Does post selection of a qubit introduce non-linearity?

Question

Problem

I have a multi-qubit state $\lvert \psi \rangle$ and an ancilla qubit $\lvert 0 \rangle$ that I use to extend my state, getting the new state $\lvert 0\rangle \otimes \lvert \psi \rangle$.

Suppose now I apply a generic unitary operation on this state and get a new state

$$ \lvert \phi \rangle = \hat{U}\big(\lvert 0\rangle \otimes \lvert \psi \rangle\big) $$

which can be rewritten as

$$ \lvert \phi \rangle = \lvert 0 \rangle \otimes\lvert a\rangle + \lvert 1 \rangle \otimes\lvert b \rangle $$

say I measure the first qubit and accept the remainder of the state only if the measurement outcome is $0$. Then, overall, I am implementing the transformation

$$ \lvert \psi \rangle \longrightarrow \lvert a\rangle $$

Question

Can this transformation be non-linear in general? In other words: is there a matrix $\hat{B}$ such that $\lvert a \rangle = \hat{B}\lvert \psi\rangle$?

Answer 1

Linearity before normalization

Yes this map is linear (up to normalization) if we don't care if the output of the map is normalized (i.e., the output is a vector but not a valid quantum state). To see it note that you can view it as a composition of three linear maps:

1. Embedding $$|\psi\rangle \to |0\rangle \otimes |\psi\rangle$$ This can be viewed as an application of the linear map $$ V = |0\rangle \otimes I. $$
2. Transformation $$|0\rangle \otimes |\psi\rangle \to |0\rangle \otimes |a\rangle + |1\rangle \otimes |b\rangle$$ This involves the application of the linear map $U$ specified in the question.
3. Projection and removal of ancilla $$ |0\rangle \otimes |a\rangle + |1\rangle \otimes |b\rangle \to |a\rangle $$ This can be viewed as action of the linear map $$ W = \langle 0| \otimes I. $$

Composing all three maps to get $WUV$ we find that $$ WUV |\psi\rangle = |a\rangle, $$ but again I stress that $|a\rangle$ is not necessarily a normalized vector.

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But what if I want a normalized vector?

Well so now we want the stronger condition that the map $$ |\psi\rangle \to \frac{|a\rangle}{\sqrt{\langle a | a \rangle}} $$ is linear. Unfortunately this is no longer the case as renormalization is an inherently nonlinear transformation.

This is relatively immediate from the definition of renormalization as the map $$ |a \rangle \to \frac{|a\rangle}{\sqrt{\langle a | a \rangle}}. $$ If this were linear then we should have that $$ 2 |a\rangle = |a\rangle + |a \rangle \to \frac{|a\rangle}{\sqrt{\langle a | a \rangle}} + \frac{|a\rangle}{\sqrt{\langle a | a \rangle}} = 2\frac{|a\rangle}{\sqrt{\langle a | a \rangle}} $$ but this can never be the case because the output of the map is always normalized and hence cannot be twice a normalized vector (which would not be normalized).

Moreover, it is clear this normalization map is not defined at the origin as you end up dividing by the norm of the $0$ vector which is $0$. Recall all linear maps satisfy $M 0 = 0$. Note that this case is relevant to the question. Imagine the unitary $U$ in the question is of the form $U = X \otimes I$ where $X$ is the Pauli X matrix. The resulting state after the application of the unitary $U$ is then $|1\rangle \otimes |\psi\rangle$ which when we perform the projection in step $3$ we end up with the $0$ vector. At which point the normalization is no longer well-defined.

Answer 2

This transformation is almost linear, but normalization makes it non-linear in general. The state update rule following a generalized measurement on some state $\rho$ is $$\rho\to \frac{A_i\rho A_i^\dagger}{\mathrm{Tr}(A_i\rho A_i^\dagger )}.$$ For a pure state, that looks like $$|\psi\rangle\langle\psi|\to \frac{A_i|\psi\rangle\langle\psi| A_i^\dagger}{\mathrm{Tr}(A_i |\psi\rangle\langle\psi| A_i^\dagger )},$$ which looks like the linear transformation $|\psi\rangle\to A_i|\psi\rangle$. However, normalization is not linear. The pure state in question transforms as $|\psi\rangle\to A_i|\psi\rangle/\sqrt{P_i(\rho)}$, where $$P_i(\rho)=\mathrm{Tr}(A_i |\psi\rangle\langle\psi| A_i^\dagger )=\langle \psi|A_i^\dagger A_i|\psi\rangle$$ is the probability of obtaining measurement result $i$.

What does it mean to not be linear? A state $p\rho+(1-p)\sigma$ does not transform into a combination of the transformations of $\rho$ and $\sigma$. To wit, $$p\rho+(1-p)\sigma\to\frac{A_i[p\rho+(1-p)\sigma]A_i^\dagger}{\mathrm{Tr}\{A_i[p\rho+(1-p)\sigma]A_i^\dagger\}}=\frac{pA_i\rho A_i^\dagger+(1-p)A_i\sigma A_i^\dagger}{p\mathrm{Tr}(A_i\rho A_i^\dagger)+(1-p)\mathrm{Tr}(A_i\sigma A_i^\dagger)},$$ which is emphatically different from $$p\frac{A_i\rho A_i^\dagger}{\mathrm{Tr}(A_i\rho A_i^\dagger)}+(1-p)\frac{A_i\sigma A_i^\dagger}{\mathrm{Tr}(A_i\sigma A_i^\dagger)}.$$

Overall, the main idea is that the updated state must be normalized, and the normalization factor is not a linear function of the state because it shows up in the denominator (i.e., even though $P_i(\rho)$ is a linear function of $\rho$, the factor $1/P_i(\rho)$ that appears is not a linear function of $\rho$). This is true for any $A_i$.