If an experiment is performed where a pair of entangled electrons is generated, and then one of the pair is introduced into a metallic body, for the sake of argument say a one centimeter cube of copper, such that it becomes part of the "electron cloud" within that metal where the copper atoms are held together with metallic bonds, does that electron retain its entangled state with the entangled electron outside of the metal? - or does becoming part of the electron cloud of the metallic body and the ensuing interactions cause a collapse of its wave function?

Further, if the metal is heated, and the entangled electron within the metal is emitted from the metal (along with countless others) via thermionic emission into a surrounding vacuum, does it still remain in its entangled condition, or does participation of the electron in thermionic emission collapse the wave function?

Thanks for your answers!

  • $\begingroup$ Entanglement is correlating the momentum, speed, direction,etc. of the two particles. When you change anything about one of the particles then the two are no longer correlated. $\endgroup$ Oct 24, 2021 at 17:15
  • 3
    $\begingroup$ @BillAlsept That's not correct. Changing a property of one of the particles doesn't undo the entanglement. As an example, consider the entangled state $|0\rangle|0\rangle+|1\rangle|1\rangle$. Apply the unitary transformation $|0\rangle\leftrightarrow|1\rangle$ to the second particle. Then the new state is $|0\rangle|1\rangle+|1\rangle|0\rangle$. Still entangled. We can't undo the entanglement of two particles by applying a unitary transformation to only one of them. $\endgroup$ Oct 24, 2021 at 17:48
  • $\begingroup$ @ChiralAnomaly The math doesn’t even attempt to explain what’s physically happening. The OP is asking if one of the electrons is physically affected by the metal what would happen. Well, physically their trajectories would be different after that and no longer correlated. You have to remember both of the electrons are physically (really) doing something. The math makes you lose focus of that. You cannot physically describe what entanglement is other than correlating them. $\endgroup$ Oct 24, 2021 at 17:56
  • 1
    $\begingroup$ @BillAlsept You're right that the math that I showed in the comment doesn't attempt to explain what's physically happening. It was not meant to be an answer to the OP's question. It was only a counterexample to the assertion in your first comment, assuming that we're talking about quantum theory. If your assertion is based on a different theory, then maybe you could clarify what theory you had in mind. Nobody is omniscient, so aside from a rote list of experimental data, we can't talk about what is "physically happening" without assuming some kind of theory. $\endgroup$ Oct 24, 2021 at 18:36
  • 1
    $\begingroup$ @ChiralAnomaly: As you might or might not be aware, Bill Alsept is going to say the things he wants to say about quantum entanglement no matter how many times and in how many ways people explain to him that those things are entirely incorrect. There is no point in arguing, but I believe it's useful to mention this in comments now and then just to warn newcomers that Bill's comments about quantum entanglement are always and everywhere wrong. $\endgroup$
    – WillO
    Oct 25, 2021 at 1:13

1 Answer 1


Of course a unitary transformation on one half of your entangled system can't disentangle it, so I'll assume that whatever you're doing to your electron is non-unitary, and hence equivalent to some observation. Let $A_1,\ldots A_n$ be the eigenstates of that observation, which we can take to ve a basis for the electron's state space, with $v_1\ldots v_n$ the corresponding eigenvalues.

Let $X_1,\ldots, X_n$ be a basis for the state space of the other electron. (I am taking the state spaces to be finite dimensional for simplicity.)

Suppose your entangled pair is initially in the state $$S=\Sigma_{i,j}\alpha_{ij}A_i\otimes X_j$$


$$T_i=\Sigma_j\alpha_{ij}A_i\otimes X_j$$ so that $S=\Sigma_iT_i$.

Let $v$ be the outcome of your observation. Then the new state of the pair is (proportional to) $$\Sigma_{v_i=v}T_i$$ which (clearly) is entangled if and only if there is more than one $i$ with $v_i=v$ and $T_i\neq 0$.


Your Answer

By clicking “Post Your Answer”, you agree to our terms of service and acknowledge you have read our privacy policy.

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