# Quantum non-unitary transformation? [closed]

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?

## closed as unclear what you're asking by Norbert Schuch, user36790, Qmechanic♦Jun 10 '16 at 6:04

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• Anti-unitary operator also preserve the probability. – AMS Jun 9 '16 at 17:30
• I believe you have to look at it from another angle. You don't expect your total probability be any different from 1, so non-unitary transformations cannot be physical. – David VdH Jun 9 '16 at 17:37
• If you think about an observable in the Heisenberg picture, applying a non -unitary transformation to it will make it non self-adjoint, therefore it will not be an observable anymore. It should not be hard to show that U(n) is the group of transformations that leave your physical observables physical observables, and therefore, "physically", you are transforming something that is "physical" into something that is "unphysical"... – PhilipV Jun 9 '16 at 17:41
• I don't understand your question. You don't apply "non-unitary transformations" in any physical setting I can think of, therefore I don't see what you mean by what is "physically" going on here. Could you give an example? – ACuriousMind Jun 9 '16 at 18:02
• Let us say that I apply the squaring operation to your age and my age, so my age $X$ maps to $X^2$ and your age $Y$ maps to $Y^2$. Then clearly the ratio of our ages, $X/Y$ is not preserved. What physically is going on here? Why, if I want to preserve the ratio, can I multiply our ages by a constant but not square them? What would constitute an answer to this question? – WillO Jun 9 '16 at 20:44

## 2 Answers

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: $$\langle \phi | {\hat a}^\dagger {\hat a} | \psi \rangle \neq \langle \phi | \psi \rangle$$

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 $|a|^2 + |b|^2 = 1$.

• If $T$ is linear, and nothing in the OP suggests otherwise, your statement that $T = aU + bA$ also means $T-aU = bA$, or that a linear transformation is identical to an antilinear one. How come? – udrv Jun 9 '16 at 19:13
• $T - aU$ is antiunitary. Both $U$ and $A$ are linear are linear. – Lawrence B. Crowell Jun 9 '16 at 20:19
• Not according to the standard definition: en.wikipedia.org/wiki/Antiunitary_operator, which you invoke yourself in your 2nd eq.: If $\psi = \lambda_1\psi_1 + \lambda_2\psi_2$, then $\langle \phi| A^\dagger A| \lambda_1\psi_1 + \lambda_2\psi_2\rangle = \lambda_1^* \langle \phi| A^\dagger A| \psi_1\rangle + \lambda_2^* \langle \phi| A^\dagger A| \psi_1\rangle$. Compare this with $\langle \phi| U^\dagger U| \lambda_1\psi_1 + \lambda_2\psi_2\rangle = \lambda_1 \langle \phi| U^\dagger U| \psi_1\rangle + \lambda_2 \langle \phi| U^\dagger U| \psi_1\rangle$ for $U$ unitary. – udrv Jun 9 '16 at 21:49