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In quantum mechanics, we generally take about "expectation values of dynamical variables". However, by the postulates of quantum mechanics, every dynamical variable in quantum theory is represented by its corresponding operator.

Is it therefore, incorrect to talk about "expectation value of an operator" (rather than "expectation value a dynamical variable")? Is is just semantics or is something more going on here?

In other words is it incorrect to write:

$$<\widehat{A} > =\int \psi \widehat{A}\psi^{*}\mathbf{ d^{3}r}$$

instead of

$$<A > =\int \psi \widehat{A}\psi^{*}\mathbf{ d^{3}r}?$$

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I feel this question has addressed the issue and dismissed it just as a matter of notation but I am more interested in knowing if it is conceptually okay to talk about the expectation value of an operator rather than a dynamical variable. –  AchiralSarkar Mar 23 at 8:11
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2 Answers 2

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Remember, operators are nothing but maps. Expectation value of an operator is pretty much defined (I guess) in general operator theory. It just turns out that in QM (Hermitian) operators correspond to dynamical variables. In general you can also calculate expectation values of operators like $L_+$ and $a^{\dagger}$ etc., which don't have any dynamical variables associated !!

On a more general note, if you want to get a representation for the operator (given some basis, which could be the eigenset of an operator), the expectation values would be the diagonal elements of the representation.

It is related to Linear algebra, This might throw some light on Operator theory. Linear operators (bounded of course) are maps defined in LVS that take one vector to another vector in the same LVS.

$$ \hat O : V \rightarrow V$$

$$ \hat O \left|\psi\right> = \left|\phi\right> $$ Then you have the inner product defined on the LVS, that takes two vectors to one complex number : $$ \left<\psi|\phi\right> : V^* \times V \rightarrow \mathbb{C} $$

Using these two, one can define the expectation value of an operator to be,

$$ <\hat O> = \left<\psi\right|\; \big(\hat O\left|\psi\right>\big) = \left<\psi|\phi\right> \in \mathbb{C} $$

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Can you please elaborate on "general operator theory". Are you talking about linear algebra in general? –  AchiralSarkar Mar 23 at 14:20
    
I have done some edits on the post, which is true to my knowledge. Note : These things are true only when the linear operators are bounded. –  user35952 Mar 23 at 14:50
    
So, we can view expectation values of an operator as nothing but as inner products? And this follows from Operator Theory? –  AchiralSarkar Mar 23 at 15:37
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Operator theory is a fancy name !! And yes, expectation values can be viewed as inner products. You can also refer to this answer, to clarify the fact how they work, let you to the probability interpretation. –  user35952 Mar 23 at 15:49
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Pardon me !! Inner product is a map between the LVS ($V$) and its dual ($V^*$), has been edited appropriately !! –  user35952 Mar 24 at 1:10
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Is it therefore, incorrect to talk about "expectation value of an operator"?

Yes, because when you write those integrals you're asking for the average of a dynamical variable $A$ whose associated operator is $\hat A$. The point is that you're asking for a number , e.g. the average position of the electron in a gaussian wavepaket. Then, when you want to calculate this expectation value $\langle x \rangle$, the associated operator shows up in the integrals. And it's not only semantics: what's the use of asking, let's say, "hey, what's the average of the Hamilton operator?" ? This would mean that you had a bunch of Hamilton operators and you wanted to know the average of those operators ... This makes no sense. ^^

You could ask for the average energy of a system: then you would use the Hamilton operator in the integrals you wrote.

So the answer to your second question (Is it incorrect to write ...) is simply yes.

EDIT: (This stuff is confusing...)

I just opened my Shankar and came across this:

enter image description here

See equation 7.3.2.: he is looking for the mean energy $\langle E \rangle$. So according to my reasoning, he writes $\langle \Psi|H|\Psi \rangle$. But then he writes $...=\langle H \rangle$, which confused me a lot. After some thinking I came up with the solution: here, he treats $H$ as an observable. And the way he treats the Hamiltonian the next section proves this: the momentum and position are treated like normal observables and not like operators. Then, the next page, he writes this:

enter image description here

So here he uses $\langle H \rangle$ as a synonym for the mean energy.

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Can you please take a look at this.Here, not only they the represent the expectation values by the first integral (with a hat) but also calculates the expectation value of the annihilation operator $\widehat{a}$ which has no corresponding dynamical variable. (Page 6 of the pdf) –  AchiralSarkar Mar 23 at 12:41
    
Hm now you made me insecure. The only thing I can tell you is this: in two books I have at home they do it the way I told you (without the hat: $\langle A \rangle$). And thinking about it it just makes sense to me to say that you want to know the expectation value of a dynamical variable $\langle A \rangle$ and the operator then turns up in the integral. Maybe some texts use the same symbol for the dynamical variable and operator? And for $\hat a$ I think it's ok to write it like this since you really want to know the e.v. of the operator itself and not of a dynamical variable. –  NoEigenvalue Mar 23 at 13:40
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I just read it in your text (p.8 at the top): "Thus, suppose we are measuring an observable $\hat A$ with a continuous range of eigenvalues". So my reasoning is still ok, and it's just the notation of your text that produces confusion. –  NoEigenvalue Mar 23 at 13:42
    
So, it is alright if we talked about expectation values of observable or operator and use both the terms interchangeably? –  AchiralSarkar Mar 23 at 14:12
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To be precise, no. The confusion arises because of notation: in your text, operator and observable have hats. In my books, the hats are dropped at some point. It then depends on the context if there should be a hat or not: if you write the symbol for an expectation value of an observable ($\langle A \rangle$) - no hats. If you want to calculate the integrals (or bra-kets) necessary for the e.v. then you have to use the according operator - so there are hats. –  NoEigenvalue Mar 23 at 14:45
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