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I understand the proof for the expectation value of $\langle\hat{Q}\rangle$, which is shown for reference. Note, if you are already familiar with this proof then there is no need to read the contents of the quote below:

$$\langle\hat{Q}\rangle=\int_{x=-\infty}^{\infty}\psi^*\hat Q\,\psi \,dx=\int_{-\infty}^{\infty}\left(\sum\limits_m a_m\phi_m\right)^*\hat Q\color{blue}{\sum\limits_n a_n\phi_n}\,dx\tag{1}$$ where the parts marked blue hold because arbitrary $\psi$ can be written as a sum of eigenvalues $\phi_n$ with expansion coefficients $a_n$ such that

$$\psi=\color{blue}{\sum\limits_n a_n\phi_n}$$ Continuing with the proof from $(1)$:

$$\int_{-\infty}^{\infty}\left(\sum\limits_m a_m\phi_m\right)^*\hat Q\sum\limits_n a_n\phi_n\,dx=\int_{-\infty}^{\infty}\sum\limits_m {a_m}^*{\phi_m}^*\sum\limits_n a_n\,\color{red}{q_n\phi_n}\,dx\tag{2}$$ where the red part arises from the fact that $\hat Q$ satisfies the eigenvalue equation: $$\hat Q\,\phi_n=\color{red}{q_n\phi_n}$$ where $q_n$ is the associated eigenvalues with the operator $\hat Q$.

Continuing the proof from $(2)$:

$$\int_{-\infty}^{\infty}\sum\limits_m {a_m}^*{\phi_m}^*\sum\limits_n a_n\,q_n\phi_n\,dx=\sum\limits_{m,n}{a_m}^*a_n\,q_n\color{#180}{\int_{-\infty}^{\infty}{\phi_m}^*\phi_n \,dx}\tag{3}$$ where the summations have been taken out the integral since they have no $x$ dependence such that the integrand comprises solely of eigenfunctions (that depend on $x$). Also, the part marked green is the Kronecker delta $\color{#180}{\large\delta_{m n}}$.

So, resuming the proof from $(3)$:

$$\sum\limits_{m,n}{a_m}^*a_n\,q_n\large\delta_{m n}=\sum\limits_n {a_n}^*a_n\,q_n=\sum\limits_n|a_n|^2q_n\tag{4}$$

Since $$\large\delta_{m n}=\begin{cases}0 & m\ne n \\ 1 & m=n\end{cases}$$

Therefore we conclude from $(4)$ that

$$\langle\hat{Q}\rangle=\sum\limits_n|a_n|^2q_n$$


I fully understand the proof that $$\langle\hat{Q}\rangle=\sum\limits_n|a_n|^2q_n$$

So now I need to proof that $$\Big\langle{\hat{Q}}^2\Big\rangle=\sum\limits_n|a_n|^2{q_n}^2$$


Here is my attempt:

I know from the eigenvalue equation for $$\hat Q\,\phi_n=q_n\phi_n\tag{A}$$ So if I operate $\hat Q$ on both sides of $(\mathrm{A})$:

$$\hat Q\left(\hat Q\phi_n\right)=\hat Q \left(q_n\phi_n\right)$$ $$\implies {\hat Q}^2\phi_n=q_n\hat Q\phi_n$$ $$\implies {\hat Q}^2\phi_n={q_n}^2\phi_n\qquad\text{using the RHS of (A)}$$

From the chain rule $$\frac{du}{dx}=\frac{dy}{dx}\frac{du}{dy}$$ operating on some function $u$, I understand from this post that the chain rule can be written as $$\frac{d}{dx}=\frac{dy}{dx}\frac{d}{dy}$$ even in the absence of a function $u$ for which it can operate on.

So does this mean I can write $${\hat Q}^2\phi_n={q_n}^2\phi_n\tag{B}$$ as $${\hat Q}^2={q_n}^2$$ where I have effectively ignored the eigenfunctions $\phi_n$.

If I now replace $\hat{Q}$ with ${\hat{Q}}^2$ in $$\langle\hat{Q}\rangle=\sum\limits_n|a_n|^2q_n$$ I will get $$\big\langle \hat{Q}^2\big\rangle=\sum\limits_n |a_n|^2{q_n}^2\tag{C}$$ as required.

