A question of separability and entanglement

[andresB]

I was reading a review on single-particle entanglement https://onlinelibrary.wiley.com/doi/10.1002/qute.202000014 and the author says this

It's probably a simple/well known result, but I'm not seeing any trivial proof at the moment. Moreover, I'm interested in the general case of arbitrary number of subsystems where the Hilbert space of the subsystems can be of any dimension (including a continuous degrees of freedom).

So, it's the result true in general for any composite system?

 

[PeterDonis]

t's the result true in general for any composite system?

Yes. It follows immediately from the fact that an entangled state, by definition, cannot be expressed as a single term of the form $\psi_1 \otimes \psi_2$, where $\psi_1$ and $\psi_2$ are states in the Hilbert spaces of subsystem 1 and subsystem 2.

 

[Jonomyster]

Here's a proof of the result - it actually seems to require that eigenstates of the observable are non-degenerate.

Let $\mathcal{H}_A$ and $\mathcal{H}_B$ be two Hilbert spaces, and let $\hat{\mathcal O}_A$ be a Hermitian operator on $\mathcal{H}_A$ representing the observable.

By the spectral theorem, let $\{ \ket{ \varphi_A^{(i)} } \}_i$ be a basis for $\mathcal{H}_A$ which has eigenvalues $\hat{\mathcal O}_A \ket{ \varphi_A^{(i)} } = \lambda_A^{(i)} \ket{ \varphi_A^{(i)} }$. Also let $\{ \ket{ \varphi_B^{(j)} } \}_j$ be a basis for $\mathcal{H}_B$. Then, a possible basis for the Hilbert space $\mathcal{H} = \mathcal{H}_A \otimes \mathcal{H}_B$ is $\{ \ket{ \varphi_A^{(i)} }\ket{ \varphi_B^{(j)} } \}_{ij}$.

To prove the theorem, consider a general state $\ket{\psi} \in \mathcal{H}$. We will show that if $\ket{\psi}$ has a definite eigenstate under $\hat{\mathcal O}_A$, then it must be possible to factorize $\ket{\psi}$ across the two Hilbert spaces, meaning that it is not entangled.

Using the basis for $\mathcal H$, we can represent $\ket\psi$ as$$
\ket\psi = \sum_{ij} c_{ij} \ket{ \varphi_A^{(i)} }\ket{ \varphi_B^{(j)} }
$$ Then, suppose that $\hat{\mathcal O}_A \otimes \hat{I}_B \ket\psi = \mu \ket\psi$. Substituting into the previous equation yields$$
\hat{\mathcal O}_A \otimes \hat{I}_B \sum_{ij} c_{ij} \ket{ \varphi_A^{(i)} }\ket{ \varphi_B^{(j)} } = \mu \sum_{ij} c_{ij} \ket{ \varphi_A^{(i)} }\ket{ \varphi_B^{(j)} }
$$Applying the operators to the LHS results in$$
\sum_{ij} c_{ij} \lambda_A^{(i)} \ket{ \varphi_A^{(i)} }\ket{ \varphi_B^{(j)} } = \mu \sum_{ij} c_{ij} \ket{ \varphi_A^{(i)} }\ket{ \varphi_B^{(j)} }
$$ Premultiplying by $\bra{ \varphi_A^{(i')} } \bra{ \varphi_B^{(j')} }$ on both sides yields
$$
c_{i'j'} \lambda_A^{(i')} = c_{i'j'} \mu
$$ If $c_{i'j'} \neq 0$, then we conclude that $\lambda_A^{(i')} = \mu$. But hang on - we could say this is for any value of $i'$. This means we have a contradiction if $c_{i'j'}$ and $c_{i''j'}$ are both nonzero for distinct eigenvalues $\lambda_A^{(i')}$ and $\lambda_A^{(i'')}$.

So, if we further suppose that all the eigenvalues $\lambda_A^{(i)}$ are distinct, then we can only have one value of $i$ for which $c_{ij}$ is nonzero, proving that the state is not entangled.

(To be honest, that proof makes the result look more complicated than it really is... I think the easiest way to see the result is just to consider the case for 2-dimensional Hilbert spaces.)

 

[PeterDonis]

it actually seems to require that eigenstates of the observable are non-degenerate.

Yes, I believe that's true. I forgot to mention that qualifier in post #2.\

I think there's an analogous result for the degenerate case, but its proof is more complicated.

 

[andresB]

Here's a proof of the result - it actually seems to require that eigenstates of the observable are non-degenerate.

Ok, got it. But what about wavefunctions?

For example, given a wavefunction $\psi(x,y)$ that can be separated by change of variables
$X =X(x,y)$ and $Y =Y(x,y)$ so that
$$\psi(X,Y)=\psi_{1}(X)\psi_{2}(Y),$$
can we say that this imply that there exist self-adjoint operators with $\psi_{1}$ and $\psi_{2}$ as eigenfunctions?

The above is the kind of things we see in, for example, the coupled Harmonic oscillators.

 

[PeterDonis]

what about wavefunctions

Wave functions are just one particular representation of quantum states. The proof that was given does not assume any particular representation; it just uses general properties of quantum states as vectors in a Hilbert space.

continuous degrees of freedom?

It looks to me like a similar proof using integrals instead of sums in order to be applicable to continuous degrees of freedom should be straightforward. Note that you can write kets for states of quantum systems with continuous degrees of freedom; they're still vectors in a Hilbert space and kets are notation for vectors in a Hilbert space. The difference is that the index on the basis kets now needs to be continuous and you need to do integrals instead of sums.

 

[Jonomyster]

It looks to me like a similar proof using integrals instead of sums in order to be applicable to continuous degrees of freedom should be straightforward.

I think it's worth seeing this written out for a simple example. Consider a single particle which can move along the real number line. The corresponding Hilbert space has a basis $\{ \ket{x} : x \in \mathbb{R} \}$. Here, the elements $\ket x$, $\ket y$, etc. represent the states "particle at $x$", "particle at $y$", etc. For continuous systems, we can use $\delta$-function normalisation:
$$
\braket{x | y} = \delta(x-y)
$$Then, a general state $\ket\psi$ can be represented as a superposition of these position states. That is,
$$
\ket\psi = \int_{\mathbb{R}} \psi(x') \ket{x'} dx'
$$ Notice what happens when we take the inner product between $\ket\psi$ and another state $\ket x$:
$$
\begin{aligned}
\braket{x | \psi} &= \int_{\mathbb{R}} \psi(x') \braket{x | x'} dx' \\[.8em]

&= \int_{\mathbb{R}} \psi(x') \delta(x-x') dx' \\[.8em]

&= \psi(x)
\end{aligned}
$$ So from this perspective, the wavefunction $\psi(x)$ can be thought of as the amount of amplitude the state $\ket\psi$ has at position $x$.

 

[PeterDonis]

The corresponding Hilbert space has a basis $\{ \ket{x} : x \in \mathbb{R} \}$.

Actually, no, those kets are not in the Hilbert space. In wave function terms, they are delta functions, and delta functions are not square integrable. To do the proof you are looking for using position eigenstates (or momentum eigenstates, for that matter) as a basis, you would need to use something like the rigged Hilbert space formalism that is described in, e.g., Ballentine.

 

[Jonomyster]

Actually, no, those kets are not in the Hilbert space. In wave function terms, they are delta functions, and delta functions are not square integrable. To do the proof you are looking for using position eigenstates (or momentum eigenstates, for that matter) as a basis, you would need to use something like the rigged Hilbert space formalism that is described in, e.g., Ballentine.

Didn't realise that, thanks!