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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.

Community
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BLAZE
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    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. – AccidentalFourierTransform Dec 08 '16 at 15:05
  • @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. – BLAZE Dec 08 '16 at 15:16
  • 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). – AccidentalFourierTransform Dec 08 '16 at 15:18
  • @AccidentalFourierTransform So my proof is incorrect? – BLAZE Dec 08 '16 at 15:19
  • I simply cannot tell. I got lost when you mentioned the chain rule. What does it have to do with anything? You dont need the chain rule at all here. – AccidentalFourierTransform Dec 08 '16 at 15:25
  • @AccidentalFourierTransform I know the chain rule is not needed here, I just used that as an example of where an operator equation is written without a function for it to act on. From that I made the connection to writing $(\mathrm{B})$ without the $\phi_n$s. You stated in your first comment that this is a correct proof. Perhaps you should delete that comment now that you 'simply cannot tell'. Regards. – BLAZE Dec 08 '16 at 15:31
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    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. – AccidentalFourierTransform Dec 08 '16 at 15:34
  • @Accidental Thanks, that makes sense. I didn't know that the differential operator and eigenvalue equation could not be manipulated in the same way. Would you please see my edit and consider making an answer to elaborate on your first comment? As I still don't understand, sorry. – BLAZE Dec 08 '16 at 16:31

2 Answers2

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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 $$ \langle f(Q)\rangle=\sum_n |a_n|^2 f(q_n) \tag{5} $$

AccidentalFourierTransform
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    @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. – AccidentalFourierTransform Dec 08 '16 at 18:01
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    @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? – AccidentalFourierTransform Dec 08 '16 at 18:44
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    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. – AccidentalFourierTransform Dec 08 '16 at 18:46
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    @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. – AccidentalFourierTransform Dec 08 '16 at 18:49
  • Okay, that was helpful thanks. But is there a rigorous proof of this that uses the same method as that in the quoted part of my question. As I am not satisfied with doing things that aren't justified. I come from more of a mathematical standpoint on this one. – BLAZE Dec 08 '16 at 18:53
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    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) – AccidentalFourierTransform Dec 08 '16 at 18:57
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    (2/2) If you are following this, then we only need to use $\langle\phi_m|\phi_n\rangle=\delta_{n,m}$. If we plug this into our formula from before, we get $\langle\psi|Q^2|\psi\rangle=\sum_{n}|a_n|^2q_n^2 $, as we wanted to show. Did you follow these steps? or is there any one in particular that you don't feel comfortable with? – AccidentalFourierTransform Dec 08 '16 at 18:59
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    Well, if $A$ is an arbitrary operator, then $\langle A\rangle\equiv \langle\psi|A|\psi\rangle$. In this case, we want to calculate $\langle Q^2\rangle$, which means that we substitute $|\psi\rangle$ into $\langle\psi|Q^2|\psi\rangle$. – AccidentalFourierTransform Dec 08 '16 at 19:07
  • I have read it very carefully and I at least think I understand you now, thank you ever so much for your time. The main confusion was that I am not familiar with Dirac's notation. So it could have been done by writing $\langle\psi|Q^2|\psi\rangle=\large\int_{-\infty}^{\infty}\phi^*Q^2\phi dx$ right? – BLAZE Dec 08 '16 at 19:26
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    @BLAZE no problem. You can write $\langle\psi|Q^2|\psi\rangle$ as an integral too, but not exactly what you wrote. The correct expression is $\langle\psi|Q^2|\psi\rangle=\int_{-\infty}^{+\infty}\ \psi^*Q^2\psi\ \mathrm dx$, with wave function $\psi(x)$, not $\phi(x)$. – AccidentalFourierTransform Dec 08 '16 at 19:28
  • By the way, just a suggestion. If I were you I would place the proof you have provided in your actual answer. It will surely get an even better score :) – BLAZE Dec 08 '16 at 19:29
  • @BLAZE maybe at some other time. Now I need to go. If you have any more questions, feel free to ask. I will answer some other day, as now I don't have any more time. Cheers :-) – AccidentalFourierTransform Dec 08 '16 at 19:31
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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.

JackI
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