Integral of partial derivative of x with respect to t

[crunchynet]

Hello Everyone,

So in other words, if you didn't understand what I'm saying from the title of this post, look at it this way:

What is the answer to this integral?

∫(partial dx)/(partial dt) * dx

According to my textbook the answer is 0 but I'm getting easily confused as to how this is the case... By the way x is never defined, you just know that it's a function of t.

God I love senior level physics. I mean I really do that's not sarcasm, I find this stuff incredibly interesting but I'm having difficulties with things like this conceptually :). If anyone could help I'd really appreciate it!

-CrunchyNet

 

[vanhees71]

This doesn't make the slightest sense! Please give a bit more context from the textbook, which I'd handle with some suspicion if it really makes such strange statements!

 

[JJacquelin]

Hi !
The context is essential. In any case, if the limits of integration are not specified the indefined integral cannot be = 0 in general, but depends to an arbitrary constant.

 

[HallsofIvy]

I strongly suspect that "x" was some specific function in this case.

It is not difficult to look at an example: if $x= st^2$ (The "s" is just thrown in as a second variable because you specificied "partial derivative". It will be treated as a constant in both derivative and integral.)
Then $\partial x/\partial t= 2t$. Since $x= st^2$, $t= \sqrt{x/s}$. $\int \partial x/\partial t dx= \int \sqrt{x/s} dx= (1/\sqrt{s})\int x^{1/2}dx= (2/3)(1/\sqrt{s})x^{3/2}+ C$.

That is certainly not 0. Could you give the precise statement in your text?

 

[arildno]

Alternatively, HallsofIvy, there are specified CONDITIONS on x(t,...), rather than an explicit formula, such as when invoking those conditions, the integral is, indeed, 0.

 

[crunchynet]

Ok so first off, thank you everyone for the replies, I don't feel so bad anymore about not getting this haha :P.

Secondly, it's become very obvious to me that the author did some serious "hand-waving" here because the book doesn't even attempt to explain how in the world that term goes away.

I will attempt to give more context hopefully without making the situation too complicated...
OK so originally this is what the equation looks like, if you recall I mentioned this is from my physics course in QM.

∫$\Psi^{*}$($\partial x$/$\partial t$) $\Psi$ dxWhere $\Psi$ is the wavefunction. The thing is though, this integral has to be zero somehow because it vanishes from the rest of the explanation! I'm beginning to think there is some serious string pulling going on here and nothing formally defined but I'd like to attempt to get a real explanation either way.

And no x is NOT explicitly defined at all.

 

[arildno]

Ah, well, you missed the whole point, didn't you?
With very weak conditions of x(t), I'm sure the very stringent conditions on the WAVE FUNCTION makes this into a true statement.
I cannot show you how (I can't any QM), but I'm sure others can.

 

[crunchynet]

I'm not so sure about that arildno but I suspect this has nothing to do with the wavefunction. The reason being is because in all other integrals like this in QM, instead of having the (partial dx)/(partial dt) term in the middle, you have something totally different like, for example say, just simply x, by itself... and the integral almost never goes to 0 in those cases. I really think the reason this whole thing goes to zero is because you can assume that $\partial x$/$\partial t$ is essentially 0 through some serious hand waving.

Can anyone clarify please?

Unless I'm missing the whole point COMPLETELY here... which would be bad :(

Ok look at this:

 

[vanhees71]

Argh! From which book is that from?

In quantum mechanics the wave function, here obviously for a one-dimensional problem, is a function of two independent variables $\psi(t,x)$. The position operator in this position representation of the state is
$$\hat{x} \psi(t,x)=x \psi(t,x).$$
The expectation value of $x$ is given by
$$\langle x \rangle=\int_{\mathbb{R}} \mathrm{d} x \psi^*(t,x) \hat{x} \psi(t,x) = \int_{\mathbb{R}} \mathrm{d} x \; x \psi^*(t,x) \psi(t,x).$$
Now you take the derivative of this expectation value with respect to time. Only the wave function depends on time, not the variable $x$ under the integral! Thus the correct statement is simply
$$\frac{\mathrm{d}}{\mathrm{d} t} \langle x \rangle= \int_{\mathbb{R}} \mathrm{d} x \; x [\dot{\psi}^*(t,x) \psi(t,x) + \psi^*(t,x) \dot{\psi}(t,x)].$$
There is no necessity to consider $\partial x/\partial t$ here, because these are two independent variables in this context anyway!

The really important thing is that you can write the latter equation in the form
$$\frac{\mathrm{d}}{\mathrm{d} t} \langle x \rangle=\frac{1}{\mathrm{i} \hbar} \int_{\mathbb{R}} \mathrm{d} x \psi^*(t,x) [\hat{H},\hat{x}] \psi(t,x), \qquad (*)$$
because of the Schrödinger equation
$$\mathrm{i} \hbar \partial_t \psi =\hat{H} \psi$$
and its adjoint
$$-\mathrm{i} \hbar \partial_t \psi^* = \hat{H} \psi^*.$$

Then (*) means that the observable corresponding to the time derivative of the position, i.e., the velocity is represented by
$$\hat{v}=\frac{1}{\mathrm{i} \hbar} [\hat{H},\hat{x}].$$
In this form, (*) known as Ehrenfest's theorem, and it shows some similarity with the canonical formalism in classical mechanics in terms of Poissong brackets!

 

[arildno]

So, it simply reduced to that a function averaged over the full time inteval isn't itself a function of time.
thought it might be something as trivial as that..

 

[crunchynet]

Yeah this is actually from my professor's slides, technically not the book verbatim but this is how she summarized it. The book actually shows something like what you described and I kind of get that though it's a bit tricky to follow.

Anywho, thanks for the replies, I guess the real problem here was my professor trying to be clever but in the end getting everyone confused. Agh this stuff is quite difficult the very first time around... sigh :/

 

[arildno]

Use the Thanks button to vanhees71 if you haven't done it already.

 

[crunchynet]

I just did, thanks for pointing it out :)