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Problem of the Week #30 - December 24th, 2012

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Chris L T521

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Jan 26, 2012
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Here's this week's problem.

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Problem: If $f(z)=\exp(z)=\sum_{n=0}^{\infty}z^n/n!$ and $A$ is a hermitian operator, show that $f(iA)$ is a unitary operator.

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Remember to read the POTW submission guidelines to find out how to submit your answers!
 
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Chris L T521

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Jan 26, 2012
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Update (additional info): After reading someone's solution to this problem, it became apparent that I didn't specify what space the operator $A$ was defined over; in fact, the text I took this question from didn't even specify the space. However, based on where this exercise was in the text, it's safe to assume that $A$ is an operator on any Hilbert space $\mathcal{H}$.
 
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Chris L T521

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Jan 26, 2012
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This week's question was correctly answered by Deveno. You can find his answer below.

Note that:$(iA)(-iA) = -(iA)^2 = (-iA)(iA)$

so that:

$f(iA)f(-iA) = \exp(iA)\exp(-iA) = \exp(iA-iA) = \exp(0) = A^0 = \text{id}$

that is:

$[f(iA)]^{-1} = f(-iA)$

now:

$[f(iA)]^H = [\exp(iA)]^H = \left[\sum_{n=1}^{\infty} \frac{(iA)^n}{n!}\right]^H = \sum_{n=1}^{\infty} \frac{[(iA)^n]^H}{n!}$

$=\sum_{n=0}^{\infty} \frac{[(iA)^H]^n}{n!} = \sum_{n=0}^{\infty} \frac{(-iA^H)^n}{n!}$ (because $\overline{i} = -i$)

$= \sum_{n=0}^{\infty} \frac{(-iA)^n}{n!}$ (since $A$ is hermetian, $A = A^H$)

$= f(-iA) = [f(iA)]^{-1}$, so $f(iA)$ is unitary.
 
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