# I The Wavefunction and eigenstates

1. Nov 14, 2016

### Higgsono

Suppose I want to measure the momentum of a quantum system. What I do is I take the momentum operator and expand my wavefunction in term of the eigenfunctions of that operator, then I operate on the wavefunction with the operator and the reusult of the measurment is that the wavefunction "collapses" into one of the eigenstates.

What if instead I want to mesure the energy? Then I do the same thing, I expand the wavefunction in eigenstates of energy and apply the energy operator.

My question.
Why can we write the wavefunction as a sum of eigenstates in two diffrent ways depending on what we want to measure? What if I want to measure both energy and momentum at the same time, how do I write the wavefunction then? And If I expand the wavefunction in eigenstates of the momentum operator where do the information about the energy of the system go? I mean, allt he information about the system should be encoded in the wavefunction. But if I expand it in its eigenstates of momentum only, then I can only gain information about the momentum of the system or what?

2. Nov 14, 2016

### BvU

[caveat]this is a student answer. For an expert answer you need orodruin or vela or vanhees71 or one of many others

You can develop any function in eigenfunctions, so an eigenstate of E can be developed in terms of p eigenfunctions and vice versa. From there you get either an exact value for the other variable (e.g. for a free particle) or a distribution of probabilities.

Depending on the commutator it can possible to find a simultaneous set of eigenfunctions ( when [a,b] = 0 ), so when you measure p of a free particle, you already have the energy ( from $p^2/(2m)$ ).

3. Nov 14, 2016

### Higgsono

Ok, so if I have two degrees of freedom A and B of my system and the operators corresponding to those variables commute is it true then that I can write my wavefunction as a product of two wavefunctions where one of them is written as a sum of eigenstates correspoding to the operator A and the other one as a sum of eigenstates corresponding to operator B?

Then if I make a measurment corresponding to the observable A, I will have a wavefunction with an eigenstate of definite eigenvalue corresponding to the operator A times the remaining part of the wavefunction which is still expanded in a sum of eigenstates corresponding to observable B. If I make a measurment again corresponding to observable B, what remains of the wavefunction is a product of two eigenstates only.

4. Nov 14, 2016

### BvU

In the free particle example the eigenvalue for the p operator of the p eigenstate $\psi(x) = e^{ikx}$ is $\hbar k$ and the eigenvale for E is $\ {\hbar^2k^2\over 2m}$.

Your question: I don't think so (product of ...) . That sounds more like separation of variables (e.g. to go from time-dependent SE to time independent).

Instead, however, you can find a set of eigenfunctions that are eigenfunctions of both operators simultaneously.

5. Nov 14, 2016

### Higgsono

But doesn't that imply that if I know the value of one of the observables that I simultaneusly know the value of the other without doing another experiment? Because if I make a measurment on one of the variables I will then automatically have the eigenstate corresponding to the other one.

6. Nov 14, 2016

### BvU

In the free particle example that is indeed the case. <H> is a function of <p> and vice versa.

I don't think you can generalize: In e.g. the hydrogen atom wave functions for the electron you can measure $n$ (the energy) but then you don't know $\ell$ (angular momentum magnitude) or $\ell_z$ (angular momentum z-component). There you do have your separation of variables (so I have to recant part of my #4 ) as you mentioned in #3.

when I google "physical meaning of zero commutator" most mention that both quantities can be measured with arbitrary precision, that they can be measured in arbitrary order in time, and such things.

7. Nov 14, 2016

### Higgsono

I think there would be a difference between measuring the observables simultaneusly at the same time or measure them at different times but still after you make the second measurment you know both quantities to arbitrary precision. If you need two measurment to determine the variables doesn't that mean that you have to write your wavefunction as a product of two by separating the variables? Because when you make the first measurment of variable A we still don't know B and therefore the wavefunction we have now is still a in a superposition of the eigenstates of the observable B. To know B we make another experiment, and now we know both A and B.

I'm not sure that this is how it works though. =p

8. Nov 14, 2016

### BvU

Why ? After measurement $a$ the thing is in an eigenstate of $a$ which is also an eigenstate of $b$. If nothing else happens, it stays there.

In the hydrogen case the eigenstates (except $n =0$) are degenerate: a bunch of states has the same energy ( $\ell$ can take on values from 0 to $n$ and $\ell_z$ from $-\ell$ to $\ell$ ). (property of the spherical harmonics) So you indeed need further measurements to determine them. And yes, they are factors in the wavefunctions. However, I don't think that this is necessarily so (but can't offhandedly quote a counter example where $a$ doesn't tell you $b$ and the wave function is not a simple product). Let's ask @Orodruin ...

9. Nov 14, 2016

### Orodruin

Staff Emeritus
Your state lives in a Hilbert space and in general you will be able to use several different bases to describe that state. I suggest starting to look at it in a finite-dimensional setting, e.g., a two- or three-level system. Many of the properties will generalise to Hilbert spaces and operators with a continuous spectrum.

