0
$\begingroup$

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?

$\endgroup$

1 Answer 1

2
$\begingroup$

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.

$\endgroup$
3
  • $\begingroup$ 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'? $\endgroup$ Commented Aug 2, 2016 at 10:07
  • 1
    $\begingroup$ @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
    Commented Aug 2, 2016 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
    Commented Aug 2, 2016 at 10:18

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service and acknowledge you have read our privacy policy.

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