# <xp> = <px>*

1. Sep 25, 2015

### Chip

First of all, I'm new here, so please bear with me if the answer to my question can be found elsewhere, but I have been working a problem and searching for an answer for two weeks now without a complete solution. In Eisberg and Resnick chapter 5, problem 15, an essential part of the problem is to show that the expectation of xp is the complex conjugate of the expectation of px. I have everything worked out except for one small thing:

In order to demonstrate that <xp> = <px>* the following indefinite integral solution must evaluate to zero:
• x*|psy|^2 evaluated from minus infinity to infinity = 0, assuming that psy is square integrable (vanishes to zero at minus infinity and infinity)
Every which way I work that problem I get that this evaluation yields a positive number except for the trivial solution psy(x) = 0. |psy(x)|^2 is nonnegative and then when multiplying by x it is negative at x<0 and positive at x> 0, and equal and opposite. Therefore, x |psy|^2 from minus infinity to infinity = 2 x |psy(x)|^2 evaluated from 0 to infinity. The answer to this can be zero only if psy is everywhere zero.

Another way to get this is to take d/dx of x*|psy(x)|^2 which yields: |psy(x)|^2 +2 x psy(x) d/dx psy(x)
Both of those terms are necessarily even functions and therefore is an even function. So it is not odd and therefore the evaluation of its integral from minus infinity to infinity is zero for psy(x) = 0 only.

I've tried to leave the term in, but only when it is zero can it be shown that <xp> + <px> yields a real result--otherwise you are left with an imaginary part/term.

Obviously I am missing something. The question is what? Thanks in advance for any help!

2. Sep 25, 2015

### ShayanJ

You need to have $\left. x|\psi(x)|^2 \right|_{-\infty}^\infty=0$. This is true because we always assume that our wave-functions vanish in a smooth way at infinity.

Also the function $2x \psi(x) \psi' (x)$ is even only if $\psi(x)$ is either even or odd(and in some exceptional cases where $\psi(x) \psi'(x)$ accidentally becomes odd!). But if $\psi(x)$ is neither even nor odd, then the said function is neither even nor odd. So, in general, its neither even nor odd.

3. Sep 25, 2015

### Staff: Mentor

Shyan is correct.

However the full explanation requires distribution theory which should be in the armoury of any applied mathematician, not just physicists:
https://www.amazon.com/The-Theory-Distributions-Nontechnical-Introduction/dp/0521558905

This also leads into Rigged Hilbert Spaces which is the correct basis for some of the otherwise mathematically 'dubious' things in QM - but first learn distribution theory.

Thanks
Bill

4. Sep 25, 2015

### Chip

If it is neither odd nor even then it is a sum of both (consider Fourier)...any odd terms vanish and the even one(s) does/do not...so we're back to my original explanation.

"This is true because we always assume that our wave-functions vanish in a smooth way at infinity" is neither correct nor complete. If the function is not everywhere zero and has an even component, it is somewhere positive and the evaluation is necessarily positive. The only way this evaluation is zero is for the original function to be odd, which as you explained is not (necessarily), or if psy is trivial.

Bhobba: how about justifying your response rather than just referencing some book?

The solution to the evaluation above is necessarily positive for any non-zero psy. If "distribution theory" changes that you ought to be able to explain why.

Last edited: Sep 26, 2015
5. Sep 25, 2015

### Chip

BTW--I can explain the derivation of the "normal" distribution better than any explanation ever given to me, all of which were flawed. I don't need to read about distribution theory right now...I need a *correct* and *complete* answer to my question.

6. Sep 26, 2015

### ShayanJ

You didn't pay attention. When you integrate by parts, you'll get two integrals one of which you want to get rid of. This integral is $\int_{-\infty}^\infty d(x|\psi(x)|^2)=\left. x|\psi(x)|^2 \right|_{-\infty}^\infty=\displaystyle \lim_{a\to \infty} a \left[ \psi(a) \psi^*(a)-\psi(-a)\psi^*(-a) \right]$.
It doesn't matter what the function does in between, its only about its behaviour at infinity and the conditions on the wave-function assure us that the limit is zero.

bhobba didn't mean probability distributions. See here!
But actually I'm not sure how distributions can be used here. I'm eager to learn.

