# Must an operator that preserves probability be unitary?

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?

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

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.
• Or $U$ could act between spaces of different dimension. – Norbert Schuch Aug 8 '19 at 19:56