Difference between operators used to represent quantum gates vs that to represent physical observables?

Question

I have learnt that informations about a physical observable property is buried in the state vector of a quantum system. To get the possible value of a property all we need to do is multiply the state with the particular matrix operator. If it yields to an eigenvalue-eigenvector equation then we can tell that the value of the particular property is the eigenvalue. If not (i.e it does not yield to an eigen equation), then we say that the list if possible values on measuring the particular property will be the set of all basis vectors corresponding to that operator with probabilities P^2 where P is the projection the state vector into the corresponding basis vector. In this case, (if I understand it correctly) the matrix multiplied by a vector equation is of absolutely no use. (please correct me if I am wrong)

But under no circumstance, does multiplying by an operator change the state vector.

But as I am learning quantum gates now, I see that even the gates are represented by operators (or matrices). But they do change the state of the quantum system. Obviously both of them are not same(since the former does not change the state). But we use matrices for both of them!

Could you please explain the difference and similarity(if any) between these two types of matrices?

Answer 1

In the standard (textbook) formulation of Quantum Mechanics, observables quantities are modelled by means of Hermitean operators on the Hilbert space $\mathcal{H}$ associated with the system under investigation, while (pure) states are identified with normalized vectors in the Hilbert space. Then, given an observable $\mathbf{A}$ and a state $|\psi\rangle$, we pair them together in order to give a real number $\omega(\mathbf{A},\psi)$ that may be interpreted as the mean value of a measure of the observable $\mathbf{A}$ on the state $|\psi\rangle$. Specifically, we have $$ \omega(\mathbf{A},\psi)\,:=\,\langle\,\psi|\mathbf{A}\psi\rangle$$ where $\langle\cdot|\cdot\rangle$ denotes the inner product on $\mathcal{H}$, and $|\mathbf{A}\psi\rangle$ is the vector that is obtained as the result of applying the Hermitean operator $\mathbf{A}$ on the vector $|\psi\rangle$.

Consequently, you see that the Hermitean operators representing observables do change the vectors in the Hilbert space, however, there is no physical interpretation for the transformed vector $|\mathbf{A}\psi\rangle$. Indeed, $|\mathbf{A}\psi\rangle$ is no longer normalized (unless $\mathbf{A}$ is the identity operator), and thus it does not represent a (pure) state. The physical interpretation is associated with the pairing $\langle\,\psi|\mathbf{A}\psi\rangle$ giving the mean value of a measurement of $\mathbf{A}$ on $|\psi\rangle$.

On the other hand, if we consider a unitary operator $\mathbf{U}$ on $\mathcal{H}$ and apply it to the state vector $|\psi\rangle$, we obtain a vector $|\mathbf{U}\psi\rangle$ that is still normalized, and thus represents a (pure) state of the system. Therefore, we may interpret the action of $\mathbf{U}$ on $|\psi\rangle$ as a state-transformation with an actual physical interpretation. The gates you are studying belong to this class of operators on $\mathcal{H}$.

Answer 2

When you do a measurement you most certainly do change the state vector (in the Schrödinger picture). The state ends up being projected onto an eigenspace of the corresponding operator.

A quantum gate, also changes the state, but unlike a projection it should be a change represented by a unitary operator.