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I have a question about finding the expectation value. Here's the question:

So I have this wave function $$|𝜙⟩=\frac{\sqrt{6}}{𝜋}\sum_{n=1}^∞ C_𝑛|n⟩ $$

where the eigenvectors |n⟩ form an orthonormal basis and:

$$C_n=\frac{(-1)^n}{n} $$

So what I need to do consider an operator |5⟩⟨2| - I don't know what that means first of all.

And I need to find the expectation value corresponding to this operator for a particle in the state |𝜙⟩.

I know that in order to calculate the expectation value:

$$ <Q> =⟨𝜙|\hat Q|𝜙⟩ $$

So my question is what does |5⟩⟨2| actually mean? And going on, what shall I do next to calculate the expectation value.

Thanks for reading.

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The operator $|5\rangle\langle2|$ is exactly what it looks like. If you apply it to any state $|\phi\rangle$ you get:

\begin{equation} (|5\rangle\langle2|)|\phi\rangle=\langle2|\phi\rangle|5\rangle \end{equation} That is, it gives you the ket $|5\rangle$ multiplied by the projection of the state $|\phi\rangle$ on $|2\rangle$.

In general, remember that an operator is defined by how it acts on the states of your vector space.

Can you now evaluate the expectation value?

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The sequence is the following $$ \langle\phi|Q|\phi\rangle=\langle\phi|5\rangle\langle 2|\phi\rangle. $$ The number 5 and 2 refer to $n=5$ and $n=2$ occupation values and so, to the states $|5\rangle$ and $|2\rangle$ respectively. Please, note that $$ \langle m| n\rangle=\delta_{mn} $$ with $m,n=1,2,\ldots$ and $\delta_{mn}=1$ for $m=n$ and = otherwise. Then, using all this, one has $$ \langle\phi|Q|\phi\rangle=\frac{6}{\pi^2}\sum_{m=1}^\infty\sum_{n=1}^\infty C_mC_n\langle m|5\rangle\langle 2|n\rangle. $$ Could you go on from here?

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  • $\begingroup$ I understand this - should it not be Cm*- so Cm is complex conjugated? $\endgroup$ – user289904 Feb 22 at 16:48
  • $\begingroup$ Your $C_n$ are real. $\endgroup$ – Jon Feb 23 at 7:08
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So first of all, your state seems to be well-defined and normalised, as, $$ \left\langle \phi\left|\phi\right\rangle \right.=1 $$ on account of, $$ \zeta(2)=\sum_{n=1}^{\infty}\frac{1}{n^{2}}=\frac{1}{1^{2}}+\frac{1}{2^{2}}+\frac{1}{3^{2}}+\frac{1}{4^{2}}+\cdots=\frac{\pi^{2}}{6} $$ https://en.wikipedia.org/wiki/Basel_problem

But you have a problem: $\left|5\right\rangle \left\langle 2\right|$ is not a self-adjoing operator, so it's not a valid observable. A self-adjoint extension of it would be: $$ \frac{1}{2}\left(\left|5\right\rangle \left\langle 2\right|+\left|2\right\rangle \left\langle 5\right|\right) $$ Once you do that, it's a simple matter of "sandwiching" the operator with the state in the <bra|ket> way and using orthogonality. Other way to interpret $\left|5\right\rangle \left\langle 2\right|$ is as a transition amplitude of going from $\left|5\right\rangle $ to $\left\langle 2\right|$. It's the words "expected value" that do not sit well with the particular operator you've proposed. I hope that was helpful, and I didn't make any idiotic mistake, to which I seem to be prone lately.

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    $\begingroup$ That the operator is not self-adjoint does not matter when computing the expectation value at all. You can, for example, calculate the expectation value of the ladder operator $a$ (or $a^\dagger$) in the framework of the quantum harmonic oscillator, although it is not self-adjoint. $\endgroup$ – Jakob Feb 22 at 16:12
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    $\begingroup$ Your point is well taken. Well, I will wait until the OP reveals more about the context. I'm not sure whether they're talking about the SHO or any other quantum system, for that matter. I wouldn't say it does not matter at all though. The OP seemed to be concerned with "what it means". I always try to encourage students to think about what they're doing, not blindly applying the mathematical formalism. My answer was in the spirit of, eg., @ACuriousMind's comment at: physics.stackexchange.com/questions/294175/… $\endgroup$ – joigus Feb 22 at 16:42
  • $\begingroup$ I completely agree! $\endgroup$ – Jakob Feb 22 at 17:01

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