I'm going through the phase estimation algorithm, and wanted to sanity-check my calculations by making sure the state I'd calculated was still normalized. It is, assuming the square of the absolute value of the eigenvalue of the arbitrary unitary operator I'm analyzing equals 1. So, does it? Assuming that the eigenvector of the eigenvalue is normalized.

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    $\begingroup$ Note that your question doesn't depend on whether the eigenvector is normalised. If you have a longer or shorter eigenvector, then that longer or shorter eigenvector has its norm changed by the same scalar factor as if it were normalised. $\endgroup$ – Niel de Beaudrap Oct 10 '18 at 8:07

Good question. The answer turns out to be Yes.
You don't even need the vector to be normalized. Watch:

Start with the definition of eigenvalues and eigenvectors:

$$ \begin{align} U|\psi\rangle &= \lambda |\psi\rangle\\ \end{align} $$

Conjugate and transpose both sides of the equation:

$$ \begin{align} \langle\psi|U^\dagger &= \langle \psi| \lambda^*. \end{align} $$

Left multiply each side of line 1 by the corresponding side of line 2.

$$ \begin{align} \langle \psi|U^\dagger\cdot U|\psi \rangle &= \langle \psi | \lambda^* \lambda |\psi\rangle \\ \langle \psi |\psi \rangle &= |\lambda |^2 \langle \psi |\psi \rangle \\ c &= |\lambda|^2 c\\ 1 &= |\lambda|^2 \end{align} $$

If $|\psi\rangle $ is normalized, it just means that $c=1$, which makes no difference in this proof because the $c$ was on both sides of the equation and can be divided out.

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    $\begingroup$ I get every step except for how $\langle \psi | U \cdot U | \psi \rangle = \langle \psi | \psi \rangle$; doesn't that only work if the matrix is hermitian as well as unitary? $\endgroup$ – ahelwer Oct 10 '18 at 1:44
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    $\begingroup$ I'm also not sure how we get from $\langle \psi | \lambda \lambda | \psi \rangle$ to $| \lambda | ^2 \langle \psi || \psi \rangle$, isn't $| \lambda | ^2$ the same as $\lambda \cdot \lambda ^*$? Where did the conjugate come from? $\endgroup$ – ahelwer Oct 10 '18 at 1:51
  • $\begingroup$ Sorry: conjugate Line 1 to get line 2. Then both your questions are answered. My answer has to be edited, but I'm on a tiny phone now (and in bed, with eyes closing in a couple seconds). Feel free to edit it if you feel up to it! $\endgroup$ – user1271772 Oct 10 '18 at 3:53
  • $\begingroup$ Actually the extremely helpful @AHusain already did it for me, I just had to click "approve edit". Thank you AHusain! $\endgroup$ – user1271772 Oct 10 '18 at 3:55

@user1271772's answer is excellent, and absolutely the right answer. I just wanted to add in some additional perspective, given recent questions regarding Hamiltonians. Many physicists start from the Hamiltonian being the underlying thing that determines evolution, and unitaries are derived as a consequence. They start from the Schrödinger equation, $$ i\frac{d|\psi\rangle}{dt}=H|\psi\rangle. $$ For a time-invariant Hamiltonian, the solution is $$ |\psi(t)\rangle=e^{-iHt}|\psi(0)\rangle, $$ where $e^{-iHt}$ is unitary because $e^{-iHt}e^{iHt}=\mathbb{I}$. Just stating this solution actually skips over the thing I really want to focus on.

We could expand a generic $|\psi\rangle=\sum_ia_i|i\rangle$, so the Schrödinger equation becomes a series of simultaneous differential equations for the $a_i$: $$ i\frac{da_i}{dt}=\sum_jH_{ij}a_j. $$ Now consider what happens to an eigenvector $|\lambda\rangle=\sum_ib_i|i\rangle$ of $H$: $$ \sum_{j}H_{ij}b_i^\star=\lambda b_i^\star. $$ We can take linear combinations of the $a_i$: $$ i\frac{d\sum_ib_i^\star a_i}{dt}=\sum_{ij}b_i^\star H_{ij}a_j=\lambda\sum_jb_j^\star a_j. $$ Hence, we see that the component $x=\sum_jb_j^\star a_j$ simply satisfies $$ i\frac{dx}{dt}=\lambda x, $$ so $x(t)=e^{-i\lambda t}x(0)$. In other words, a state initially created as an eigenvector $|\lambda\rangle$ stays in that state and just acquires a phase over time $e^{-i\lambda t}|\lambda\rangle$. Hence, eigenvectors of $H$ are also eigenvectors of the unitary $U$, with eigenvalues $e^{-i\lambda t}$, and these have modulus 1.


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