In quantum mechanics, in the context of symmetry transformations, it is often said that for a transformation $T$ to conserve probabilities it must be unitary.

But by performing any (even non-unitary) transformation on the system we are just taking account for the fact we are looking at it in a different way (i.e. using different basis vectors).

By looking at the system in a different way I cannot see how we could change the probabilities of a measurement (as long as these probability where calculated correctly, which may need the introduction of a matrix into the scalar product).

Am I correct? If so why is it said that only unitary matrices conserve probabilities and if not why not?


The probability to detect state $\psi$ in state $\phi$ is given by $$ P(\psi,\phi) = \frac{\lvert \langle \psi,\phi\rangle\rvert^2}{\langle \psi,\psi\rangle\langle \phi,\phi\rangle}$$ and by Wigner's theorem every sensible physical transformation $T$ (that should a priori be thought of to act on rays in Hilbert space rather than individual vectors) that preserves this probability in the sense that $P(\psi,\phi) = P(T\psi,T\phi)$ holds for all states can be given by a unitary operator on the Hilbert space of states.

Arbitrary operators do not preserve the probability, which is most evident for the non-invertible ones. If you want to modify the scalar product to make other invertible transformations preserve it, then you're effectively choosing a new scalar product such that the transformation in question becomes unitary, which means this doesn't add anything useful.

  • $\begingroup$ But by saying that a non-unitary transformation doesn't preserve probabilities, is this not saying the equivalent of 'by simply measuring things w.r.t. a basis that is not orthonormal we can change the predictions of quantum mechanics'? $\endgroup$ – Quantum spaghettification Aug 2 '16 at 10:07
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    $\begingroup$ @Quantumspaghettification I don't know what "measuring things w.r.t. a basis that is not orthonormal" means. What we measure are observables, which by definition are self-adjoint operators, which by the spectral theorem have orthonormal eigenbases. $\endgroup$ – ACuriousMind Aug 2 '16 at 10:12
  • $\begingroup$ @Quantumspaghettification Only orthogonal states can be distinguished by measurements. If you had a non-orthogonal basis (you can't, see comment above), say formed by the states $|0\rangle$ and $\frac{|0\rangle+|1\rangle}{\sqrt{2}}$, the latter state has a 50% probability of being equal to the former. $\endgroup$ – Bosoneando Aug 2 '16 at 10:18

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