Standard deviation for a determinate state of observable Q

[armis]

In general, identical measurements on identically prepared systems do not yield reproducible results; however, some states are determinate, for a particular observable, in the sense that they always give the same result. For a determinate state of observable Q, the standard deviation is zero:

0=\sigma^{2}_{Q}=\langle(\hat{Q}-{\langle}Q{\rangle})^2\rangle=\langle\psi\mid(\hat{Q}-{\langle}Q{\rangle})^2\psi\rangle


J.Griffiths

There are two things I don't understand here:
Firstly, why in the standard deviation expression we use the Q operator instead of the Q value itself? Writting Q seemed more natural and then in the next expression I would just stick the corresponding operator. Which would lead to the same result or is it just two errors cancelling each other?
Secondly, why in the last expression we are not using an operator of the quantity that was in the previous expression under the "average" brackets? Is it an operator of itself like coordinate x is for example?

 

[malawi_glenn]

Expectation value of a number is just a number, you want the operator Q to act on the state to find its expectation value.

So for instance <psi|<Q> |psi> = <Q>
Nothing "happened" since <Q> is just a number.

but what is <Q> ? that you must evaluate.

You could recapitulate what standard deviation and expectation value is in theory of statistics in math.

 

[armis]

Expectation value of a number is just a number, you want the operator Q to act on the state to find its expectation value.

So for instance <psi|<Q> |psi> = <Q>
Nothing "happened" since <Q> is just a number.

This I know

but what is <Q> ? that you must evaluate.

Is there a difference? I think there is none. Feel free to correct me

Let me clarify. First of all I don't understand why in this expression
\sigma^{2}_{Q}=\langle(\hat{Q}-{\langle}Q{\rangle})^2\rangle
we use the operator Q instead of the value Q.
For example if we make repeated measurements on identically prepared systems we may get a bunch of different Q values, we take their average and hence we can calculate the standard deviation. So why use an operator?

 

[malawi_glenn]

difference between what?

Okay let's make an example.

Caclulate the standard deviation for the Energy for this wave function:

\Psi _E = 2\psi _{E1} -3\psi _{E2} + 4\psi _{E3}

Where \psi _{Ei} is the energy eigenstate to the Hamiltonian with energy i MeV.

What is the "value of E" here? It is meaningless to talk about it before one has operated on the wavefunction with operator.

You want to calculate the standard deviation, not measure it.

 

[armis]

Cool, I like this kind of approach :)

Well, I think it would look like this, right?

\sigma^{2}_{H}=\langle(H-{\langle}H{\rangle})^2\rangle=\langle\psi\mid(\hat {H}-{\langle}H{\rangle})^2\psi\rangle

Notice I didn't write the hat on H in the second expression ( the part I don't understand )

Then I would stick in the wave equation you gave me and start multiplying the members. The orthogonal eigenstates would give zero while others would sum up to a number thus I will end up with a certain value. Is that correct?

 

[malawi_glenn]

no you must "hat" them to denote that they are operators.

&lt;(\hat{H} - &lt; \hat{H} &gt;)^2 &gt; = &lt;\hat{H} \hat{H} &gt; - <br /> &lt; \hat{H} &gt;^2

Then you can start with:
&lt;\hat{H}&gt; = \int (2\psi _{E1}^* -3\psi _{E2}^* + 4\psi _{E3}^*)\hat{H}(2\psi _{E1} -3\psi _{E2} + 4\psi _{E3}) =
...

You must learn what is meant by &lt; \hat{Q} &gt;

It is the expectation value for the operator \hat{Q}

What if Q operator was the momentum operator in position space for instance?

You must also learn what the relation between operators and states is.

 

[armis]

So what you are saying is that I can already write the operator within the "average" brackets like this <br /> &lt; \hat{Q} &gt; <br />?
I thought I was not allowed to do that and had to write <br /> &lt; Q &gt; <br />
instead and only once I am trying to compute the expectation value I sandwich the corresponding operator between the wave functions.

 

[malawi_glenn]

The average of a value is simply the value, the average of an operator is totally different.

 

[omephy]

What is the the average of an operator ?