3
$\begingroup$

Let $\mathcal{H}$ be the separable Hilbert space associated to some quantum system, and let $\langle\cdot,\cdot\rangle :\mathcal{H}\times\mathcal{H}\rightarrow\mathbb{C}$ denote it's inner product. The projectivisation $\mathbb{P}\mathcal{H}$ of $\mathcal{H}$ then corresponds to the set of pure states of the system. The transition probability between two states $\Psi,\Phi\in\mathbb{P}\mathcal{H}$ is given by $$ T(\Psi,\Phi):=|\langle\psi,\phi\rangle| $$ where $\psi$ and $\phi$ are arbitrary unit vectors in the rays $\Psi$ and $\Phi$ respectively. In almost any elementary introduction to quantum mechanics, a symmetry is defined as a bijective map $s:\mathbb{P}\mathcal{H}\rightarrow\mathbb{P}\mathcal{H}$ which preserves transition probabilities i.e. $$ T(s\Psi,s\Phi)=T(\Psi,\Phi). $$ This is the definition of a symmetry for which Wigner's Theorem is relevant. The intuition behind it is clear, since a symmetry must preserve the physically relevant structures (i.e. probability amplitudes) of the theory. However, transition probabilities are only a subset of these structures.

More generally, each physical observable is identified with a self-adjoint operator $O:\mathcal{H}\rightarrow\mathcal{H}$. Denote the set of such observables by $\mathcal{O}$. Then the Born rule states that if a system is in the state $\Psi\in\mathbb{P}\mathcal{H}$, the probability that a measurement of the observable $O$ will produce a value within the Borel-measurable subset $U\subset \mathbb{R}$ is given by \begin{equation} \textrm{Prob}_{\Psi,O}(U):=\langle\psi,P_{O}(U)\psi\rangle, \end{equation} where $P_{O}:\mathcal{H}\rightarrow\mathcal{H}$ denotes the projection-valued-measure associated to the self-adjoint operator $O$ by the spectral theorem. This motivates the definition of a quantum symmetry as a pair of bijective maps, $s_{1}:\mathbb{P}\mathcal{H}\rightarrow\mathbb{P}\mathcal{H}$ and $s_{2}:\mathcal{O}\rightarrow\mathcal{O}$, which preserve the above probability measure. That is, for all $O\in\mathcal{O}$ and $\Psi\in\mathbb{P}\mathcal{H}$ \begin{equation} \textrm{Prob}_{\Psi,O}=\textrm{Prob}_{s_{1}(\Psi),s_{2}(O)}. \end{equation}

My question is whether these two definitions of a symmetry transformation can be shown to be equivalent in some sense, and if so, whether one must impose any constraints on the maps $s_{1}$ and $s_{2}$ to show this. In other words, does the preservation of transition probabilities guarantee the preservation of Born probabilities?

$\endgroup$
  • $\begingroup$ Crossposted to physics.stackexchange.com/q/199794/2451 $\endgroup$ – Qmechanic Aug 11 '15 at 23:53
  • 1
    $\begingroup$ I'm sorry, I didn't realise that cross-posting was bad etiquette. I've deleted the post on the physics site. $\endgroup$ – Alex W Aug 12 '15 at 0:02
  • $\begingroup$ I'm not sure i totally understand your question. It seems to me that the link you are looking for is provided by Wigner's theorem. Furthermore, once you have defined the action of the symmetry on states you get a corresponding one by transposition on the observables through $\omega(sO) = (s^{-1}\omega)(O)$. This then leads to the "second definition". $\endgroup$ – Phoenix87 Aug 14 '15 at 10:58
  • $\begingroup$ Sorry, I'll try to be clearer. It seems to me that the second definition is the most natural since it makes reference to all Born probabilities, not only transition probabilities. There's an obvious isomorphism between the elements of $\mathbb{P}\mathcal{H}$ and the rank-one projectors acting on $\mathcal{H}$; since the latter are self-adjoint, the restriction of $s_2$ to these operators must agree with $s_1$ if the maps are to be compatible. Then the second definition implies the first. But the converse doesn't seem obvious to me: why necessarily do we need $s_2:O\mapsto s_{1}^{-1}Os_1$? $\endgroup$ – Alex W Aug 14 '15 at 14:09

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Browse other questions tagged or ask your own question.