2
$\begingroup$

In the fundamental postulate of quantum mechanics (page 18), there exists a one-to-one correspondence like that:

$\text{closed quantum system $\leftrightarrow$ Hilbert space ; quantum state $\leftrightarrow$ a ray in Hilbert space } $

but here the ray can carry some additional phase factor $e^{i\phi}$? What's the observable influence due to the phase factor in state vector? I mean if you take inner product the redundant factor will disappear.

$\endgroup$
  • 1
    $\begingroup$ What do you mean by "influence"? $\endgroup$ – ACuriousMind Feb 22 '17 at 9:07
  • $\begingroup$ The phase factor shows the time evolution of phase of the system. It appears even for a stationary state. It doesn't show any influence on the vector $\endgroup$ – UKH Feb 22 '17 at 9:36
  • $\begingroup$ @ACuriousMind I mean is there any 0bservable effects? $\endgroup$ – Jack Feb 22 '17 at 9:55
  • 1
    $\begingroup$ Everything would be so easy if there was the one-to-one correspondence you are describing. Sadly, there are many very strong suggestions that this should not be the case. The existence of uncountably many inequivalent irreducible representations of the canonical commutation relations for quantum fields is one of such suggestions. Another is the fact that not every quantum state can be represented, in a given (irreducible) representation, as a ray in Hilbert space (or as a density matrix, actually). $\endgroup$ – yuggib Feb 22 '17 at 10:14
2
$\begingroup$

A global phase has absolutely no measurable consequences. Experimental values, such as average values and other moments, do not depends on this overall phase. One often keep track of this phase to simplify calculations. For instance, when rotating a state it is often quite convenient to express the final result without constantly eliminating the overall phase.

On the other hand relative phases matter quite critically as they enter through interference terms.

Note that, when formulating QM in terms of density operator, the overall phase of a state $\vert \psi\rangle$ automatically drops out (at least implicitly) of the expression for the density operator $\hat \rho = \vert\psi\rangle\langle \psi\vert$. This formulation is fully equivalent to the Schr$\ddot{\hbox{o}}$dinger formulation in the case of the so-called pure states.

$\endgroup$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.