I'm working through Susskind's "Quantum Mechanics" book (TTM series), which I quite like.


In Lecture 7 (Chapter 7), he studies a 2-spin system. A single spin has eigenvectors:

$$|u\rangle=\begin{pmatrix}1\\0\end{pmatrix},~~ |d\rangle=\begin{pmatrix}0\\1\end{pmatrix}$$

and then a 2-spin state has eigenvectors:

$$|uu\rangle=\begin{pmatrix}1\\0\\0\\0\end{pmatrix},~~ |ud\rangle=\begin{pmatrix}0\\1\\0\\0\end{pmatrix},~~ |du\rangle=\begin{pmatrix}0\\0\\1\\0\end{pmatrix},~~ |dd\rangle=\begin{pmatrix}0\\0\\0\\1\end{pmatrix}$$

Alice studies the first with an operator $\sigma$ and Bob the second with an operator $\tau$ (these are really product operators of single-spin $\sigma_z$ with the identity $I$: $\sigma_z\bigotimes I$ and $I\bigotimes\sigma_z)$:

$$\sigma = \begin{pmatrix} 1 & 0 & 0 & 0 \\ 0 & 1 & 0 & 0 \\ 0 & 0 & -1 & 0 \\ 0 & 0 & 0 & -1 \end{pmatrix} ~~~ \tau = \begin{pmatrix} 1 & 0 & 0 & 0 \\ 0 & -1 & 0 & 0 \\ 0 & 0 & 1 & 0 \\ 0 & 0 & 0 & -1 \end{pmatrix}$$

Now for the interesting stuff.

We can have a product state where the two spins ("subsystems") are independent (no entanglement):

$$\psi ~=~ (a_1|u\rangle+a_2|d\rangle)\bigotimes(b_1|u\rangle+b_2|d\rangle)$$


where the $a_i$ and $b_i$ are separately normalized to $1$ so that if we calculate the expectation for either spin the other does not factor in at all. For example $\langle\psi|\sigma|\psi\rangle=a_1^2-a_2^2$ with no appearance of the $b_i$.

Then Susskind says that most randomly chosen coefficients of the $|uu\rangle...$ (normalized) will not factorize as in $(1)$. Then they are entangled. And an example of a maximally entangled state is the singlet state:

$$|S\rangle=\frac{1}{\sqrt 2}(|ud\rangle-|du\rangle)$$

Now $\langle S|\sigma|S\rangle=0$ so you have zero information about the individual spins. However, you have information about correlated measurements, because $\langle S|\tau\sigma|S\rangle=-1$ where by matrix multiplication

$$ \tau_z\sigma_z = \begin{pmatrix} 1 & 0 & 0 & 0 \\ 0 & -1 & 0 & 0 \\ 0 & 0 & -1 & 0 \\ 0 & 0 & 0 & 1 \end{pmatrix} $$

Susskind then discusses how you can test whether a state is entangled or not (and how much entangled) by computing the correlation of operators $A$ and $B$, or checking the eigenvalues of single-state density matricies ($\rho_{2x2})$, which should be $\{1,0,0,0...\}$, or checking if the state coefficients $\{0,\frac{1}{\sqrt 2},-\frac{1}{\sqrt 2},0\}$ can factorize as in $(1)$ (they can't).

Question (rewritten after helpful answers by tparker and Emilio Pisanty)

Aren't these entanglement tests all relative to the chosen 4x4 operators, $\sigma_z$ and $\tau_z$, which reflect a particular choice of dividing the state into subsystems?

