# I Measuring entangled state of degenerate eigenstates

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1. Mar 21, 2016

### maxverywell

Let's suppose that we have an entangled state of two systems $A$ and $B$:
$$\frac{1}{2}\left(|\psi_1 \phi_1\rangle+|\psi_2 \phi_2\rangle \right)$$
where $|\psi \rangle$ and $|\phi \rangle$ are energy eigenstates of $A$ and $B$ respectively. However the eigenstates$|\phi_1\rangle$ and $|\phi_2\rangle$ are degenerate:
$$\hat{H}_B|\phi_1\rangle=E|\phi_1\rangle$$
$$\hat{H}_B|\phi_2\rangle=E|\phi_2\rangle$$
What will be the state of the system $AB$ after measuring the energy of $B$ and finding the value $E$?
My guess is:
$$\frac{\left(|\psi_1\rangle+|\psi_2 \rangle \right)}{\sqrt{2}} \frac{\left(|\phi_1\rangle+\phi_2\rangle \right)}{\sqrt{2}}$$

Last edited: Mar 21, 2016
2. Mar 21, 2016

### Jilang

Has the measurement revealed anything about the state that you didn't already know?

3. Mar 21, 2016

### jfizzix

If the state is
$\frac{1}{\sqrt{2}}\Big(|\psi_{1}\rangle|\phi_{1}\rangle +|\psi_{2}\rangle|\phi_{2}\rangle\Big)$
And
$\hat{H}_{B}|\phi_{1}\rangle= E|\phi_{1}\rangle$ and $\hat{H}_{B}|\phi_{2}\rangle =E|\phi_{2}\rangle$
then measuring the energy of B without touching A has the effect of applying the operator $\big(I\otimes\hat{H}_{B}\big)$ onto the state, giving us
$\big(I\otimes\hat{H}_{B}\big)\frac{1}{\sqrt{2}}\Big(|\psi_{1}\rangle|\phi_{1}\rangle +|\psi_{2}\rangle|\phi_{2}\rangle\Big)$.
As a result, the state us unaltered, up to a constant factor:
$\frac{1}{\sqrt{2}}\Big(\big(I\otimes\hat{H}_{B}\big)|\psi_{1}\rangle|\phi_{1}\rangle +\big(I\otimes\hat{H}_{B}\big)|\psi_{2}\rangle|\phi_{2}\rangle\Big)$
$=E\frac{1}{\sqrt{2}}\Big(|\psi_{1}\rangle|\phi_{1}\rangle +|\psi_{2}\rangle|\phi_{2}\rangle\Big).$

4. Mar 21, 2016

### maxverywell

I think that measuring energy and acting on the state with the Hamiltonian operator is not the same thing. For example, if the state is a superposition of energy eigenstates, then this state is also an eigenstate of the Hamiltonian, and acting with the Hamiltonian operator on this state doesn't change it. However, measuring the energy will change the state ("collapse it") to one of the eigenstates that compose the initial superposition state.

In my example It seems wrong that measuring the energy of the system B won't have any effect on its state or the state of the whole system AB. Think that if the states $\phi_1$ and $\phi_2$ weren't degenerate (i.e. $E_1 \neq E_2$), then after the measurement the state of the system B would be $\phi_1$ or $\phi_2$, therefore the state of the AB becomes $\psi_1 \phi_1$ or $\psi_2\phi_2$, depending on the outcome of the measurement ($E_1$ or $E_2$). In the degenerate case, the measurement of the energy cannot distinguish this two states in principle, thus creating a superposition of them.

5. Mar 22, 2016

### maxverywell

Now that I have rethought it, I think that jfizzix is right, but cannot delete my last post.