General expectation value

Question

I have a basic question related to finding expectation values of an operator $\hat{Q}$

We know that the expectation of $\hat{Q}$ (in the position space) is given by

$$\langle Q \rangle=\int {\Psi^* Q\Psi \,\mathrm dx} \tag 1$$

How do we know that the above equation is valid for all $\hat Q$?

I understand that expectation of $x$ is given by

$$\langle x \rangle=\int {\Psi^* (x)\Psi\, \mathrm dx}$$

so differentiating this wrt to $t$ we get the momentum expectation value $\langle p\rangle$, the author just derives $\langle p\rangle$ and tells for a general $\hat Q$ the expectation value is given by the eq(1)

One reason I can think of (which is probably wrong) is that any physical operator can be expressed as a combination of $x$ and $p$.

How do we know eq(1) gives the expectation value for any $Q$? Is there something I am missing out?

Any help appreciated!!

Answer 1

To be an observable an operator has to have real eigenvalues for that operator. And eigenvectors of that operator with different eigenvalues have to be orthogonal. And every state needs to be (a limit of a sequence of) linear combinations of eigenvectors of that operator. The set of eigenvectors with a fixed particular eigenvalue is called an eigenspace.

Next we need to say what a measurement does. A measurement takes a state $\Psi$ and orthogonally projects $(\Psi\mapsto P_\lambda \Psi)$ the state onto one of the eigenspaces (the eigenspace of eigenvalue $\lambda$) of that operator $\hat Q$ and it does so with probability $\langle P_\lambda\Psi,P_\lambda\Psi\rangle/\langle \Psi,\Psi\rangle$.

Now we know what the expectation value should be. It should be $$\sum_\lambda \lambda \langle P_\lambda\Psi,P_\lambda\Psi\rangle/\langle \Psi,\Psi\rangle$$ in other words each value weighted by the probability of getting that value.

If the state was already normalized then $\langle \Psi,\Psi\rangle=1$ so you get $\sum_\lambda \lambda \langle P_\lambda\Psi,P_\lambda\Psi\rangle$ and because of orthogonality of different eigenspaces you get $\sum_\lambda \lambda \langle \Psi,P_\lambda\Psi\rangle$ and because $P_\lambda\Psi$ is an eigenvector of $\hat Q$ with eigenvalue $\lambda$ you get it equals $\sum_\lambda \langle \Psi,\hat QP_\lambda\Psi\rangle.$ Next, use linearity (if $\hat Q$ is continuous) to get $\langle \Psi,\hat Q \sum_\lambda P_\lambda\Psi\rangle$ and now since $\Psi$ is a (limit of a) sum of eigenvectors, $\Psi=\sum_\lambda P_\lambda\Psi$ so you get $\langle \Psi,\hat Q\Psi\rangle$ for the expectation value.

Answer 2

The general "scheme" goes like this (I skip "annoying" details).

1. The expected value of $\hat Q$ over a state $|\Psi\rangle $ is formally defined as $\langle\Psi|\hat Q|\Psi\rangle$.

2. For any complete set of states $|a\rangle$ for some label $a\in A$, you can "resolve" the identity operator $I$ as $I =\sum_a |a\rangle \langle a|$, the sum being over the set of labels $A$. If $a$ is a continuous set then the sum is actually a formal integral, see this and this (for the "position" case).

3. Use the identity: $\langle\Psi|\hat Q|\Psi\rangle = \langle \Psi | I \hat Q I |\Psi\rangle = \sum_{a, a'} \langle\Psi|a\rangle \langle a|\hat Q |a'\rangle \langle a'|\Psi\rangle$.

4. Define the "wave function" as the complex number $\Psi(a) = \langle a |\Psi\rangle$. If $a$ is the position label $x$, then this is the usual wave function, see this question. The notation $c_a = \langle a |\Psi\rangle$ is also used, to stress that those are just coefficients of a linear combination: $|\Psi\rangle = \sum_a c_a |a\rangle$.

5. Using the above points: $\langle\Psi|\hat Q|\Psi\rangle=\sum_{a, a'} \Psi^*(a) \langle a|\hat Q |a'\rangle \Psi(a')$. You may stop here or make a convenient choice of basis: the one where $ \langle a|\hat Q |a'\rangle$ is diagonal.

6. If the states $|a\rangle$ are eigenvectors of $\hat Q$, namely $\hat Q |a\rangle=Q_a |a\rangle $ where $Q_a$ is a real number, then $\langle a|\hat Q |a'\rangle=Q_a \delta_{aa'}$. Therefore: $\langle\Psi|\hat Q|\Psi\rangle=\sum_{a } \Psi^*(a) Q_a \Psi(a')=\sum_{a } Q_a |\Psi(a )|^2$.

7. Clearly, from the above points we immediately have that $\langle\Psi|f(\hat Q)|\Psi\rangle=\sum_{a } f(Q_a) |\Psi(a )|^2$, see e.g. this question.