# Probability density that "seems" complex instead of real

1. Aug 23, 2014

### fluidistic

1. The problem statement, all variables and given/known data
Hello guys,
I am extremely worried because I do not understand something. My statistical mechanics course somehow follows Reichl's book for some parts.
For a system of N particles, Reichl's define $\rho (\vec X^N, t)$ as the density of probability on the phase space. $\vec X^N$ is a point in a 6N dimensional phase space. (page 288).
A few pages later he gets that $\rho (\vec X^N, t)=e^{-i\hat L ^N t}\rho(\vec X ^N ,0)$.
Then my lecture notes go on and skipping several steps they reach that $\rho (\vec X^N, t)=\sum _{j=1}^N e^{-i \lambda _j t} c_j f_j (\vec X ^N)$ where the lambda _j's are the eigenvalues of the Liouville's operator and the f_j's are the eigenvectors of such operator. c_j's are constants (could be complex numbers I suppose).
But when I look at that expression I cannot see how rho can be real.
I took the special case of N=2 and wrote the first 2 terms of the series.
2. Relevant equations

Euler's equation.

3. The attempt at a solution
I believe that for rho to be real for all t, every single term of the series must be real but I'm not 100% sure.
So, by using N=2 I got the condition that $r_1 \sin (\theta _1 -t \lambda _1)=r_2 \sin (\theta _2 -t \lambda _2)$ but in my opinion that is impossible to be satisfied for all t's. Mainly because every lambda is different and so the 2 sines have a different frequency...
Anyway I also had a look a Fourier expansion of a function but in general the series starts from -j up to j in such a way that it's easy to see that all complex terms vanish. And when the series starts from j=1, everything is real. So overall there is no ambiguity that the function is real.
But in my case it's different... and I see no way how rho could be real.

Any thoughts will be appreciated. This boggles me a lot.

2. Aug 23, 2014

### AlephZero

I don't know anything about statistical mechanics, but one way the expression could be real is if all the L's are pure imaginary (i.e. their real parts are zero).

That sort of thing can certainly happen in other branches of applied math. For example the "elementary" formulation of simple harmonic motion gives you the differential equation $\ddot x + \omega^2 x= 0$ where $\omega$ is the real natural frequency. But if you forget about what $\omega$ means physically and just consider it as an arbitrary differential equation with constant coefficients, you would take the eigenvalues as $\lambda_1 = i \omega$ and $\lambda_2 = -i\omega$.

Another possibility is that your function $\rho$ really is complex, but when you calculate some physically meaningful quantity, the expression will always contain something like $\rho^*\rho$, which will always be real.

Last edited: Aug 23, 2014
3. Aug 23, 2014

### Staff: Mentor

I would expect this option.

4. Aug 23, 2014

### fluidistic

Thanks AlephZero. Unfortunately all the lamba's are strictly real according to my lecture notes.
It's because they are the eigenvalues of a Hermitian operator (Liouville's operator) which is defined as $\hat L ^N=-i\hat H ^N$ where $\hat H ^N$ is the Hamiltonian of the N particles. (Reichl's page 292).
What I had in mind was to associate rho with |psi|^2 in QM, because both are supposed to be densities of probability and so real valued. So I'll never get $\rho^*\rho$.
Now if rho is complex valued then I completely lose faith in my lecture notes.
Because the conclusion of the notes is that a classical system will "oscillates" forever.

5. Aug 23, 2014

### fluidistic

A quote from the book:

6. Aug 23, 2014

### AlephZero

Another option would be that he really means the probability is the real part of $\rho$.

Again by analogy with solving ODE's, if the coefficients in a linear ODE are real, the general form of the solution comes out naturally as a complex function, but the real and imaginary parts of it both satisfy the ODE individually. For the simple harmonic motion example again, you can take the general solution as $A \cos \omega t + B \sin \omega t$ where A and B are real, or as the real part of $C e^{i\omega t} + D e^{-i\omega t}$ where C and D are complex. Most of the math is simpler using the complex form.

