Significance of the Exchange Operator commuting with the Hamiltonian

In summary: I guess this is why I brought up the position operator. Its eigenbasis spans L2 Hilbert space.In summary, Griffiths claims that P and H are compatible observables, and hence we can find a complete set of functions that are simultaneous eigenstates of both. That is to say, we can find solutions to the Schrodinger equation that are either symmetric (eigenvalue +1) or antisymmetric (eigenvalue -1) under exchange. However, he claims that the existence of a common eigenbasis is necessary for this to be true.
  • #1
sophiatev
39
4
In an Introduction to Quantum Mechanics by Griffiths (pg. 180), he claims that

"P and H are compatible observables, and hence we can find a complete set of functions that are simultaneous eigenstates of both. That is to say, we can find solutions to the Schrodinger equation that are either symmetric (eigenvalue +1) or antisymmetric (eigenvalue -1) under exchange"

I understand why commuting (or as he calls them, compatible) observables share a common eigenbasis. What I don't see is why P, the exchange operator, and H, the Hamiltonian, need to commute for the second sentence to be true. If P is an observable, then supposedly it is a Hermitian operator whose eigenstates span the L2 Hilbert space. It's true that for some Hermitian operators (like the position one, for example), the eigenstates of the operator do not themselves lie in L2 Hilbert space. That being said, I can imagine other ways of showing that the eigenstates of P lie in L2 Hilbert space and thus constitute valid solutions to Schrodinger's equation. I feel like there's something else he's trying to show here, but I'm not exactly sure what. What is he trying to show by stating the P and H commute?
 
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  • #2
sophiatev said:
What I don't see is why P, the exchange operator, and H, the Hamiltonian, need to commute for the second sentence to be true.

The only eigenvalues of P are +1 and -1. So any state that is an eigenstate of both P and H must have one of those two eigenvalues when operated on by P.
 
  • #3
PeterDonis said:
The only eigenvalues of P are +1 and -1. So any state that is an eigenstate of both P and H must have one of those two eigenvalues when operated on by P.
Sure, I get that much. And I get that since the common eigenbasis is composed of vectors that are eigenvectors of both P and H, every vector in that basis when operated on by P will yield either +1 or -1. But I still don't understand what the significance is of that common eigenbasis existing in the first place.
 
  • #4
sophiatev said:
I still don't understand what the significance is of that common eigenbasis existing in the first place.

What do you mean by "the significance" of the common eigenbasis existing? You say you understand why commuting observables have a common eigenbasis, and you say you understand why any state in the common eigenbasis of P and H has eigenvalue +1 or -1 when operated on by P. What else is there to understand? What's the problem?
 
  • #5
PeterDonis said:
What do you mean by "the significance" of the common eigenbasis existing? You say you understand why commuting observables have a common eigenbasis. What don't you understand? What's the problem?
I don't understand why it was necessary to show the existence of the common eigenbasis to show that
"we can find solutions to the Schrodinger equation that are either symmetric (eigenvalue +1) or antisymmetric (eigenvalue -1) under exchange". Why does this directly follow from the existence of the common eigenbasis? I guess this is why I brought up the position operator. Its eigenbasis spans L2 Hilbert space, and so we can construct any solution to the Schrodinger equation using its eigenvectors. But its eigenvectors don't themselves lie in L2 Hilbert space, so they aren't solutions. There seems to be something significant about the fact that the eigenbasis in this case consists of eigenvectors of H as well. I just don't see the connection between the eigenbasis being composed of eigenvectors of H (and P) and the existence of eigenvectors of P that are solutions to Schrodinger's equation.
 
  • #6
sophiatev said:
I don't understand why it was necessary to show the existence of the common eigenbasis to show that
"we can find solutions to the Schrodinger equation that are either symmetric (eigenvalue +1) or antisymmetric (eigenvalue -1) under exchange".

Is the problem that the phrase "solutions to the Schrodinger equation" was used without qualification, instead of "solutions to the Schrodinger equation that are eigenstates of the Hamiltonian"? If so, I believe that by "Schrodinger equation" here Griffiths means the time-independent Schrodinger equation, i.e., the eigenvalue equation for H. So any solution of this equation is automatically an eigenstate of H.
 
  • #7
sophiatev said:
I guess this is why I brought up the position operator. Its eigenbasis spans L2 Hilbert space

No, it doesn't, because, as you yourself say just a little further on, it's eigenfunctions aren't even in the L2 Hilbert space. You can't span a Hilbert space with functions that aren't even in the space. Some sources will be sloppy and talk as though you can, but that's sloppy and incorrect.
 
  • #8
sophiatev said:
I don't understand why it was necessary to show the existence of the common eigenbasis to show that
"we can find solutions to the Schrodinger equation that are either symmetric (eigenvalue +1) or antisymmetric (eigenvalue -1) under exchange".

How else would you prove this? It can't simply be assumed. The argument using the exchange operator is probably the simplest proof, but you need some sort of proof.
 
  • #9
PeterDonis said:
If so, I believe that by "Schrodinger equation" here Griffiths means the time-independent Schrodinger equation, i.e., the eigenvalue equation for H. So any solution of this equation is automatically an eigenstate of H.
Ahhh, I see. So is the claim that since all the eigenstates of H are solutions to the time-independent Schrodinger, the common eigenbasis necessarily consists of solutions to the (time-independent) Schrodinger equation?
 
  • #10
sophiatev said:
So is the claim that since all the eigenstates of H are solutions to the time-independent Schrodinger, the common eigenbasis necessarily consists of solutions to the (time-independent) Schrodinger equation?

That's the way I'm reading it, yes.
 

Related to Significance of the Exchange Operator commuting with the Hamiltonian

1. What is the significance of the Exchange Operator commuting with the Hamiltonian?

The Exchange Operator commuting with the Hamiltonian means that the two operators share the same eigenstates. This is important in quantum mechanics because it allows for the prediction of the energy levels and behavior of a system.

2. How does the Exchange Operator commute with the Hamiltonian?

The Exchange Operator commutes with the Hamiltonian because it is a symmetry operator, meaning it does not change the state of the system when applied. This is due to the fact that the Hamiltonian, which represents the total energy of the system, is invariant under exchange of identical particles.

3. What is the physical significance of the Exchange Operator?

The Exchange Operator represents the exchange of identical particles in a quantum system. This is a fundamental concept in quantum mechanics, as it allows for the prediction of the behavior of particles and their interactions.

4. How does the Exchange Operator affect the quantum state of a system?

The Exchange Operator does not change the quantum state of a system, as it is a symmetry operator. However, it does affect the energy levels and behavior of the system, as it commutes with the Hamiltonian and shares the same eigenstates.

5. What are some real-world applications of the Exchange Operator commuting with the Hamiltonian?

The Exchange Operator commuting with the Hamiltonian has important applications in fields such as quantum chemistry, where it is used to predict the behavior of electrons in molecules. It also plays a crucial role in the study of superconductivity and other quantum phenomena.

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