1
$\begingroup$

One property of the unitary operator is that it preserves the norm of the state-vectors: $$ \langle \Psi | U^\dagger U | \Psi \rangle = \langle \Psi | \Psi \rangle $$ If $U$ is unitary.

Is the inverse statement also true? For an operator that will satisfy above equation for all $|\Psi \rangle$, will that operator be unitary?

$\endgroup$
  • $\begingroup$ Hint: look up the so-called 'polarization identity'. $\endgroup$ – Emilio Pisanty Aug 8 '19 at 19:33
  • 3
    $\begingroup$ Not necessarily, it could be isometric (the identity above) but not surjective when the Hilbert space has infinite dimension. $\endgroup$ – Valter Moretti Aug 8 '19 at 19:34
  • $\begingroup$ Don’t anti-unitary operators also preserve probability? $\endgroup$ – G. Smith Aug 8 '19 at 19:38
  • $\begingroup$ Yes but they are anti- linear, here the question regards "operators", i.e. , linear maps. $\endgroup$ – Valter Moretti Aug 8 '19 at 19:45
  • $\begingroup$ @ValterMoretti that means, for surjective operator, the proposition holds? $\endgroup$ – Quantumwhisp Aug 8 '19 at 21:04
2
$\begingroup$

If $U:H\to H$, with $H$ a Hilbert space, is linear it turns out that it preserves the norm (in particular it is injective) if and only if it preserves the scalar product. The proof is based on the so-called polarization identity as Emilio suggested. However this does not mean that the operator is unitary, since the condition $UU^*=I$, corresponding to surjectivity, is not necessarily valid if the space has not finite dimensione. If the dimension is finite, injectivity and surjectivity are equivalent as is known from elementary linear algebra.

$\endgroup$
  • 1
    $\begingroup$ Or $U$ could act between spaces of different dimension. $\endgroup$ – Norbert Schuch Aug 8 '19 at 19:56
  • 1
    $\begingroup$ Yes, but usually (not always however) "unitary" is referred to an operator from a Hilbert space to itself... $\endgroup$ – Valter Moretti Aug 8 '19 at 19:59
  • $\begingroup$ Sorry for typos, I am using my phone... $\endgroup$ – Valter Moretti Aug 8 '19 at 20:01
  • $\begingroup$ Exactly, that's what I mean: A linear map acting between two different (finite) dimensional Hilbert spaces which preserves the norm will not be unitary. (Just to say that such a case does not require infinite dimensional Hilbert spaces - not particularly deep.) $\endgroup$ – Norbert Schuch Aug 8 '19 at 20:04
  • $\begingroup$ Sure, it is however also possible to construct an isometric linear map from a Hilbert space into itself which is not surjective. It happens if the space is infinite dimensional. $\endgroup$ – Valter Moretti Aug 8 '19 at 20:08

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.