In the case of a two- or three-level system, the operators can be represented by matrices and you can write down the state in terms of the eigenvectors of either matrix. The idea in the general case is the same.

It really does not matter what basis you write things down in. The only thing going to the eigenbasis of the quantity you want to measure does for you is to simplify your computations. Again, I suggest doing this in a two- or three-level system before going to the continuous case.

You can always write the wave function in whatever eigenbasis you want - it is still going to be the same function. Just as a vector is always going to be the same regardless of what basis you chose to work in.

See above. It is still the same state, just expanded in a different basis.

10. Nov 16, 2016

### peregoudov

Higgsono,
you are right that if A and B can be measured simultaneously but only A is really measured then the value of B is not necessarily defined but may remain uncertain. But you have misleading concept that wavefunction should therefore factorize somehow.

Let's take, say, $Y_{11}=i\sqrt{3/8\pi}\,\sin\theta\,e^{i\phi}$ which is eigenfunction of (commuting)
$$L^2=-\frac1{\sin\theta}\frac\partial{\partial\theta}\sin\theta\frac\partial{\partial\theta}-\frac1{\sin^2\theta}\frac{\partial^2}{\partial\phi^2}$$
and $L_z=-i\partial/\partial\phi$ with eigenvalues of $l(l+1)=1\cdot(1+1)$ and $m=1$ correspondingly. Then, if you omit $\theta$ dependence $e^{i\phi}$ is still eigenfunction of $L_z$, but if you omit $\phi$ dependence $\sin\theta$ is not eigenfunction of $L^2$.

Let start with the beginning. The common way to analize a wavefunction is to represent it as a linear combination of some basic states. Basic states should 1) be orthogonal, 2) form a full set, i. e. any wavefunction can be expressed as their linear combination.

It is often convenient to choose eigenstates of some variable A as basic states. Then you can distinguish basic states with eigenvalues $a$ of variable A. It may happen however that several eigenstates of A have the same eigenvalue (degeneracy). Then eigenvalue of A alone is not sufficient to distinguish basic states.

Now we consider the second variable B, commuting with A, so they have common eigenstates. It may happen that eigenstates of A with degenerate eigenvalue $a$ all have different eigenvalues of B. Then we are able to distinguish basic states by eigenvalues of A and B.

But it may also happen that degeneracy (possibly reduced) remains and we still have several eigenstates with the same eigenvalues $a$ and $b$. I guess by now you know what to do: we consider the third variable C, commuting with A and B. On this way we finally come to complete set of variables A, B, C, ... such that either unique eigenstate exists for given set of eigenvalues $a$, $b$, $c$, ... or no state at all. For example, if we consider the point on sphere, $L^2$ and $L_z$ form the complete set: for any $l\geq0$ and $-l\leq m\leq l$ there exists unique eigenstate of $L^2$ and $L_z$ with eigenvalues $l(l+1)$ and $m$, and no state exists for, say, $l=1$, $m=5$.

What happens when we measure A and find particular $a$ eigenvalue? Wavefunction reduces to subspace of $a$. If eigenstate with $a$ is unique then values of B, C, ... are also known. But If $a$ is degenerate then wavefunction still remains linear combination of eigenstates of, say, B and the value of B is uncertain. Further measurement of B reduces wavefunction to subspace of particular $b$ eigenvalue. Value of B becomes known but in the case of degeneracy C, ... are still uncertain. And so on.

You see no factorization of wavefunction is needed.

11. Nov 16, 2016

### BvU

[I'm in my student role] Thanks, peregoudov, that's indeed what I was searching for.

And now I 'm inclined to agree with Higgsono post #7: Measuring $n$ first leaves you with a probability distribution of e.g. $\ell$. And measuring $\ell$ first and then $n$ would then be a different process.

Or wouldn't it ? The joint $n, \ell$ probability distribution would still come out the same, whatever pathological state you begin with.

12. Nov 16, 2016

### Staff: Mentor

If you want to measure two things, A and B, at the same time, then you need to build an observable O from which you will be able to deduce the two things you want to measure. The observable O therefore has to commute with both A and B and there must be no degeneracy of the eigenvalues of O with respect to those of A and B (each eigenvalue of O must be associated with only one eigenvalue of A and one of B), and I think this is only possible if A and B commute. In that case, then the order of measurement would be irrelevant.

If A and B don't commute, then I do not think it is possible to construct an observable O that can measure both simultaneously.

13. Nov 16, 2016

### Staff: Mentor

This is correct, it isn't.

14. Jan 31, 2017

### peregoudov

Sorry for bringing up an old topic. Yes, intermediate state is different depending on what you measure first. But it is nothing more than usual conditional probability. Final probabilities are the same whatever sequence of measurements you choose. In fact if all observalbles you are working with commute there is no difference between quantum mechanics and (classical) probability theory.