7. Aug 23, 2014

### fluidistic

Well I think Reichl really means density of probability and I have no problem in seeing that rho is real valued if I look at the expressions in the book. The problem arises in my lecture notes.
Both the book and my lecture notes reach the equation $\rho (\vec X ^N, t)=\exp (-it \hat L ^N ) \rho (\vec X ^N ,0)$. Here all is fine, because $\hat L ^N$ is worth $-i \hat H$ so that the exponential is real.
Here my lecture notes depart from the book. They state that $\hat L^N$ is Hermitian (the book also states this) and that $\hat L^N f_j=\lambda _j f_j$ and the fact that the operator is Hermitian implies that the lambda are real. So far I trust it.
Then it's stated that the $f_j$ form an orthogonal basis in the Hilbert space "where rho can be defined". He then writes $\rho (\vec X ^N ,0)= \sum _{j=1}^N c_j f_j(\vec X^N)$. Therefore $\hat L \rho (\vec X ^N,0)=\sum _{j=1}^N c_j \lambda _j f_j(\vec X ^N)$ which yields $\rho (\vec X^N ,t)=\sum _{j=1}^N \exp (-it \lambda _j)c_j f_j(\vec X ^N)$ which is the expression I'm having troubles with.
I am not sure where the sloppy step is.
Is it the step where he assumed that the f_j's form a basis in a Hilbert space which would imply that rho is complex valued?
Seems like the notes were dealing with classical mechanics but suddenly switched to quantum mechanics while Reichl's book stayed with classical mechanics.

8. Aug 24, 2014

### nrqed

The hamiltonian is hermitian so how can $\hat L ^N$ be hermitian? Are you sure that $\hat L ^N$ is not actually anti hermitian?
Well, your expression $\rho (\vec X^N ,t)$ is a wave function so there is no reason it should be real. Then it does sound like the actual expression for the probability density should be the modulus squared of this.

9. Aug 24, 2014

### fluidistic

99% sure it's Hermitian. Not only Reichl states it at page 292, but it's given as an exercise in my course, to show that it's Hermitian. I haven't solved the problem yet, haven't really thought how to approach it yet. That's why I'm not 100% sure.

Yes. I have the feeling that my prof. initially associated rho to |psi|^2 first, just like the book does. However when my prof. departed from the book by saying that rho could be defined in a Hilbert space he slipped because the way he wrote rho, it can be associated with psi in QM and not with |psi|^2. So in a way he wasn't consistant with the definition of rho which was supposed to be a density of probability.
His conclusion that a classical system will come back to its original state is still right though, but the way he showed it isn't, I believe. At least that's the impression I get by using Reichl's book.

Last edited: Aug 24, 2014
10. Aug 24, 2014

### Telemachus

You have defined $\hat L^N=-i\hat H^N$, and you have $\rho( X^N, t)=e^{-i\hat L^N}\rho( X^N, 0)$, if you replace the definition for the Liouville operator, you get a real exponential with the Hamiltonian acting on it. It is not imaginary, the probability density decays exponentially.

Edit. I'm sorry, I see that was not the case.

Last edited: Aug 24, 2014
11. Aug 24, 2014

### Telemachus

If you have that $\displaystyle \hat H^N |\psi >=E^N | \psi >$, then: $\displaystyle e^{\hat H^N} | \psi >=e^{E^N} | \psi >$ so I think that what you must have in the exponential should be the real eigenvalues. But it is confusing, because if you think in the eigenvalues of the Liouville operator instead of the eigenvalues of the Hamiltonian it looks like you would have something complex in the exponential. If you think of it as an expansion in the hamiltonian eigenfunctions instead of the Liouville eigenfunctions, then it is clear that the exponential terms it has to be real for that expansion.

I'm not sure what happens when you express this in the basis of the Liouville operator eigenfunctions, I think that you should try working with the equality between the Hamiltonian and the Liouville operator, and trying to go from one base of eigenfuctions to the other, because it is clear to me that in the base of the Hamiltonian eigenfunction that exponential will be real naturally. Perhaps you get something real when you have the explicit form for the eigenfunctions, or the c_j's. But it is clear from the definition, or making the expansion in the basis of eigenfunctions for the Hamiltonian that rho is real.

