Resulting state vector interpretation

[Dale]

If we have an operator A which operates on some state described by vector x the result is a new vector y
A |x> = |y>

My question is: is the new vector y considered to be a different state vector in the same vector space as x or is it considered to be a vector in an entirely different vector space? If the former then what does that new vector represent (I would assume a new state of the system after the operator)? If the latter then what does that new vector space represent?

 

[meopemuk]

If we have an operator A which operates on some state described by vector x the result is a new vector y
A |x> = |y>

My question is: is the new vector y considered to be a different state vector in the same vector space as x or is it considered to be a vector in an entirely different vector space? If the former then what does that new vector represent (I would assume a new state of the system after the operator)? If the latter then what does that new vector space represent?

The answer depends on the type of operator A.

If A is a Hermitian operator of observable (e.g., A = H, the Hamiltonian), then vector |y> has no physical intepretation.

However, if A is a unitary operator, then physical interpretation is available. For example, if A = exp(-iHt) = the operator of time translation, then |y> = exp(-iHt)|x> is the state vector of the system after time t has elapsed.

 

[Civilized]

If A is a Hermitian operator of observable (e.g., A = H, the Hamiltonian), then [in general the] vector |y> has no physical intepretation [at least none which follow directly from the standard postulates of quantum mechanics].

I filled in the [] phrases because it is a frequently used fact that if |0> and |1> are the spin down and spin up eigenstates of the pauli operator Z, then X |0> = |1> and X |1> = |0> where X is the pauli x operator. This is more than a coincidence and it plays a big role in e.g. quantum spin chains, but it follows more from the nature of the su(2) algebra spanned by the pauli operators than it does from QM itself.

 

[Fredrik]

My question is: is the new vector y considered to be a different state vector in the same vector space as x or is it considered to be a vector in an entirely different vector space?

It's in the same vector space. The only exception I can think of is creation and annihilation operators, which change the number of particles. And even in that case, it's a matter of how you look at it. Both the |x> and the |y> are in the Fock space, but |x> is in the subspace of n-particle states while |y> is in the subspace of (n+1)-particle states (or n-1).

 

[Dale]

Thanks! I found these very helpful.