Hi I just want to confirm my interpretation of the following question:

Let the quantum state be given as $$|\psi_0 \rangle = [\sqrt{2}|\phi_1 \rangle + \sqrt{3}|\phi_2 \rangle + | \phi_3 \rangle + |\phi_4 \rangle]/ \sqrt{7}$$ and $$\hat{H}|\phi_n \rangle = n^2 \epsilon_0 |\phi_n \rangle~~\text{and}~~\hat{A}| \phi_n \rangle = (n+1)a_0| \phi_n \rangle.$$

Consider the following sequence of measurements: Suppose the measurement of the energy yields $4 \epsilon_0$. We then measure the energy again followed by a measurment of $A$. What state and measurement do we get for each step:

Proposed answer: Since the energy yields $4 \epsilon_0$, we are in state $| \phi_2 \rangle$. Then a measurement of the energy yields: $$\hat{H}| \phi_2 \rangle = 4 \epsilon_0 |\phi_2 \rangle,$$ hence we measure $4 \epsilon_0$ and we are in the state $4 \epsilon_0 |\phi_2 \rangle$. If we now measure $A$ then we get $$\hat{A}\hat{H}|\phi_2 \rangle = 4 \epsilon_0 \hat{A} |\phi_2 \rangle = 12 \epsilon_0 a_0 |\phi_2 \rangle.$$ Hence we are in the state $12 \epsilon_0 a_0 |\phi_2 \rangle$ with measurement of $A$ given by $12 \epsilon_0 a_0$.

Is this fine?

  • 1
    $\begingroup$ There is no difference between the state $|\phi_2\rangle$ and the state $4\epsilon_0|\phi_2\rangle$. $\endgroup$ – WillO Jun 7 '16 at 15:46
  • $\begingroup$ @WillO Okay but how is there no difference? If you project onto the position eigenvector $\langle x |$ you get $$\phi_2(x) = \langle x | \phi_2 \rangle$$ and $$4 \epsilon_0 \phi_{2}(x) = 4 \epsilon_0 \langle x | \phi_2 \rangle = 4 \epsilon_0 \phi_2(x).$$ How are these not different? $\endgroup$ – Alex Jun 7 '16 at 15:51
  • 1
    $\begingroup$ @Alex you are supposed to normalize when calculating observables, or probability densities. $\endgroup$ – Jahan Claes Jun 7 '16 at 15:52
  • $\begingroup$ @JahanClaes Oh yes I agree now. I just realized that after sending that comment :) $\endgroup$ – Alex Jun 7 '16 at 15:54

You're wrong in step 2. We don't go to $4\epsilon_0|\phi_2\rangle$. We go to $|\phi_2\rangle$. Measuring an observable projects you into an eigenstate of that observable, but it does NOT mulitply your state by the eigenvalue of that observable.

In most cases, as @WillO said, multiplying by the eigenvalue isn't even a measurable operation, since you're supposed to normalize your eigenstate when calculating observables. However, imagine if you had a state with energy zero, $|\phi_0\rangle$. Then if you mulitply the state by it's energy, you get an undefined ket, $0|\phi_0\rangle=0$, that cannot be normalized and is thus not part of our state space.

  • 1
    $\begingroup$ I'd have said this a little differently ---- of course you go to $4\epsilon_0|\phi_2\rangle$. You also go to $|\phi_0\rangle$, to $\pi|\phi_0\rangle$, to $\sqrt{137}|\phi_0\rangle$ and to $-196883|\phi_0\rangle$, because these are all different names for the same thing. $\endgroup$ – WillO Jun 7 '16 at 15:56
  • $\begingroup$ @WillO absolutely. I just wanted to make the point that if $\epsilon_0=0$, you get garbage. $\endgroup$ – Jahan Claes Jun 7 '16 at 17:37

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.