Linear combination of states with Pauli's principle

[Salmone]

If I have two identical particles of $1/2$ spin, for Pauli's exclusion principle all physical states must be antysimmetrical under the exchange of the two particles, so $\hat{\Pi}|\alpha\rangle=-|\alpha\rangle$. Now, let's say for example this state $\alpha$ is an Hamiltonian eigenfunction and we have a superposition of different eigenstates, what about the antysimmetric condition? I thought this applies: since the exchange operator $\Pi$ is linear and all the states must be antysimmetric, even a states which is a superposition of eigenstates need to be antysimmetric so it must be a superposition of antysimmetric eigenstates. Example: $|\psi\rangle=\frac{1}{\sqrt{2}}|\alpha_1\rangle+\frac{1}{\sqrt{2}}|\alpha_2\rangle$ I apply the exchange operator: $\Pi|\psi\rangle=\Pi(\frac{1}{\sqrt{2}}|\alpha_1\rangle+\frac{1}{\sqrt{2}}|\alpha_2\rangle)=\Pi\frac{1}{\sqrt{2}}|\alpha_1\rangle+\Pi\frac{1}{\sqrt{2}}|\alpha_2\rangle=-\frac{1}{\sqrt{2}}|\alpha_1\rangle-\frac{1}{\sqrt{2}}|\alpha_2\rangle=-(\frac{1}{\sqrt{2}}|\alpha_1\rangle+\frac{1}{\sqrt{2}}|\alpha_2\rangle)=-|\psi\rangle$ and so it is antysimmetric. Is this right?

 

[PeroK]

More generally, if two states are eigenstates of a given operator with the same eigenvalue then any superposition of those states is also an eigenstate of the operator with the same eigenvalue. In other words, eigenspaces of a linear operator are sub-spaces (i.e. closed under linear vector operations).

 

[Salmone]

More generally, if two states are eigenstates of a given operator with the same eigenvalue then any superposition of those states is also an eigenstate of the operator with the same eigenvalue. In other words, eigenspaces of a linear operator are sub-spaces (i.e. closed under linear vector operations).

Sorry I don't understand, what do you mean with this? And in general, I was referring to superposition even of states that are not eigenstates with the same eigenvalue.

 

[PeroK]

Sorry I don't understand, what do you mean with this?

It's standard QM terminology. What don't you understand?

And in general, I was referring to superposition even of states that are not eigenstates with the same eigenvalue.

All the states you use appear to be antisymmetric (i.e. eigenstates of the exchange operator with eigenvalue $-1$).

 

[Salmone]

It's standard QM terminology. What don't you understand?

All the states you use appear to be antisymmetric (i.e. eigenstates of the exchange operator with eigenvalue $-1$).

Ok, so what I wrote is correct? And in general, what I am supposed to do when dealing with a superposition is to check that every state of the superposition is antisymmetric?

 

[PeroK]

Ok, so what I wrote is correct?

Yes.

And in general, what I am supposed to do when dealing with a superposition is to check that every state of the superposition is antisymmetric?

That doesn't follow. In general, any state can be expressed in any basis. In particular, an antisymmetric state can be expressed in any basis and in a superposition of non-antisysmmetric states.

It's the whole state you need to check.

That said, if every state in the superposition is antisymmetric, then the state itself must be. That's the argument in posts #1 and #2.

 

[Salmone]

Yes.

That doesn't follow. In general, any state can be expressed in any basis. In particular, an antisymmetric state can be expressed in any basis and in a superposition of non-antisysmmetric states.

It's the whole state you need to check.

That said, if every state in the superposition is antisymmetric, then the state itself must be. That's the argument in posts #1 and #2.

Ok I understand. I thought it was impossible to express an antisymmetric state as a superposition of non-antisymmetric states. Lastly, do you confirm that the exchange operator is linear?

 

[PeroK]

Ok I understand. I thought it was impossible to express an antisymmetric state as a superposition of non-antisymmetric states.

For example, we have the anti-symmetric singlet state:$$\ket {00} = \frac 1 {\sqrt 2}\big (\ket {\uparrow \downarrow} - \ket {\downarrow \uparrow} \big )$$In general, the properties of a vector do not apply to the basis vectors. They cannot, as the basis vectors span the space and cannot have every property themselves.

Lastly, do you confirm that the exchange operator is linear?

You should be able to prove that.

In general, "operator" means "linear operator" in QM, unless otherwise stated!

 

[Salmone]

For example, we have the anti-symmetric singlet state:$$\ket {00} = \frac 1 {\sqrt 2}\big (\ket {\uparrow \downarrow} - \ket {\downarrow \uparrow} \big )$$In general, the properties of a vector do not apply to the basis vectors. They cannot, as the basis vectors span the space and cannot have every property themselves.

You should be able to prove that.

In general, "operator" means "linear operator" in QM, unless otherwise stated!

Ok I understand you. But the state $|\uparrow \downarrow\rangle$ doesn't have a defined symmetry under exchange of the two particles, what I mean was: if the state $|\psi\rangle$ is a superposition of states with a well-defined symmetry (symmetric or antisymmetric), if we want $|\psi\rangle$ to be antisymmetric we need that all the states of the superposition are antisymmetric. I can't imagine a counter-example. To be clearer on what I mean: $|\psi\rangle=|\alpha_1^s\rangle+|\alpha_2^s\rangle+|\alpha_3^a\rangle$ with $a,s$ indicating respectively antisymmetric and symmetric states, $|\psi\rangle$ cannot be antisymmetric. I don't know if this is right.

 

[PeroK]

Ok I understand you. But the state $|\uparrow \downarrow\rangle$ doesn't have a defined symmetry under exchange of the two particles, what I mean was: if the state $|\psi\rangle$ is a superposition of states with a well-defined symmetry (symmetric or antisymmetric), if we want $|\psi\rangle$ to be antisymmetric we need that all the states of the superposition are antisymmetric. I can't imagine a counter-example. To be clearer on what I mean: $|\psi\rangle=|\alpha_1^s\rangle+|\alpha_2^s\rangle+|\alpha_3^a\rangle$ with $a,s$ indicating respectively antisymmetric and symmetric states, $|\psi\rangle$ cannot be antisymmetric. I don't know if this is right.

Well, a superposition of symmetric states must be symmetric. But, most states are neither symmetric nor antisymmetric.

You're leading yourself astray somewhat by these arguments. If we have:
$$\ket \psi = \ket {\psi_1} + \ket {\psi_2}$$Then, in general, the properties of $\ket \psi$ are not the same as the properties of $\ket {\psi_1}$ and $\ket {\psi_2}$.

Insisting that if $\ket \psi$ is (anti-)symmetric, then both $\ket{\psi_1}$ and $\ket{\psi_2}$ are (anti-)symmetric is very wrong.