I am considering the problem of two coupled harmonic oscillators. Ignoring factors of two, the Hamiltonian for this system is $$H=p_1^2+p_2^2 +k x_1^2+kx_2^2 +k(x_2-x_1)^2$$ One can do a nice coordinate transformation and gets: $$H=\hat{p}_1^2+\hat{p}_2^2 +k \hat{x}_1^2+3k\hat{x}_2^2$$

This is just two uncoupled harmonic oscillators and the solution of the Schrödinger equation with this Hamiltonian is just that of two independent harmonic oscillators. If I now look at the ground state I get of course two unentangled gaussian wavefunctions. But if I now do the reverse of the coordinate transformation I get unseperable wave functions for the ground state. But how can entanglement be basis dependent? Or does it not make sense to talk about entanglement here because these are only pure states?

EDIT: I want to add another question here. Is the ground state in the basis of the new coordinates the ground state in the original basis? Or is this basis independent. If not is there a way to transform the number operators?


1 Answer 1


Entanglement always depends on your decomposition of Hilbert space. If you locally change basis within one mode, nothing happens to your entanglement, but if your basis change is actually changing what you refer to as the two modes of your Hilbert space, you will produce entanglement.

This coordinate transformation causing entanglement shows up in many places. For example, if you solve the hydrogen atom quantum mechanically, you normally do so in the centre of mass coordinate system. But if you go back to the coordinate system in which the electron and proton can both move through free space, you'll find their momenta to be entangled.

As another example, if you are talking about polarization, a state that is separable in the horizontal/vertical mode decomposition is in general entangled in the diagonal/antidiagonal mode decomposition. There are going to be some states that don't become entangled when you do this kind of mode transformation, but most states that are separable in one decomposition of your Hilbert space will be entangled in another decomposition.

So the main physical question with entanglement is how you choose to decompose your Hilbert space! And that's really a fundamental question that still needs to be answered. For practical purposes, sometimes entanglement is useful, so that can help you choose your preferred decomposition to quantify the usefulness of a state for some particular task.

  • $\begingroup$ That's really interesting. Does that mean, that sometimes some systems are "useful" in some basis? In this example of coupled harmonic oscillators can one use the entanglement in the bases in which they are entangled? $\endgroup$
    – eeqesri
    Commented Feb 28, 2022 at 14:42
  • 1
    $\begingroup$ @eeqesri absolutely. I always think of NOON states, which are optimal for interferometry. These states are a superposition of being [$N$th excited state of the first mode and ground state of the second mode] and [ground state of the first mode and $N$th excited state of the second mode]. If you are trying to sense a relative phase acquired between these two modes, then NOON states are optimal. But if the relative phase is acquired between the two modes in the transformed coordinates, the original NOON state is no longer optimal! Instead, a NOON state in the new coordinates is optimal $\endgroup$ Commented Feb 28, 2022 at 15:21

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