7. Sep 26, 2015

### Staff: Mentor

Because the detail cant be done in a post - not even an overview.

But here is a link - although I dont think its as good as that book which I have a copy of and is very good:
http://www.mat.univie.ac.at/~stein/lehre/SoSem09/distrvo.pdf

Its written in the language of pure mathematics - my suggested text is less formal - but just as rigorous. That's a very unusual combination - perfect for those into applied rather than pure math.

Here is a link on Rigged Hilbert Spaces:
http://physics.lamar.edu/rafa/webdis.pdf [Broken]

But its very advanced and you need a good deal of background in functional analysis.

Thanks
Bill

Last edited by a moderator: May 7, 2017
8. Sep 26, 2015

### Staff: Mentor

Its how you can make sense of such integrals even for functions that do not behave well at infinity and things that aren't really functions at all eg the Dirac Delta Function.

Distribution Theory is not the theory of probability distributions - its something else entirely - although has application to probability theory eg see Bochners Theorem you will find in the book I suggested.

Thanks
Bill

9. Sep 26, 2015

### Chip

"You didn't pay attention"...I did pay attention. Your limit evaluates in the best case (again, when the original wave function is odd only) to zero times infinity which is undetermined, not zero. You're not making things better, but worse.

I understood the integration by parts. It is the one term that remains a problem. It is zero only when psy everywhere zero.

10. Sep 26, 2015

### ShayanJ

Yes, its $\infty \times 0$ and its a indeterminate form. But it doesn't mean it doesn't have a value, it just means it can have any value and you should calculate the value for the particular problem at hand using the properties of the function. For example if the wave-function varies like $\frac 1 x$ at infinity, then its norm squared varies like $\frac{1}{x^2}$ at infinity and if you multiply it by x and take the limit at infinity, you'll have $\displaystyle \lim_{x\to \infty} \frac 1 x$ which is zero. This is the sort of condition I'm talking about. I just don't remember the exact conditions and can't find them anywhere.

11. Sep 26, 2015

### Chip

Dirac Delta makes perfect sense to me. It is not meaningfully different from a normal distribution with a standard deviation that approaches zero.

Let's try a different approach: how about someone provide a concrete psy function for which the said term above evaluates to zero? You don't need to refer me to multiple books to do that do you?

Last edited: Sep 26, 2015
12. Sep 26, 2015

### ShayanJ

Take the ground state of the harmonic oscillator $\psi(x) \propto e^{-ax^2}$. You can check that $\displaystyle \lim_{x\to\pm\infty} x\exp(-ax^2)=0$.

13. Sep 26, 2015

### Chip

For that ground state harmonic oscillator example...x times the square of that function (which is proportional to x e^-2ax^2) is odd...and when that is the integral solution and is evaluated from minus infinity to infinity will be positive..back to my original point.

14. Sep 26, 2015

### ShayanJ

You're not paying attention. You don't need to integrate anything because the integrand is an exact differential. So you just need to evaluate it at the boundaries. Which means you should calculate $\displaystyle \lim_{x\to \infty} 2 x e^{-2 a x^2}$ which is equal to zero.

15. Sep 26, 2015

### Chip

You're right...I wasn't paying attention that time. Indeed that limit evaluates to zero, and the function is also zero at x = 0. It seems that in this case (with a Gaussian integral) it works because of the square of the independent variable in the exponential (and of course the negative exponential). I guess wave functions always take on forms such that it works out this way? I'm not sure I could have ever gotten that out of chap 5 of Eisberg and Resnick.

16. Sep 26, 2015

### Chip

Oh, and I forgot to say 'thanks'!

17. Sep 26, 2015

### andrewkirk

Perhaps I'm missing something, but the above all looks more complex than I thought would be needed to prove $\langle PX\rangle = \langle XP\rangle^*$. Ground states and quantum harmonic oscillators are much more advanced (at least, in the order in which I learned QM) than the concepts needed to do the proof.

