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!!

  • 1
    $\begingroup$ Which author? Which book? $\endgroup$ – Ryan Unger Feb 26 '16 at 5:02
  • $\begingroup$ The equation is valid provided $|\Psi\rangle$ is stationary state of $\hat{Q}$ $\endgroup$ – user36790 Feb 26 '16 at 5:02
  • $\begingroup$ @ocelo7 Griffiths, Introduction to QM $\endgroup$ – Oswald Feb 26 '16 at 5:04
  • $\begingroup$ By $\int {\Psi^* (x)\Psi\, \mathrm dx}$, do you mean $\int {\Psi^* (x)\, x\,\Psi(x)\, \mathrm dx}$? If so, that's some terrible notation there. $\endgroup$ – Emilio Pisanty Feb 26 '16 at 10:19
  • $\begingroup$ @EmilioPisanty Yes, even Griffiths uses the same notation, even I found it quite confusing in the beginning $\endgroup$ – Oswald Feb 26 '16 at 10:23

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.


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