Instead of a subdivision based on the two spins, we can subdivide based on $|S\rangle$ and the triplet states $|T_1\rangle=\frac{1}{\sqrt 2}(|ud\rangle+|du\rangle),~~|T_2\rangle=\frac{1}{\sqrt 2}(|uu\rangle+|dd\rangle)$ and $|T_3\rangle=\frac{1}{\sqrt 2}(|uu\rangle-|dd\rangle)$. Let's change basis with a similarity matrix $P=(|T_3\rangle~|T_2\rangle~|T_1\rangle~|S\rangle)$. In this new basis, $|S\rangle...|T_3\rangle$ are basis vectors and

$$ A=\tau_z\sigma_{z,new basis} = \begin{pmatrix} 1 & 0 & 0 & 0 \\ 0 & 1 & 0 & 0 \\ 0 & 0 & -1 & 0 \\ 0 & 0 & 0 & -1 \end{pmatrix} =\begin{pmatrix} 1 & 0\\ 0 & -1 \end{pmatrix} \bigotimes \begin{pmatrix} 1 & 0\\ 0 & 1 \end{pmatrix} $$

$$ B=\tau_y\sigma_{y,new basis} = \begin{pmatrix} 1 & 0 & 0 & 0 \\ 0 & -1 & 0 & 0 \\ 0 & 0 & 1 & 0 \\ 0 & 0 & 0 & -1 \end{pmatrix} =\begin{pmatrix} 1 & 0\\ 0 & 1 \end{pmatrix} \bigotimes \begin{pmatrix} 1 & 0\\ 0 & -1 \end{pmatrix} $$

We consider the new basis vectors as product vectors isomorphic to single spins which can each be in states labeled $|+\rangle$ and $|-\rangle$ (so as not to confuse with $|u\rangle$ and $|d\rangle$) and we get that

$$ |S\rangle = |{--}\rangle,~~~~ |T_1\rangle = |{-+}\rangle,~~~~ |T_2\rangle = |{+-}\rangle,~~~~ |T_3\rangle = |{++}\rangle $$

Since $A$ and $B$ are of the form of product operators, we can let them define a new subdivision of the full system. Each new subsystem no longer corresponds to an electron at a specific location, as in the original division. A and B can be thought to operate on one label each (A on the first + or -, B on the second).

With this new subdivision, each of $|S\rangle...|T_3\rangle$ are not entangled.

To conclude

Entanglement is in the eye of the beholder (4x4 operator, or subsystem division). Yes?

  • $\begingroup$ So you are proposing expanding the arbitrary state as a superposition of the singlet and triplet states instead of the 2-spin states? I guess you would need some measuring device that only determines the singlet/triplet state without measuring any individual spins? $\endgroup$ Commented Sep 10, 2018 at 14:09
  • $\begingroup$ Yes. I'm sure there are practical issues building equipment. But at an abstract level, QM allows us to create any observables, and define operators in their eigenvector basis. Entanglement is not an intrinsic property of a state. Entanglement usually is discussed with the "subsystems" being chosen as two spins that are physically removed, because that's what happens in the interesting EPR/Bell case. I'm thinking: a state is just a state, and whether or not it appears entangled depends on how you interact with it (observe it, operate on it). $\endgroup$
    – johndecker
    Commented Sep 10, 2018 at 14:19
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    $\begingroup$ So to word it a little differently, you are proposing it should be possible to take a state where one superposition does not have sets of coefficients whose squares sum to $1$, but if we were to express the same state as a superposition using a different basis, we would find that a partition of the coefficients exists such that the sum of squares within each partition does sum to $1$. If this is what you are saying, I think you are correct, but this is not my central area of study. Hopefully someone with more experience in this can weigh in. $\endgroup$ Commented Sep 10, 2018 at 15:21
  • $\begingroup$ @AaronStevens, essentially yes. (Though I'm not sure I'd call the parts superpositions; perhaps "factor substates." $\endgroup$
    – johndecker
    Commented Sep 10, 2018 at 15:41
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    $\begingroup$ You have this phrase ..."then neither |S⟩ nor |T⟩ is entangled, but pure states." You seem to imply that an entangled state cannot be a pure state. This is WRONG. $\endgroup$
    – wcc
    Commented Sep 10, 2018 at 19:39

2 Answers 2


I think I understand your question, but I don't understand Aaron Stevens's comments at all, which you claim to be a valid rephrasing, so it's possible that I'm not actually understanding your question correctly. With that caveat:

Your basic idea is right, but your statements aren't quite mathematically precise enough to be completely correct. (For one thing, you're using the words "entangled" and "pure" as if they were mutually exclusive, but they're not - the maximally entangled state that you describe is both entangled and pure.) Yes, whether a state has internal entanglement does indeed depend on how you factor the Hilbert space into subsystems.

