# State and measurement after a sequence of measurements

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?

• There is no difference between the state $|\phi_2\rangle$ and the state $4\epsilon_0|\phi_2\rangle$. – WillO Jun 7 '16 at 15:46
• @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? – Alex Jun 7 '16 at 15:51
• @Alex you are supposed to normalize when calculating observables, or probability densities. – Jahan Claes Jun 7 '16 at 15:52
• @JahanClaes Oh yes I agree now. I just realized that after sending that comment :) – 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.
• 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. – WillO Jun 7 '16 at 15:56
• @WillO absolutely. I just wanted to make the point that if $\epsilon_0=0$, you get garbage. – Jahan Claes Jun 7 '16 at 17:37