Invariance of a system under symmetry operations

[sophiatev]

TL;DR Summary

Trying to understand the precise mathematical reason as to why a quantum number being conserved means that a system is invariant under the corresponding symmetry operation

I'm trying to understand the precise reason we claim that a value being conserved means that the system in question is invariant under the corresponding symmetry transformation. Take parity for example. If the parity operator satisfies the commutation relation $[P, H] = 0$ for a given Hamiltonian H, we can show that the expected value of P for a given wave function does not change over time, meaning "parity is conserved". Now from what I understand, if a wave function $\psi$ is an eigenfunction of H that corresponds to a non-degenerate eigenvalue, since $[P, H] = 0$ it must also be an eigenfunction of P. And furthermore, since P is unitary and unitary operators retain the normalization of $\psi$, the action of P on $\psi$ can at most add an overall phase factor of +1 or -1, meaning that its eigenvalues must satisfy $|\lambda|$ = 1. In this specific case, $P\psi = \psi' = \pm \psi$, such that $\psi'$ is physically indistinguishable from $\psi$. In this case, I can see why the system is invariant under a parity transformation, since the corresponding wave function is physically indistinguishable from the original. However, I'm not sure why this would be true in the case of degenerate eigenvalues. If $\psi$ is now an eigenfunction of H that corresponds to a degenerate eigenvalue, since $[P, H] = 0$ it must be composed of a linear combination of eigenfunctions of P, i.e. $\psi = \alpha |a> + \beta |b>$, where $|a>$ and $|b>$ are eigenfunctions of P. So now if P acts on $\psi$, $\psi$ will collapse to either $|a>$ or $|b>$. In this case, wouldn't the resulting wave function be distinguishable from the original? What does the "system being invariant under the action of the parity operator" mean in this case?

 

[fresh_42]

I moved your thread into the QM forum, but it is at its core a topic of classical physics. Only your example makes it QM. So do you know the answer to your question if you cancel the words quantum numbers?

 

[sophiatev]

Sorry, I'm not sure I understand what you mean by "cancel the words quantum numbers". I don't know what the answer to my question would be, since I'm still not clear as to what the precise definition of a system being invariant is. Based on my example, it doesn't seem to mean that applying an operator gives us back a physically indistinguishable wave function. But the textbooks I read seem to imply that a system being invariant means it should not be possible to distinguish the mirror image of a process (post-parity operator) from the original. The two seem to be at odds, and I'm trying to reconcile them.

 

[fresh_42]

Quantum numbers are only one example. Classical examples are momentum or energy and others.

The first question to answer is: What does invariance mean? What is invariant and under which operations? The answers are: The differential equations which describe a system (Lagrangian) are invariant under certain coordinate transformations (depending on the system, e.g. time arrow). This is the mathematical background which has been proven by Emmy Noether a century ago. The physical part is: If something doesn't change, if we change the frame aka coordinates, then it is a conserved quantity, since it doesn't change.

Here is an example: (problem #23)
https://www.physicsforums.com/threads/math-challenge-november-2018.960003/page-3#post-6091091
and here a solution in a pdf: (left most)
https://www.physicsforums.com/threads/solution-manuals-for-the-math-challenges.977057/

Explaining it with mathematical rigor is far more to do. See e.g.
https://www.amazon.com/dp/0387950001/?tag=pfamazon01-20

 

[sophiatev]

The physical part is: If something doesn't change, if we change the frame aka coordinates, then it is a conserved quantity, since it doesn't change.

I guess this is what I'm confused by. It seems like something in my case does change. Our original wave function is $\psi = \alpha |a> + \beta |b>$. We change the frame via P by applying P to $\psi$ via $P \psi = \psi'$. This collapses our wave function to either $|a>$ or $|b>$. Since P and H commute I think that the energy of $\psi$ should be invariant, i.e. $\psi'$ and $\psi$ yield the same energy eigenvalue when acted on by H. (I started out by assuming $\psi$ is an eigenfunction of H). But it seems like $\psi$ did change to an entirely different wave function.

 

[fresh_42]

I have to fold on the QM part, since chances are that I say something wrong. Let's call for Mr. Wolf ... uhm
@Orodruin or others.

