Solving the Indefinite Integral for <xp>=<px>*

[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!

 

[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.

 

[bhobba]

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/dp/0521558905/?tag=pfamazon01-20

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

 

[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.

 

[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.

 

[ShayanJ]

If it is neither odd nor even then it is a sum of both (consider Fourier)...the odd term(s) vanishes 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).

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.

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.

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.

 

[bhobba]

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

Because the detail can't be done in a post - not even an overview.

But here is a link - although I don't 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

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

Start with Distribution Theory.

Thanks
Bill

 

[bhobba]

But actually I'm not sure how distributions can be used here. I'm eager to learn.

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

 

[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.

 

[ShayanJ]

"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 is zero.

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.

 

[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?

 

[ShayanJ]

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

 

[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.

 

[ShayanJ]

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.

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.

 

[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.

 

[Chip]

Oh, and I forgot to say 'thanks'!

 

[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).

 

[bhobba]

Perhaps I'm missing something, but the above all looks more complex than I thought would be needed to prove $\langle PX\rangle = \langle XP\rangle^*$.

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

 

[andrewkirk]

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.

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.

 

[bhobba]

What axiom is that?

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

Thanks
Bill

 

[andrewkirk]

never mind - I shouldn't hit send so quickly

 

[bhobba]

My bad - not yours.

Here is how its done in distribution theory. If an operator A is defined on a test space then its defined on its dual (the space of distributions) in the following obvious way <x|(Ay)> = <(xA)|y> - here y is from the test space and x from the space of distributions. So if you prove it on the test space its automatically true on its dual. Proving it on the test space is trivial if you assume its of compact support - although so Fourier transforms apply that's usually extended to so called open support functions because the Fourier transform of an open support function is also and open support function.

To the OP - the above is likely giberish - you need to read the first reference I gave.

Thanks
Bill

 

[Chip]

Thank you everyone for your input. I finally have clarity on this problem. (btw Bhobba: I am going to look into the book you recommended after all, so thank you for the recommendation).

I didn't show all of the process I used to demonstrate that <XP> = <PX>*. I just inquired about one term that I used in that process: the evaluation of |psy|^2 x from minus infinity to infinity. It must equal zero, and I now understand precisely why this is necessarily true as long as psy is square integrable. Under such conditions, this function will always evaluate to zero both from minus infinity to zero, and from zero to infinity. So the answer is always zero given a square integrable psy. In the derivation process I used, which involved basic calculus, this term must be zero to show <XP> = <PX>*.

As far as I'm concerned, my original query is resolved. Thanks again!

 

[Chip]

Re: relating a Dirac Delta to a Gaussian distribution of infinitesimally small standard deviation, I by no means have rigorous proof of the idea...perhaps I am just being "intuitive". In any event, both functions are zero everywhere except infinitesimally large at a single value where both functions integrate to one. That's close enough for me.

 

[stevendaryl]

I just inquired about one term that I used in that process: the evaluation of |psy|^2 x from minus infinity to infinity. It must equal zero, and I now understand precisely why this is necessarily true as long as psy is square integrable.

Sorry for prolonging this thread after it has already started winding down, but what you're claiming doesn't sound correct.

You're saying $\int_{-\infty}^{+\infty} |\psi(x)|^2 x dx = 0$. That is not true, in general. To give a counter-example, Let $\psi(x)$ be defined as follows:

$\psi(x) = A e^{-k (x-a)^2}$

(where $A$ is chosen so that $\int_{-\infty}^{+\infty} |\psi(x)|^2 dx = 1$). then

$\int_{-\infty}^{+\infty} |\psi(x)|^2 x dx = a$, not zero.

 

[Chip]

No, the function is already integrated. Look up to the very first line of the first response (from Shyan) to my original post (I suppose I really ought to learn LaTex).

 

[stevendaryl]

No, the function is already integrated. Look up to the very first line of the first response (from Shyan) to my original post (I suppose I really ought to learn LaTex).

Oh, sorry.

 

[vanhees71]

Isn't this another example for the fact that quite often the Dirac formalism is simpler. We just have to use that $\hat{x}$ and $\hat{p}$ are self-adjoint. Now we define the operator $\hat{A}=\hat{x} \hat{p}$, which is not self-adjoint but obeys $\hat{A}^{\dagger}=\hat{p}^{\dagger} \hat{x}^{\dagger}=\hat{p} \hat{x}$. Now we only have to prove that for any state $\langle \hat{A}^{\dagger} \rangle=\langle A \rangle^*$. This is simple: Any state is represented by a trace-class positive semidefinite self-adjoint operator, the Statistical operator $\hat{R}$, with $\mathrm{Tr} \hat{R}=1$. With a complete set of eigenvectors $|j \rangle$ with eigen values $P_j$ of $\hat{R}$ we then easily conclude
$$\langle \hat{A}^{\dagger} \rangle=\mathrm{Tr}(\hat{R} \hat{A}^{\dagger})=\sum_j P_j \langle j|\hat{A}^{\dagger} j \rangle = \sum_j P_j \langle \hat{A} j|j \rangle = \sum_j P_j \langle j|\hat{A} j \rangle^*=\left (\sum_j P_j \langle j|\hat{A} j \rangle \right)^* = \langle \hat{A} \rangle^{*}.$$
In the pre-last step we have used that $P_j \in \mathbb{R}$.