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Consider the following circuit, where $F_n$ swaps two n-qubit states.

quantum swap circuit

If the inital state is $|0\rangle \otimes |\psi\rangle \otimes |\phi\rangle = |0\rangle|\psi\rangle|\phi\rangle$, the state before measurement is (unless I'm wrong):

$$\frac{1}{2}\left(|0\rangle \left(|\psi\rangle|\phi\rangle + |\phi\rangle|\psi\rangle\right) + |1\rangle \left(|\psi\rangle|\phi\rangle - |\phi\rangle|\psi\rangle\right)\right)$$

How to calculate the post measurement distribution for the first qubit, in terms of $|\psi\rangle$ and $|\phi\rangle$?

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While DaftWullie's answer gives you everything you need to calculate the answer in this particular case, I'd like to focus on a particular approach which is helpful in situations like yours, where you have an $n$ qubit state state$\def\ket#1{\lvert#1\rangle}\def\bra#1{\!\langle#1\rvert}$ $$ \ket{\Psi} = \ket{0}\ket{\alpha} + \ket{1}\ket{\beta}\,,$$ where $\ket{\alpha}$ and $\ket{\beta}$ are not necessarily normalised vectors on $n-1$ qubits. (Notice that at least one of $\ket{\alpha}$ and $\ket{\beta}$ must be sub-normalised in this case if $\ket{\Psi}$ has norm 1.) We can then ask: given such a $\ket{\Psi}$, what distribution do we expect on $\ket{0}$ and $\ket{1}$?

'Normalising' your superpositions

If you had a very slightly different representation for $\ket{\Psi}$, of the form $$ \ket{\Psi} = u_0 \ket{0}\ket{\alpha'} + u_1 \ket{1}\ket{\beta'}\,,$$ where $\ket{\alpha'}$ and $\ket{\beta'}$ were indeed normalised, then you'd probably be comfortable with this: you'd just recognise that the probability of '0' is $\lvert u_0 \rvert^2$ and the probability of '1' is $\lvert u_1 \rvert^2$. But we can obtain this just by considering the norms of $\ket{\alpha}$ and $\ket{\beta}$, and computing $$ u_0 = \sqrt{\langle \alpha \vert \alpha \rangle}\,,\qquad u_1 = \sqrt{\langle \beta \vert \beta \rangle} $$ and (if both $u_0$ and $u_1$ are non-zero) defining the normalised versions $\ket{\alpha'} \propto \ket{\alpha}$ and $\ket{\beta'} \propto \ket{\beta}$ by $$ \ket{\alpha'} = \tfrac{1}{u_0} \ket{\alpha}\,,\qquad\ket{\beta'} = \tfrac{1}{u_1} \ket{\beta}\,. $$

Short-cutting to the measurement probabilities

But actually, the states $\ket{\alpha'}$ and $\ket{\beta'}$ are beside the point: what you actually wanted are $u_0$ and $u_1$, or more precisely, $$\Pr\!\big[\,0\,\big] = \lvert u_0 \rvert^2 = \langle \alpha \vert \alpha \rangle\,,\qquad \Pr\!\big[\,1\,\big] = \lvert u_1 \rvert^2 = \langle \beta \vert \beta \rangle\,.$$ So you can just compute those inner products without even worrying about representing the state $\ket{\Psi}$ in one particular way or another, and in particular without giving any thought as to whether or which of $\ket{\alpha}$ or $\ket{\beta}$ is normalised.

Example.

In your particular case, you have: $$ \ket{\alpha} = \frac{1}{2} \bigl( \ket{\psi}\ket{\phi} + \ket{\phi}\ket{\psi} \bigr) , \qquad \ket{\beta} = \frac{1}{2} \bigl( \ket{\psi} \ket{\phi} - \ket{\phi} \ket{\psi} \bigr); $$ then computing the probability of obtaining either '0' or '1' is just a question of computing some inner products, and in particular will give interesting results when $\ket{\psi}$ and $\ket{\phi}$ are either orthogonal (in which case $\ket{\alpha}$ and $\ket{\beta}$ are both clearly maximally entangled, normalisation aside) or parallel (in which case $\ket{\beta}$ is clearly zero).

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Let's start from the state $$ |\Psi\rangle=\frac12\left(|0\rangle(|\psi\rangle|\phi\rangle+|\phi\rangle|\psi\rangle)+|1\rangle(|\psi\rangle|\phi\rangle-|\phi\rangle|\psi\rangle)\right). $$ There are a couple of ways to do the calculation. If you want to be formal, which typically leads to fewer mistakes, you identify the measurement operators on a single spin $$ P_0=|0\rangle\langle 0|\otimes\mathbb{I}^{2n}\qquad P_1=|1\rangle\langle 1|\otimes\mathbb{I}^{2n} $$ and you evaluate the probabilities of the two outcomes as $$ p_i=\langle\Psi|P_i|\Psi\rangle $$

Slightly less formally, but equivalent, you can collect the terms for $|0\rangle$ and $|1\rangle$, much as you have, but make sure the state on the other qubits is normalised. $$ |\Psi\rangle=\frac12\left(\sqrt{2+2|\langle\psi|\phi\rangle|^2}|0\rangle\frac{|\psi\rangle|\phi\rangle+|\phi\rangle|\psi\rangle}{\sqrt{2+2|\langle\psi|\phi\rangle|^2}}+\sqrt{2-2|\langle\psi|\phi\rangle|^2}|1\rangle\frac{|\psi\rangle|\phi\rangle-|\phi\rangle|\psi\rangle}{\sqrt{2-2|\langle\psi|\phi\rangle|^2}}\right). $$ Then you can read of the probability amplitude for finding the state in $|0\rangle$ or $|1\rangle$, and take the mod-square to get the probability

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A general way you can predict the measurement outcomes is to calculate the density matrix as:

$\rho = |\Psi \rangle \langle \Psi|$ and calculate the partial trace over qubits $B$ and $C$: $\rho_A = \rm{Tr}_{B,C} \left(\rho \right)$.

You then have a 2x2 density matrix where the two diagonals tell you the probabilities of getting $|0\rangle$ or $|1\rangle$ respectively.

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  • $\begingroup$ @NieldeBeaudrap: I deleted the old answer and kept only the density matrix part, which you said was correct: "Inasmuch as you are basically saying that it is possible to represent the state as a density matrix and then take the marginal on the first qubit, yeah." $\endgroup$ – user1271772 Oct 22 '18 at 18:50

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