# Resulting state vector interpretation

1. Jun 23, 2009

### Staff: Mentor

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?

2. Jun 23, 2009

### meopemuk

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.

3. Jun 23, 2009

### Civilized

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.

4. Jun 23, 2009

### Fredrik

Staff Emeritus
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).

Last edited: Jun 23, 2009
5. Jun 23, 2009

### Staff: Mentor

Thanks! I found these very helpful.

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