Quantum non-unitary transformation?

Question

Let us say that I apply a non-unitary transformation $\def\ket#1{| #1\rangle} \def\braket#1#2{\langle #1|#2\rangle} \hat A$ to the ket's: $$\ket{\psi} \rightarrow \hat A \ket{\psi}$$ $$\ket{\phi}\rightarrow \hat A \ket{\phi}$$ Clearly in this case the probability: $$P=|\braket{\phi}{\psi}|^2$$ Will change. What physically is going on here? i.e. why for unitary operators we can perform such a transformation but for non-unitary operators we can't?

Answer 1

I see two sides to your question, in addition to what was already pointed out in comments:

1. For a transformation to preserve the scalar product for any pair of kets $|\psi\rangle$, $|\phi\rangle$ in the Hilbert space, it is sufficient that it be an orthogonal transformation: Say $U$ is such a transformation. If $\lbrace |\psi_n\rangle \rbrace_n$ is an orthonormal basis set, then $U$ must be such that
   $$ \langle \psi_m | U^\dagger U | \psi_n \rangle = \langle \psi_m | \psi_n \rangle \equiv \langle \psi_m | I | \psi_n \rangle = \delta_{mn},\;\;\;\forall\;m,n $$ which means $$ U^\dagger U = I $$ and so $U$ is orthogonal. Note that unitary transformations also satisfy $U U^\dagger = I$.

2. When you apply a non-orthogonal or non-unitary transformation $V$, the scalar product of the transformed kets amounts to $$ \langle \phi | V^\dagger V | \psi \rangle \neq \langle \phi | \psi \rangle $$ But note that this is actually a matrix element for the hermitian operator, and so potentially the observable, $V^\dagger V$. One ubiquitous example: the annihilation operator $\hat a$ for a many-particle system. It is not hermitian, does not preserve the scalar product, but the scalar product of the transformed kets gives a matrix element of the occupation number, which is not the corresponding amplitude: $$ \langle \phi | {\hat a}^\dagger {\hat a} | \psi \rangle \neq \langle \phi | \psi \rangle $$

Answer 2

A nonunitary operator can be thought of as the sum of unitary and antiunitary operators. So let me let $T = aU + bA$, a, b = normalization constants, and we evaluate $$ \langle T\phi|T\psi\rangle = \langle\phi|T^\dagger T|\psi\rangle. $$ The product is $T^\dagger T = U^\dagger U + U^\dagger A + A^\dagger U + A^\dagger A$. The unitary operator is easy $U^\dagger U = 1$. For $A^\dagger A$ we have $$ \langle\phi|A^\dagger A|\psi\rangle = \langle\psi|\phi\rangle. $$ For $A^\dagger U$ we have $$ \langle\phi|A^\dagger U|\psi\rangle = \overline{\langle\phi|AU|\psi\rangle} $$ and similarly $$ \langle\phi|U^\dagger A|\psi\rangle = \overline{\langle\psi |UA^\dagger|\phi\rangle} = \langle\psi |AU|\phi\rangle $$ The probability will then be a rather long summ $$ P = \langle T\phi|T\psi\rangle = |a|^2|\langle\psi|\phi\rangle|^2 + b^2|\langle\phi|\psi\rangle|^2 = (|a|^2 + |b|^2)|\langle\psi|\phi\rangle|^2 $$ where $|a|^2 + |b|^2 = 1$.