 

[Orodruin]

So now if P acts on $\psi$, $\psi$ will collapse to either $|a>$ or $|b>$. In this case, wouldn't the resulting wave function be distinguishable from the original? What does the "system being invariant under the action of the parity operator" mean in this case?

There is a difference between $P$ acting on a state and a measurement of the eigenvalue of $P$ on a state.

 

[PeterDonis]

if P acts on $\psi$, $\psi$ will collapse to either $|a>$ or $|b>$.

No, it won't. If $|a>$ is an eigenfunction of $P$ with eigenvalue $+1$, and $|b>$ is an eigenfunction of $P$ with eigenvalue $-1$ (the only two possibilities), then $P \left( \alpha |a> + \beta |b> \right) = \alpha |a> - \beta |b>$. That is obvious from linearity.

 

[PeterDonis]

What does the "system being invariant under the action of the parity operator" mean in this case?

The system is not invariant under $P$ in this case, since it is by construction a superposition of even and odd parity states. So $P$ applied to this system's state does not give a physically equivalent state, because the even and odd parts are acted on differently.

 

[sophiatev]

There is a difference between $P$ acting on a state and a measurement of the eigenvalue of $P$ on a state.

No, it won't. If $|a>$ is an eigenfunction of $P$ with eigenvalue $+1$, and $|b>$ is an eigenfunction of $P$ with eigenvalue $-1$ (the only two possibilities), then $P \left( \alpha |a> + \beta |b> \right) = \alpha |a> - \beta |b>$. That is obvious from linearity.

Ah, of course, sorry guys. This is indeed obvious, not sure how I confused the two.

The system is not invariant under $P$ in this case, since it is by construction a superposition of even and odd parity states. So $P$ applied to this system's state does not give a physically equivalent state, because the even and odd parts are acted on differently.

Okay, so this seems to confirm my suspicion then. Not sure this is in your wheelhouse, but do you know then why particle physics textbooks claim that the strong/electromagnetic interactions are "parity invariant"? This seems to only be true if every state we're talking about is an eigenstate of P, and this seems to be a stronger condition than just saying that $[P, H] = 0$. And yet Henley and Garcia's Subatomic Physics, chapter 7 seems to claim (though I could be seriously misinterpreting something) that a system being invariant under some transformation is equivalent to saying that the observable associated with that transformation is conserved. But here seems to be an example where $[P, H] = 0$ means parity is conserved in the sense that its expected value doesn't change in time, and yet we can easily find wavefunctions that aren't invariant under a parity transformation.

 

[PeterDonis]

here seems to be an example where $[P, H] = 0$ means parity is conserved in the sense that its expected value doesn't change in time, and yet we can easily find wavefunctions that aren't invariant under a parity transformation.

Yes, we can find wave functions that aren't invariant when the operator $P$ is applied to them; but the expectation value of $P$ is still the same. The expectation value being the same doesn't mean the wave function has to be the same, or even physically indistinguishable; there are other ways of physically distinguishing wave functions besides taking the expectation value of $P$.

 

[sophiatev]

The expectation value being the same doesn't mean the wave function has to be the same, or even physically indistinguishable

Right, I guess this is my whole point. This diagram is an example of what I'm confused about. Henley and Garcia claim that parity invariance means that an electron cannot have an electric dipole moment:

By performing a parity inversion about the midplane, we see that the parity inverted particle has an electric dipole moment oppositely directed relative to the spin of the particle, whereas the magnetic dipole moment remains parallel to the spin. Thus, if parity is conserved, the particle cannot have an electric dipole moment since you can tell the mirror picture from the original one.

But as we established, $[P, H] = 0$ does not mean that that the resulting wave function has to be physically indistinguishable (unless it's an eigenstate of P). So on what grounds are they saying that parity conservation means we shouldn't be able to distinguish the system pre and post parity transformation?

 

[PeterDonis]

as we established, $[P, H] = 0$ does not mean that that the resulting wave function has to be physically indistinguishable (unless it's an eigenstate of P).

More precisely, what we established is that it has to be an eigenstate of both $H$ and $P$, or, equivalently, it has to be an eigenstate of $H$ that is non-degenerate (which is sufficient to show that it must then also be an eigenstate of $P$).

