# Annihilation Operator on Entangled State - How Does This Not Break No-Communication Theorem?

If I have the state

$$|\psi\rangle=\frac{1}{\sqrt{2}}\left(|2\rangle_{A}|0\rangle_{B}+|0\rangle_{A}|2\rangle_{B}\right)$$

and apply the annihilation operator to B

$$a_{B}|\psi\rangle=|\phi\rangle=\frac{1}{\sqrt{2}}\left(a_{B}|2\rangle_{A}|0\rangle_{B}+a_{B}|0\rangle_{A}|2\rangle_{B}\right)$$ $$|\phi\rangle=\frac{1}{\sqrt{2}}\left(0+\sqrt{2}|0\rangle_{A}|1\rangle_{B}\right)$$ $$|\phi\rangle=|0\rangle_{A}|1\rangle_{B}$$

So, if the state of A is measured after the annihilation operator for B is applied to $$|\psi\rangle$$, A can only be measured to be in state $$|0\rangle$$, whereas if B is not annihated, then A can be measured to be in $$|0\rangle$$ or $$|2\rangle$$ with equal probability.

Where have I gone wrong?

For an example of a physical process I believe corresponds to this, imagine the case of a box where you know that you have either 2 muons, or 2 electrons in the box. And you have a method to remove 1 electron from the box, but that won't remove any muons. You'd therefore start in state $$|\psi\rangle$$ above, where state A is number of muons and state B is number of electrons.

Removing an electron I believe would be the same as applying $$a_{B}$$, so you would get to state $$|\phi\rangle$$. Hence when you observed the particle content of the box, you would always find 0 muons and 1 electron. However this clearly isn't the case, and you should find either 2 muons and 0 electrons, or 0 muons and 1 electron with equal probability.

Where have I gone wrong?

• What physical process corresponds to applying the annihilation operator? Are you thinking of a measurement of the number observable $(a_B)^\dagger a_B$? If so, then this could be a good example of why it's important to think of an observable as a collection of projection operators labelled by eigenvalues. One of the eigenvalues happens to be zero, but that doesn't mean the resulting state is the zero-vector. But I'm not sure if this is really the right interpretation of your question. – Chiral Anomaly Jan 28 at 0:37
• The annihilation operator $a_B$ is not hermitian. Hence it is not an observable. – Thomas Fritsch Jan 28 at 0:42
• @Jack Suppose that the initial state were the vacuum state (no electrons an no muons). What would happen if you removed an electron from this state? – Chiral Anomaly Jan 28 at 1:33
• @Jack We can move an electron from one box to another, or we can convert the electron to something else, but there is no physical process that starts with a state having $N$ electrons (and nothing else) and ends up with $N-1$ electrons (and nothing else). – Chiral Anomaly Jan 28 at 1:40
• @Jack Right. Physical processes correspond to unitary operators, which preserve inner products between state-vectors and, in particular, preserve the norms of the state-vectors. – Chiral Anomaly Jan 28 at 1:42

(For a quick understanding of why this is the case, suppose that you could do that, and instead of that superposition you were just given the vacuum state $$|\psi⟩ = |0⟩_B$$. Then your procedure would leave you with no state at all, and you'd be completely unable to describe the subsequent evolution of your system! Keep in mind that any physical operation needs to conserve probability, and yours doesn't.)
What you can do is to perform a projective-measurement procedure on the B side, where the $$|2⟩_B$$ state gets sent to $$|1⟩_B$$ and triggers a "particle present" click, whereas the vacuum $$|0⟩_B$$ state gets sent to some ancilla state $$|\mathrm a⟩_B$$ and triggers a "particle not present" click.