But you're missing a key point, which is that the Hilbert spaces for a composite system is a tensor product of the individual systems' Hilbert spaces, not a direct sum. The Hilbert space $\mathcal{H}_{AB} = \{ |uu\rangle, |ud\rangle, |du\rangle, |dd\rangle \}$ for a two-spin system is the tensor product $\mathcal{H}_{AB} = \mathcal{H}_A \otimes \mathcal{H}_B$, where $\mathcal{H}_A$ and $\mathcal{H}_B$ are both isomorphic to the Hilbert space $\{ u, d \}$ for a single spin. So we can meaningfully talk about operator that only act on one subsystem. But the set of linear combinations of the $|S\rangle$ and $|T\rangle$ states forms the direct sum $\{|S\rangle\} \oplus \{|T\rangle\}$, so we can't think of the $|S\rangle$ and $|T\rangle$ states as subsystems that operators can act on independently.

Sometimes, a composite system's Hilbert space can be written as a tensor product in two inequivalent ways. This really does correspond to two different valid ways to divide the complete system into subsystems, and whether or not the subsystems are entangled can indeed depend on that division. (But this is not quite the same thing as basis dependence, because it turns out that the entanglement is independent of the basis one chooses for each subsystem. Once one chooses a division of the complete system into physical subsystems, then any change of basis within one subsystem will not affect the entanglement.)

We can't see this with your two-spin example, but we can see it if we consider a system of three spins $A$, $B$, and $C$, whose Hilbert space is $\mathcal{H}_A \otimes \mathcal{H}_B \otimes \mathcal{H}_C = \{ uuu, uud, udu, udd, duu, dud, ddu, ddd \}$. Consider the state $$\frac{1}{\sqrt{2}} (|u_A d_B\rangle - |d_A u_B\rangle) \otimes |u_C\rangle = \frac{1}{\sqrt{2}}(|udu\rangle - |duu\rangle).$$ In this state the spins A and B are maximally entangled, but the spin C is not entangled with either of them. One person might only have experimental access to operators that act on either (a) the A and B spins or (b) the C spin. This person would naturally consider the A and B spins together as comprising a single subsystem, and the C spin as comprising a separate subsystem. They would therefore naturally factor the Hilbert space as $\mathcal{H} = \mathcal{H}_{AB} \otimes \mathcal{H}_C$, and say that the state is not entangled. They would not observe any unusual correlations between spins in "separate subsystems".

But someone else might have experimental access to a different set of operators, which can only act on either (a) the A spin, or (b) the B and C spins. This second person would naturally consider the A spin as comprising a single subsystem, and the B and C spins together as comprising a separate subsystem. They would therefore naturally factor the Hilbert space as $\mathcal{H} = \mathcal{H}_A \otimes \mathcal{H}_{BC}$, and say that the state is entangled (in fact, maximally entangled). They would observe perfect correlations between (what they describe as) "separate subsystems".

But again, once you specify a particular tensor factorization of your Hilbert space into fixed subsystems, then the entanglement between the subsystems is both basis- and observer-independent.