Are we allowed to use the following?
\begin{align*}
&\langle \Omega \rangle =\langle \psi | \Omega | \psi \rangle\\
&P:f(p)\to pf(p)\textrm{, as shorthand for }\big(\forall p\in\mathbb{R}:\langle p|\psi\rangle=f(p)\big)\Rightarrow \big(\forall p\in\mathbb{R}:\langle p|P|\psi\rangle=pf(p)\big)\\
&P:f(x)\to -i\hbar\frac{df(x)}{dx}\\
&X:f(x)\to xf(x)\\
&X:f(p)\to i\hbar\frac{df(p)}{dp}\\
&\langle x|p\rangle = \frac{e^{i\hbar xp}}{\sqrt{2\pi}}\\
&(ab)^*=a^*b^*\\
&\int_{-\infty}^\infty\delta(x-c)f(x)dx=f(c)
\end{align*}
If so, the proof is a pretty straightforward sequence of steps, and not very long, and the only integration needed is of the type
$$\int_{-\infty}^\infty \delta(x'-x)x'e^{i\hbar p x'}dx'$$

or is justifying one of these formulas what the above discussion is about?

Also, it doesn't make sense to me to say that the Dirac Delta is basically just like a normal distribution. It's not even a distribution in the way that word is used for the normal distribution. Comparisons of the Dirac Delta to a shrinking normal distribution are intuition pumps only, and have no mathematical validity (or at least I've never seen them presented with any mathematical rigour).

18. Sep 26, 2015

### Staff: Mentor

There is no need to prove it at all because its an axiom of an inner product vector space which is the assumed QM space.

Its only done in beginner texts in developing the ideas.

At that stage simply assume the functions you are dealing with are zero at some very large positive and negative value. These are the only physically realisable ones anyway.

The difficulty arises when you want to deal with functions that are not that nice. They are not physically realisable but are introduced for mathematical convenience eg the eigenstates of momentum used in the above:
http://quantummechanics.ucsd.edu/ph130a/130_notes/node138.html

It most definitely is not well behaved at infinity, such integrals do not exist. To make sense of this you need distribution theory and at a more advanced level Rigged Hilbert Spaces.

To the OP do not get caught up with this to start with - simply assume they are approximations to a function that is physically realisable and are zero at large values. Later look into distribution theory to see mathematically how it is resolved.

The math above uses distribution theory which a beginning student may not know. The Dirac Delta Function is a distribution as the term is defined in distribution theory - its also called a generalised function. It can be handled with absolute rigour - which is done in the textbook I mentioned in my first post. As I said there every applied mathematician needs to know it.

Thanks
Bill

Last edited: Sep 26, 2015
19. Sep 26, 2015

### andrewkirk

What axiom is that?
Are you thinking of the axiom that $\langle \psi|\theta\rangle = \langle \theta|\psi\rangle^*$?

That is not the same as $\langle PX\rangle = \langle XP\rangle^*$ which is, written out more fully:
$$\langle \psi|PX|\psi\rangle = \langle \psi|XP|\psi\rangle^*$$

Indeed, I would have thought that for arbitrary linear operators $\Omega,\Lambda$, it is not necessarily the case that
$\langle \Omega\Lambda\rangle = \langle \Lambda\Omega\rangle^*$. I got the feeling that my proof implicitly relied on the fact that $P$ and $X$ are conjugate, although I may have misled myself on that and would be happy to be corrected.

Edited addition: I'm assuming here that what the OP wants to prove is that $\langle PX\rangle = \langle XP\rangle^*$, not just that $\langle p|x\rangle = \langle x|p\rangle^*$. In the latter case it would indeed be just instancing an axiom. The lack of formatting in the OP makes it hard to be sure what the question is. But the references to 'psy' (psi?) make me lean towards guessing the former rather than the latter.

Last edited: Sep 26, 2015
20. Sep 26, 2015

### Staff: Mentor

You are right - its different. Your proof is correct in distribution theory.

Thanks
Bill