wrt $x$, then $t$ is essentially a constant and the normal rule of parts applies with partial derivatives instead of ordinary derivatives, where appropriate.

Also, a technical point: it's not actually enough that $\Psi(x, t) \rightarrow 0$ as $x \rightarrow \pm \infty$. You need the stronger condition that $x\Psi \frac{\partial \Psi^*}{\partial x} \ \rightarrow \ 0$. But, generally, physically viable wave-functions will have this property and tend exponentially to zero for large $x$.

 

[Demystifier]

To be honest with, I have no idea of what it means by "you can peel a derivative off one factor in a product, and slap it onto the other one".

It's a poetry. If you don't understand poetry, here is a quote of Paul Dirac:
"In science one tries to tell people, in such a way as to be understood by everyone, something that no one ever knew before. But in poetry, it's the exact opposite."

 

[Tony Hau]

It's a poetry. If you don't understand poetry, here is a quote of Paul Dirac:
"In science one tries to tell people, in such a way as to be understood by everyone, something that no one ever knew before. But in poetry, it's the exact opposite."

By the way, the poet is David J. Griffiths :)

 

[PeroK]

By the way, the poet is David J. Griffiths :)

It's time for a Haiku, I guess:

Quantum Mechanics.
A statistical heaven,
Or classical hell?

 

[Demystifier]

It's time for a Haiku, I guess:

Quantum Mechanics.
A statistical heaven,
Or classical hell?

To be or not to be?
Ontological interpretation,
Or a Copenhagen one?

 

[vanhees71]

I think this proof is overly complicated. One can just use the Schrödinger equation
$$\mathrm{i} \hbar \partial_t \psi=\hat{H} \psi, \quad -\mathrm{i} \hbar \partial_t \psi^* = \hat{H} \psi^*.$$
From this you get
$$\mathrm{d}_t \langle x \rangle=\int_{\mathbb{R}^3} \mathrm{d}^3 x \frac{1}{\mathrm{i} \hbar}[\psi^* x \hat{H} \psi - x \hat{H} \psi^* \psi]=\frac{1}{\mathrm{i} \hbar} \langle [\hat{x},\hat{H}] \rangle=\frac{1}{m} \langle p \rangle.$$
In the last steps I've used the self-adjointness of $\hat{H}$ and $[\hat{x},\hat{H}]=1/(2m) [\hat{x},\hat{p}^2]=1/m \hat{p}$.