Is this proof valid and if it's not valid what is the correct way to derive $(\mathrm{C})$?


EDIT:

In response to the first comment which mentions that with arbitrary operator $\hat O$ $$\langle\hat{O}\rangle=\sum\limits_n|a_n|^2o_n$$ for any operator so I can simply put $$\hat O=\hat Q^2$$ and somehow $$\langle\hat{Q}^2\rangle=\sum\limits_n|a_n|^2{q_n}^2$$ I do not understand the logic of this approach. More specifically why does letting $$\hat O=\hat Q^2$$ allow me to write $$\langle\hat{Q}^2\rangle=\sum\limits_n|a_n|^2{q_n}^2\,?$$ If anyone could please elaborate by giving some more details to this I will be most grateful, thanks.


EDIT 2:

I have been given a good answer now by 2 users especially AccidentalFourierTransform. I acknowledge now that eigenfunctions can never be ignored like I was doing before. I know that the $|a_n|^2$ are probabilities, but does the $a_n$ have some particular relationship to the eigenvalue equation $${\hat Q}^2\phi_n={q_n}^2\phi_n\,?$$ If so, this could be the source of my confusion and I would like to know how they are related?

Thanks again.

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    $\begingroup$ Your proof of $\langle\hat{Q}\rangle=\sum\limits_n|a_n|^2q_n$ is fine. In fact, it can be generalised to $\langle\hat{O}\rangle=\sum\limits_n|a_n|^2o_n$. With this, you dont need to repeat the proof for $\hat Q^2$. It suffices to take $\hat O=\hat Q^2$. In other words, your first proof does already contain the second one as a subcase; no need to repeat it twice. $\endgroup$ – AccidentalFourierTransform Dec 8 '16 at 15:05
  • $\begingroup$ @Accidental Thanks for your response; So does this really mean that I can write the eigenvalue equation $(\mathrm{B})$ for $\big\langle \hat{Q}^2\big\rangle$ without the $\phi_n$s? This seems mathematically incorrect to me. $\endgroup$ – BLAZE Dec 8 '16 at 15:16
  • $\begingroup$ No, you can't. $(B)$ is fine as written. You cannot drop the $\phi_n$'s. (Some people will do it anyway, for some reason that is beyond me). $\endgroup$ – AccidentalFourierTransform Dec 8 '16 at 15:18
  • $\begingroup$ @AccidentalFourierTransform So my proof is incorrect? $\endgroup$ – BLAZE Dec 8 '16 at 15:19
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    $\begingroup$ but an operator equation and an eigenvalue equation are not the same thing. Let $I$ be the identity operator. For any $v$ you have $Iv=v$ and so you can write $I=1$. Now let $A$ be some operator that happens to have a certain eigenvalue $a$, with eigenvector $v$: $Av=av$. You cannot write $A=a$, because $A$ may have other eigenvectors different from $a$. In other words, $Av=av$ holds only for $v$, and therefore you cannot drop the $v$. You can only drop the vector if the equation holds for all vectors, and not only for one vector in particular. $\endgroup$ – AccidentalFourierTransform Dec 8 '16 at 15:34
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Let $O$ be an arbitrary operator with eigenvectors $|n\rangle$ and eigenvalues $o_n$: $$ O|n\rangle\equiv o_n|n\rangle \tag{1} $$

Then, as proven in the OP, the expectation value of $O$ is $$ \langle O\rangle=\sum_n |a_n|^2 o_n \tag{2} $$

For example, if $O=Q$ then $o_n=q_n$ and $$ \langle Q\rangle=\sum_n |a_n|^2 q_n\tag{3} $$

On the other hand, if $O=Q^2$ then $o_n=q_n^2$ and $$ \langle Q^2\rangle=\sum_n |a_n|^2 q^2_n \tag{4} $$

For generally, if $O=f(Q)$ then $o_n=f(q_n)$ and $$ \langle f(Q)\rangle=\sum_n |a_n|^2 f(q_n) \tag{5} $$

