Does the Hamiltonian time-evolution operator actually change the state of the system?

According to my understanding of things, the time evolution operator in QM looks something like this,

$$U = \exp(-iHt/\hbar)$$

Which acts on the state vector / wave-function of the system to effectively step forward in time.

I notice that this is a unitary operator. Given that one of the postulates of quantum mechanics is that two states are identical if they differ only by a phase factor (a unitary complex scalar?), then surely this operation does not actually change the state of the system?

Am I missing something here?

• A unitary operator is very different from a unitary scalar. Think about the operator in matrix form... – valerio Jun 8 '16 at 9:16

$\hat U$ is an operator, and an operator is very different from a scalar.

Just think about this: every operator can be expressed as a matrix in some basis and every state as a vector. So the difference between

$$\exp \left(-\frac{i \hat H t}{\hbar} \right) \mid \psi \rangle$$

and

$$\exp(-i \phi) \mid \psi \rangle$$

where $\phi$ is a real number, is the same difference that exists between

$$\hat A \ \vec v$$

and

$$a \ \vec v$$

Where $\hat A$ is a matrix and $a$ is a number.

• Ah right I see what you're saying. So that expression doesn't simple multiply psi, it acts on it to produce a new state. The matrix formulation of QM always made more sense to me somehow but I suppose it becomes tricky to represent operators as matrices when you start dealing with infinite-dimensional Hilbert spaces. – JeneralJames Jun 8 '16 at 9:45
• Yes, it indeed gets tricky in infinite-dimensional spaces (you have to dig a little bit into functional analysis and some questions about continuity and convergence arise), but the principle is always the same. As you said, an operator doesn't simple multiply the state vector, it acts on it to produce a new state :-) – valerio Jun 8 '16 at 10:00
• @JeneralJames Operators are different than corresponding matrices. These matrices are a collection of the expectation values when operated by two states $H_{mn}=\left\langle{\Psi_m}|H|{\Psi_n}\right\rangle$, while operators are pure quantum mechanical entity – hsinghal Jun 8 '16 at 10:12