  • $\begingroup$ I like the answer, but I am feeling a little uneasy with the statement that observers have the freedom to factorize Hilbert space...In your example, what prevents the observers from completely factorizing to $\mathcal{H}_A \otimes \mathcal{H}_B \otimes \mathcal{H}_C$? Can you provide an example for the set of observables that "forces" the the observer to conclude with $\mathcal{H}_A \otimes \mathcal{H}_{BC}$? $\endgroup$
    – wcc
    Commented Sep 10, 2018 at 19:51
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    $\begingroup$ @IamAStudent I think it all comes down to what the observer can measure. If they can only measure each spin separately, then it would be in their best interest to use the factorization you have suggested. If they can only measure A and B together or C by itself, then the first suggested factorization in the answer is more useful. $\endgroup$ Commented Sep 10, 2018 at 20:05
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    $\begingroup$ tparker, don't worry about my comments of the question. I think your paragraph starting with "Sometimes, a composite system's Hilbert space can be written as a tensor product in two inequivalent ways." is really what I was trying to get at, and I think this is the best part that sufficiently answers the OP's question. $\endgroup$ Commented Sep 10, 2018 at 20:11
  • $\begingroup$ @IAmAStudent Great question. There's a big subtlety that I swept under the rug in my answer - I was only considering bipartite entanglement within pure states. We could also consider multipartite entanglement or bipartite entanglement within mixed states (which, thanks to the purification theorem, are actually mathematically equivalent concepts). In this case the concept of entanglement becomes much more complicated and subtle. The observer is certainly free to factorize the Hilbert space further, but I didn't want to get into that story. $\endgroup$
    – tparker
    Commented Sep 10, 2018 at 21:11
  • $\begingroup$ @IAmAStudent Aaron Stevens' comment below yours is exactly correct. Mathematically, you're free to factor your Hilbert space in many different ways, which might formally differ on whether the state is entangled. But the physically natural way to do so is to group together subsystems which it's experimentally feasible to measure all at once (without losing quantum coherence). $\endgroup$
    – tparker
    Commented Sep 10, 2018 at 21:14

Entanglement is in the eye of the beholder (4×4 operator, or subsystem division). Yes?

Yes, but that is a pretty useless observation.

The formal definition of an entangled state of a bipartite quantum system with state space $\mathcal H = \mathcal H_A\otimes \mathcal H_B$ is as follows:

  • a separable state is one whose density matrix can be separated as a sum of tensor products of individual density matrices, i.e. if $\rho\in \mathcal B(\mathcal H)$ is the density matrix of the system, $\rho$ is separable if and only if there exist density matrices $\rho_{A,i}\in \mathcal B(\mathcal H_A)$ and $\rho_{B,i}\in \mathcal B(\mathcal H_B)$ and weights $p_i\geq 0$ such that $$ \rho = \sum_i p_i \rho_{A,i}\otimes \rho_{B,i}. $$
  • an entangled state is any state that is not separable.

For clarity, entanglement is an intrinsic property of the state, together with the partition of the state space into tensor factors.

If you're willing to re-factorize your total state space into some other tensor-product factorization, then a state that's entangled in the $A$, $B$ bipartite scheme is indeed liable to be seen as separable in some alternative $A'$, $B'$ factorization.

However, if you're able to re-factorize your total state space in such a way, then that tells you that your initial split into parties wasn't very meaningful to begin with. In real-world scenarios, we use entanglement as a relevant concept for bipartite systems where the tensor-product factorization of the state space (i.e. the splitting of the system into the two "parties" alluded to in "bipartite") is fixed from the context and cannot be changed easily. If you see it used in a context where that's not the case (ahem) then any conclusions drawn from the entanglement are correspondingly weakened.

One useful way to see this is by noting that the theory of entanglement is, very often, best thought of as a resource theory. Resource theories are great ways to analyze situations where you have one class of operations which is easy to implement but which might be insufficient to achieve some pre-specified goal. Other good examples are thermodynamics (where the operations are energy-conserving processes, and the resource is entropy) and gaussianity (where the operations are linear optical operations); in entanglement, the class of free operations is that of Local Operations and Classical Communication, generally abbreviated as LOCC, and it is obviously tied strictly to a splitting of the system into parties which can operate 'locally' and which can communicate classically.

Resource theories, of course, are only useful when the resource they describe is actually valuable, and when their restricted operations are in fact hard to implement: just as the study of thermodynamics is pretty useless if you have a magical black box that can inject and remove energy from any part of your system at your command, the study of entanglement is pretty meaningless if you have free access to non-LOCC unitary operations that cut across the A-to-B split.