So now the question becomes: what kinds of states of electrons are Henley and Garcia talking about? I don't have their book so I can't check, but in Ballentine, for example, in section 18.1, he presents an argument for why a non-degenerate eigenstate of $H$ for a system with $j$ particles (he doesn't specify what kind of particle) cannot have a nonzero average electric dipole moment (i.e., a nonzero expectation value for the electric dipole moment). The argument is as follows:

First, the electric dipole moment operator is $\mathbf{d} = \Sigma_j q_j \mathbf{Q}_j$, where $q_j$ and $\mathbf{Q}_j$ are the charge and position operator of the $j$th particle. This operator can be shown to have odd parity, i.e, $P \mathbf{d} P^{-1} = - \mathbf{d}$, or, equivalently, $\mathbf{d} P = - P \mathbf{d}$. If we now consider two states, $\psi$ and $\psi^\prime = P \psi$, we find that their expectation values for $\mathbf{d}$ are related by:

$$
\langle \psi | \mathbf{d} | \psi \rangle = \langle \psi | P^{-1} P \mathbf{d} P^{-1} P | \psi \rangle = \langle \psi^\prime | P \mathbf{d} P^{-1} | \psi^\prime \rangle = - \langle \psi^\prime | \mathbf{d} | \psi^\prime \rangle
$$

Then, consider a stationary state $\psi$ of $H$, i.e., $H \psi = E \psi$, where $E$ is a non-degenerate eigenvalue. Assume $H$ is invariant under $P$, i.e., $P H P^{-1} = H$, or, equivalently, $[P, H] = 0$. Then, if $P \psi = \psi^\prime$, we have $H \psi^\prime = H P \psi = P H \psi = P E \psi = E P \psi = E \psi^\prime$. In other words, $\psi^\prime$ is an eigenstate of $H$ with eigenvalue $E$.

But we specified that $E$ is non-degenerate, which means $\psi^\prime$ must be physically the same state as $\psi$, i.e., $\psi^\prime = c \psi$, where $c$ is some complex number. But since we also have $\psi^\prime = P \psi$, if $\psi^\prime$ is physically the same state as $\psi$, then $\psi$ must be an eigenstate of $P$, which means that $c$ must be one of the eigenvalues of $P$, i.e., $c = \pm 1$.

Putting together all of the above, the expectation value of $\mathbf{d}$ for $\psi$ is:

$$
\langle \psi | \mathbf{d} | \psi \rangle = - \langle \psi^\prime | \mathbf{d} | \psi\prime \rangle = - c^2 \langle \psi | \mathbf{d} | \psi \rangle = - \langle \psi | \mathbf{d} | \psi \rangle
$$

This can only be satisfied if $\langle \psi | \mathbf{d} | \psi \rangle = 0$.

The above argument works fine for $j = 1$, so it is sufficient to show that a non-degenerate eigenstate of $H$ for a single electron can't have a nonzero expectation value for $\mathbf{d}$. Notice, though, that this only works if $\psi$ is a non-degenerate eigenstate of $H$, as Ballentine explicitly notes. He gives a counterexample for the degenerate case: the first excited state of the hydrogen atom, $n = 2$, is fourfold degenerate, and linear combinations of the different orbital states can have a nonzero expectation value for $\mathbf{d}$.

Ballentine also remarks right after this that a necessary condition for a state to have a nonzero expectation value for $\mathbf{d}$ is for it to be a linear combination of even and odd parity states. Notice that that is exactly the kind of state we used earlier as a counterexample for the degenerate case, and the reason Ballentine gives for why this kind of state is a counterexample is exactly the reason I gave earlier: the parity operator acts differently on the even and odd components.

 

[sophiatev]

Notice, though, that this only works if ψψ\psi is a non-degenerate eigenstate of HHH, as Ballentine explicitly notes.

Ahh, okay, this makes a lot more sense to me.

More precisely, what we established is that it has to be an eigenstate of both HHH and PPP, or, equivalently, it has to be an eigenstate of HHH that is non-degenerate

So to be physically indistinguishable, we do need it to be an eigenstate of both, as I initially thought.
Thank you so, so much for the extremely thorough walkthrough of this proof, I can't emphasize how much I appreciated it and your help.

 

[PeterDonis]

Thank you so, so much for the extremely thorough walkthrough of this proof, I can't emphasize how much I appreciated it and your help.

You're welcome! I'm glad it was helpful.