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    $\begingroup$ @BLAZE it is actually simple: here $o_n$ is the eigenvalue of $O$. If the eigenvalue of $Q$ is $q$, then the eigenvalue of $Q^2$ is $q^2$ (you proved this yourself in the OP, when you said ${\hat Q}^2\phi_n={q_n}^2\phi_n$. Therefore, if $O=Q^2$ then $o=q^2$. Do you see it now? or not yet? If not, I'll try to explain it some other way. $\endgroup$ – AccidentalFourierTransform Dec 8 '16 at 18:01
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    $\begingroup$ @BLAZE If I got you right, you understand the proof of $\langle Q\rangle =\sum |a_n|^2q_n$, right? well, the proof of $\langle Q^2\rangle=\sum|a_n|^2q_n^2$ is exactly the same, but replacing $Q$ with $Q^2$. In other words, follow the proof in the quote step by step, but every time you see a $Q$, write $Q^2$. The only difference is that instead of $\hat Q\,\phi_n=\color{red}{q_n\phi_n}$ you have $\hat Q^2\,\phi_n=\color{red}{q^2_n\phi_n}$, but everything remains unchanged. Do you follow this or not? $\endgroup$ – AccidentalFourierTransform Dec 8 '16 at 18:44
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    $\begingroup$ BTW: the $a_n$ are the coefficients of $|\psi\rangle$, and therefore they contain the information about the properties of $\psi$. They are unrelated to the eigenvalue equation. If you change the system into $\psi'$, the amplitudes change into $a'_n$, but the eigenvalue equation for $Q$ stays the same. $\endgroup$ – AccidentalFourierTransform Dec 8 '16 at 18:46
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    $\begingroup$ @BLAZE no, there is nothing that justifies that. In general you cannot replace an operator into other. What I am trying to make you understand is that in this case, the algebraic manipulations that lead to $\langle Q\rangle=\sum_n |a_n|^2q_n$ are valid for $Q^2$ too. In other words, everything the quoted proof does is also valid for $Q^2$. Step by step, the proof is also valid for $Q^2$. The proof may not be valid for some other operator, but for $Q^2$ it is. $\endgroup$ – AccidentalFourierTransform Dec 8 '16 at 18:49
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    $\begingroup$ Ok, let me see if I can fit it into a comment. Let $|\psi\rangle=\sum a_n|\phi_n\rangle$. We want to calculate $\langle\psi|Q^2|\psi\rangle$. If we plug $|\psi\rangle=\sum a_n|\phi_n\rangle$ into this expression, we get $\langle\psi|Q^2|\psi\rangle=\sum_{n,m}a_n a_m^* \langle\phi_m|Q^2|\phi_n\rangle$. So far so good? If yes, then we use $Q^2|\phi_n\rangle=q^2_n|\phi_n\rangle$. Still with me? Plugging this into our expression from before, we get $\langle\psi|Q^2|\psi\rangle=\sum_{n,m}a_n a_m^* q_n^2 \langle\phi_m|\phi_n\rangle$. (1/2) $\endgroup$ – AccidentalFourierTransform Dec 8 '16 at 18:57
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In Dirac's notation, if your operator $\hat Q$ (from now on, I will omit the hat over the operator Q) is an hermitian operator, the operator is equal to his conjugate transpose and you can say that: $$ \left \langle Qv| w \right \rangle = \left \langle v| Qw \right \rangle $$ and this is true if the Q operator is associated with an observable quantity. Therefore, can write: $$ \left \langle \psi | Q^2 | \psi \right \rangle = \left \langle \psi | Q \cdot Q| \psi \right \rangle = $$ Making one operator work on the bra and the other on the key, using the eigenvalues equation with q as the eigenvalue, and using linearity property: $$ = \left \langle Q\psi | Q \psi \right \rangle = \left \langle q\psi | q \psi \right \rangle = q^2 \cdot \left \langle \psi | \psi \right \rangle = q^2 $$ This demonstration, valid for a $\psi$ that is a single state, can be generalized for a superposition of states remembering that: $$\left \langle \psi_i | \psi_j \right \rangle =0 $$ if i is different from j. In this way you can obtain the equation you have to demonstrate.

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