That doesn't mean that you can't talk about entanglement in such a situation, like e.g. the spins of two electrons which are in bound states in the same atom or molecule, but if the re-factorization is physically possible in anything like a reasonable sense, then the conclusions that stem from the presence of the entanglement will be correspondingly trivialized.

But more importantly, if you look at real-world usage, it is always of the form

this system is entangled with that system.

Under your re-factorization, the first part of that sentence to lose its meaning is not "entangled", it's "system".

(The answer below addresses a specific interpretation of v6 of the question, which was, frankly, much more interesting than the current version. I'm keeping it around because of that.)

What Susskind provides is known as an entanglement witness, and here you do get some amount of "eye of the beholder" behaviour. Generically, an entanglement witness is some operator $A$ such that its expectation value in state $\rho$, $\mathrm{Tr}(\rho A)$, will satisfy $$ \mathrm{Tr}(\rho A) \geq 0 \qquad \forall\text{ separable }\rho, $$ so that $$ \mathrm{Tr}(\rho A) < 0 \implies \rho\text{ is entangled}. $$ However, most entanglement witnesses are imperfect: that is, for any given entanglement witness $A$, there will typically be entangled states $\rho$ for which $\mathrm{Tr}(\rho A) \geq 0$, so that $A$ cannot detect the entanglement of that particular entangled state.

Nevertheless, for any given entangled state, there will always be at least one entanglement witness that can certify that it is entangled.

In other words, the definition of entanglement is independent of the operators used to detect its presence, but typically those operators will have a limited scope in which entangled states they can detect.

And if that makes it sound like entanglement is a tricky object to detect and characterize, then... yes, pretty much.

  • $\begingroup$ I think - although I am not positive - that the freedom to tensor-factorize the Hilbert space in different ways (with different resulting evaluations of entangled vs. not for the same state) was the core of the OP's question. $\endgroup$
    – tparker
    Commented Sep 10, 2018 at 21:42
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    $\begingroup$ I think we're all trying to say that entanglement is defined relative to a factorization of the space. The question isn't that confused and it's simply asking whether or not unseparable states become seperable when you change basis; the answer to that is "yes". $\endgroup$
    – DanielSank
    Commented Sep 13, 2018 at 15:31
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    $\begingroup$ @DanielSank Nitpick: when you change tensor factorization, not when you change "basis". A tensor factorization of a Hilbert space is a completely basis-independent concept. Choosing such a factorization makes certain bases more natural to work with (bases whose basis vectors are product states with respect to that factorization), but strictly speaking "tensor factorization" and "basis" are completely independent concepts. $\endgroup$
    – tparker
    Commented Sep 13, 2018 at 16:44
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    $\begingroup$ I disagree with your claim that there is always an overwhelmingly natural/useful way to factorize the Hilbert space. It's true that a real-space factorization is often the most useful in practice, but sometimes the "particle-basis" factorization is useful instead - this is basically how "first quantization" works. Consider a wavefunction $\psi(x,y) = \phi_A(x) \phi_B(y)$ for two distinguishable particles $A$ and $B$. ... $\endgroup$
    – tparker
    Commented Sep 13, 2018 at 17:03
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    $\begingroup$ Such a wavefunction is a product state with respect to the particle "basis" (see caveat above for why quotation marks), but is entangled in the position "basis" - if you measure $n = 0, 1, \text{ or } 2$ particles in the left-hand half of the system, then you instantly know that there are exactly $2 - n$ particles in the right-hand half of the system. Not as exciting as a violation of the Bell inequalities, but still counts as entanglement. Indeed the Born-Oppenheimer and Hartree-Fock approximations in DFT and quantum chemistry both consist of assuming no entanglement in the particle "basis" $\endgroup$
    – tparker
    Commented Sep 13, 2018 at 17:05

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