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How to obtain $U$ for the amplitude dampening channel?

The behaviour of the amplitude damping channel on the system can be represented by $U$, a unitary matrix which has the following effect:

  • $U|00\rangle = |00\rangle$
  • $U|11\rangle = |11 \rangle$
  • $U|01\rangle = \sqrt{1-p}|01\rangle + i \sqrt{p}|10\rangle$
  • $U|10\rangle = i\sqrt{p}|01\rangle + \sqrt{1-p}|10\rangle$

We can then apparently write $U$ as follows: $$U = |00\rangle \langle 00| + \sqrt{1-p} |01\rangle \langle 01| + i \sqrt{p}|01\rangle \langle 10| + i\sqrt{p}|10\rangle \langle 01| + \sqrt{1-p}|10\rangle \langle 10| + |11\rangle \langle 11|$$

I see that each ket of the right hand side of the above four equations is being multiplied by the bra of the state being multiplied by $U$ left hand side, but I don't understand then why they are added together. I partially understand intuitively it is to give the correct positions in the matrix, but I don't understand mathematically how we obtain this result.

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Quantum mechanics is a linear theory. So, pretty much by definition, you can add things up. (That's a little glib.)

More precisely, unitaries describe basis transformations. You start with one orthonormal basis, and you are mapped to another orthonormal basis.

A term like $|\phi\rangle\langle\psi|$ basically says "if the initial state is $|\psi\rangle$, change it into $|\phi\rangle$", which you can see instantly if you calculate $$ (|\phi\rangle\langle\psi|)|\psi\rangle $$ but, as I say, it's not enough to know about a single term. You need to know how every state transforms. Let's say I have some initial state $|\Psi\rangle$, which I can decompose in terms of an orthonormal basis $\{|\psi_i\rangle\}$. Then $$ |\Psi\rangle=\sum_i\alpha_i|\psi_i\rangle. $$ If the unitary is supposed to describe a basis change $|\psi_i\rangle\rightarrow|\phi_i\rangle$ for all $i$, then the linearity of quantum means that $$ |\Psi\rangle\longrightarrow\sum_i\alpha_i|\phi_i\rangle, $$ and you are simply looking for a way to describe that transformation. The trick is that because you're using an orthonormal basis, if you were to apply $|\phi_j\rangle\langle\psi_j|$, it only affects the specific term $|\psi_j\rangle$, and not any of the other basis elements ($\langle\psi_j|\psi_i\rangle=\delta_{ij}$). So that means you can just add them up, and each term will act independently to construct the full transformation that you want.

Moreover, it really does give you a unitary: if I've constructed $$ U=\sum_i|\phi_i\rangle\langle\psi_i| $$ then $$ UU^\dagger=\sum_{i,j}|\phi_i\rangle\langle\psi_i|\psi_j\rangle\langle\phi_i|=\sum_i|\phi_i\rangle\langle\phi_i|=I. $$

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  • $\begingroup$ So in this case, are we saying that $|\psi_{1}\rangle = |00\rangle$, $|\psi_{2}\rangle = |01\rangle$, $|\psi_{3}\rangle=|10\rangle$ and $|\psi_{4}\rangle = |11\rangle$ and then, say for example, $|\phi_{3}\rangle= i\sqrt{p}|01\rangle + \sqrt{1-p}|10\rangle$. Additionally, wouldn't then the $\alpha_{i}$'s change when it undergoes the basis change because it introduces new co-efficients (i.e. $i\sqrt{p}\alpha_{3}$)? $\endgroup$
    – am567
    Commented Nov 13 at 10:31
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    $\begingroup$ Yes, that's right for the identification of the $|\psi_i\rangle$ and $|\phi_i\rangle$. No, the $\alpha_i$ don't change; the basis changes. Alternatively, you can write the new state in terms of the old basis. Then the $\alpha_i$s would change. The two things are equivalent, but you don't do both. $\endgroup$
    – DaftWullie
    Commented Nov 13 at 11:26

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