Last edited: Aug 24, 2014
12. Aug 24, 2014

### nrqed

Ok, but then how do you reconcile this with the statement

$\hat L ^N=-i\hat H ^N$ where $\hat H ^N$ is the Hamiltonian ?

It sounds like your conclusion is correct. Hopefully you can ask him so that he can clarify the situation for the class.

13. Aug 25, 2014

### fluidistic

I do not think I'll ever get something real for all t. Because the only dependence on t is in the exponentials. The eigenfunctions of the Liouville's operator (at least as written in my notes) do not depend on time.

I've no idea. Apparently here's a proof of the Hermiticity: http://www.nyu.edu/classes/tuckerman/mol.dyn/lectures/lecture_9/node2.html.

Yeah I will try...

14. Aug 25, 2014

### Telemachus

But by definition $\rho$ was a real valued function, and that should hold for the series expansion. I've studied from Reichl for Statistical Mechanics too, and I followed most of the deductions, and examples. I didn't noted this before, but what I can tell you is that I've found some mistakes in the book. I can't tell you right now where are those mistakes because I gave my notes to a friend of mine, it was silly mistakes anyway, but its a possibility that there is a mistake in the definition of the Liouville operator. As nrqed showed, it doesn't look like an hermitian operator, because H is hermitian, and when you multiply it by i, you get complex numbers in the diagonal, right?

15. Aug 25, 2014

### fluidistic

Yeah now that I look more into the book I'm getting more confused. I think I'll just asume for now that the probability density of a classical system will oscillate in time due to the Liouville's equation.

16. Aug 25, 2014

### nrqed

Ok, thanks. Now I see what this L is. But then it cannot be equal to i times the Hamiltonian.
Where did you see this identification ($L =- i H$)? It is not in that derivation. Is it in the book? In the class notes?

17. Aug 25, 2014

### fluidistic

Both in my lecture notes and the book (page 292).

Edit: Hmm actually he uses $\hat H^N$ and then a curved hat H and it's extremely confusing to me what he's doing/defining on page 291.
He speaks about a Poisson bracket for what seems to be a sum of Poisson brackets and to me it seems he mixes H for the curved H.

Last edited: Aug 25, 2014
18. Aug 25, 2014

### nrqed

Hi again,
I managed to get my hands on a copy of Reichl. But on pages 291 and 292, he is doing classical mechanics, not quantum mechanics. His H is the operator that implements the Poisson bracket. Can you tell me where in the book he generalizes these results to quantum mechanics?

19. Aug 25, 2014

### fluidistic

Ok nice.
Well the part of the book I was reading never generalized this to quantum mechanics (as I said at the end of post #7), he stayed with Classical Mechanics.

I had a feeling I could make an analogy with Reichl's $\rho(\vec X^N , t)$ to the quantum mechanics $|\Psi (x,t)|^2$ because both are supposed to be densities of probability and therefore real.
However the way my prof. wrote down $\rho(\vec X^N , t)$, it seems it could be associated to $\Psi(x,t)$ instead of its modulus squared, right after he said that he could define rho into a Hilbert space.

20. Aug 26, 2014

### fluidistic

No wonder I was confused on page 291 about H and the curved H (http://librarum.org/book/10896/309 [Broken]).
The book I own in my hands is a different edition (1st, year 1980) and that page is page 192 and he's consistantly using the curved H, unlike the newer version which mixes both H confusingly to me!

Anyway if the operator L is Hermitian, it must have real eigenvalues and even what Reichl gets seems complex valued now to me: $\rho (\vec X^N, t)=e^{-i\hat L ^N t}\rho(\vec X ^N ,0)$. You might tell me to replace L by -iH but I won't do that. I'll keep it as the book did. The argument used by the book regarding this expression is that it oscillates in time because L is Hermitian and has real eigenvalues. He is clearly saying that the probability density will oscillate in time unlike the Fokker-Planck equation.

So I am asking, as a student, is rho really a density of probability?

Last edited by a moderator: May 6, 2017