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In a seminar, I heard that the unitary aspect of representations was important physically, because in quantum mechanics unitarity is closely tied to the conservation of probability. Could someone explain why this is so?

After a bit of thought, one vague connection I can think of is that a matrix generated by a Hermitian matrix, $$U =e^{iH},$$ where $H$ is Hermitian is bound to be unitary . As Hermitian matrices correspond to observables, this must have some signficiance. Though, as I said, I still have the faintest idea.

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3 Answers 3

up vote 11 down vote accepted

Probabilites are (squares of) probability amplitudes, which can be obtained as inner products of vectors on Hilbert space: $\langle X|Y \rangle$. Under a transformation U, the ket transforms as

$$|Y\rangle \rightarrow |Y'\rangle = U |Y\rangle$$

and the bra as

$$\langle X| \rightarrow \langle X'| = \langle X| U^{\dagger}$$

So if U is unitary, the probability amplitude $\langle X|Y \rangle$ is maintained under the transformation, since $U^{\dagger}$ = $U^{-1}$ and hence $\langle X'|Y' \rangle = \langle X|Y \rangle$.

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+1, so what you are saying is that unitarity is a way to guarantee that the space-time is really isotropic under Poincare transformations, with the special case of understanding the propagator as the time displacement operator –  lurscher Nov 17 '11 at 18:15

This answer is good if by "conservation of probability", you mean dynamical conversation but let me try to parse the phrase "unitary aspect of representations was important physically" in another way.

Schrodinger's wave mechanics and Heisenberg's matrix mechanics both satisfy $[Q,P] = i \hbar$ but do so in totally different Hilbert spaces. You may ask, who is right? -- or, how many other representations of this relation are there? Well, the Stone-von Neumann theorem tells you that all representations of $[Q,P] = i\hbar$ are unitarily equivalent. That is, there exists a unitary $U$ taking one representation to the other: $[U Q U^\dagger, UPU^\dagger] = i\hbar$.

So the Schrodinger and Heisenberg representation are unitarily equivalent -- but you still want to know which one is the right one. Well, it turns out that they both are. In any representation, the physical predictions are manifest as probabilistic predictions for the outcome of an event $E$ (represented as a positive operator) given the state of the system is $|\psi\rangle$, viz. $$\Pr(E|\psi) = \langle\psi|E|\psi\rangle.$$ Now consider the predictions of unitarily equivalent representation: $$\Pr(UEU^\dagger|U\psi) = \langle\psi|U^\dagger UEU^\dagger U|\psi\rangle = \langle\psi|E|\psi\rangle.$$ So unitarily equivalent representations are physically indistinguishable.

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+1, but not saying anything about why the propagator needs to be unitary (past amplitudes and future amplitudes have the same normalization constant) –  lurscher Nov 17 '11 at 18:12
@Chris: It's always good to have a "dynamical conversation" ;-) –  Mattia Nov 8 '13 at 12:26

The important thing in QM is the average of a certain observable if you use unitary transformations the average doesn't change at all. As a special case you can take the average of the Identity operator, this should give you 1 for any unitary transformation.

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