# Non-separability : basis dependent

1. Feb 18, 2015

### jk22

If we consider the singlet state (0,1,-1,0)/sqrt2 then it is easy to see that the unitary block transformation : A=RoR^-1with R a rotation of 45 degrees gives the vector 1/2(-1,1,-1,1) which is separable. Thus entanglement disappears in that basis.

2. Feb 18, 2015

### atyy

3. Feb 18, 2015

### jk22

Or : the nonseperability goes from the wavefunction to the operator

4. Feb 19, 2015

### TrickyDicky

The basis-independent example you give is not of entanglement but of entropy(basis-independent quantum phase transitions). I am not aware of any definition of basis-independent entanglement, the mathematical definition seems to be basis-dependent(and preparation/measurement dependent) by construction: while the tensor product is basis-independent for factorisable or elementary tensor products, it is not for non-elementary(entangled) tensors.

5. Feb 19, 2015

### Demystifier

Can you explain why do you call (0,1,-1,0)/sqrt2 singlet, and why do you call 1/2(-1,1,-1,1) separable?

6. Feb 19, 2015

### TrickyDicky

Basically by choosing the operator, one can prepare different states, make them separable or not. This is quite evident for bipartite systems of 2-dimensional Hilbert spaces like spin or polarization that always have the option to be constructed as factorisable tensor products.
One always decides the kind of operation that relates quantum systems based on the operator demanded by the specific physical problem to act on their states. For other operators like for certain hamiltonians for instance one can never construct factorisable tensor products. This is precisely the situation for entangled particles in a composite particle as can be seen in a concurrent featured thread https://www.physicsforums.com/threads/does-quantization-always-work.797848/

7. Feb 19, 2015

### kith

Kind of. If you want to view the unitary transformation as a change of basis, your transformation sends the product basis vectors to entangled basis vectors. This doesn't mean that the entanglement is gone because you still can't write your state as a tensor product of states of the individual Hilbert spaces. The textbook notion of entanglement is not basis dependent.

I think it basically is enough to answer your question this way, but since you were talking about the state as "singlet", let me add that the situation is less clear for indistinguishable particles. atyy's second link suggests a different definition of entanglement, while Arnold Neumeier has a FAQ entry arguing that we should use the term entanglement only if the particles can be at least effectively distinguished in experiments.

Last edited: Feb 19, 2015
8. Feb 19, 2015

### atyy

The link I gave was about phase transitions, but the entanglement entropy between two subsystems can also be taken as a basis-independent measure of entanglement. If the entanglement entropy is non-zero, that means that the state is not a product state in any basis.

9. Feb 19, 2015

### atyy

Let me add these references to the discussion:
http://arxiv.org/abs/1109.0036
http://arxiv.org/abs/1406.7304

10. Feb 19, 2015

### TrickyDicky

But one thing is the measurement of the degree of entanglement of a bipartite system, for wich you can use entropy in a basis-independent way like in your example where it measures maximal entanglement, that is thru entrpy measure you can ascertain independently of choice of basis that a system is maximally entangled, and another thing is the phenomenon of entanglement wich itself is basis-dependent in general as seen in the mathematical definition.

11. Feb 19, 2015

### Strilanc

I think you applied the operations incorrectly.

The initial state of the system is:

$\begin{bmatrix}0 & 1 \\ -1 & 0\end{bmatrix}$

I've laid out the vector in a square to more easily work with the two qubits. The horizontal direction represents one qubit, the vertical direction is the other, and the entries being along a diagonal is an easy-to-spot indication of entanglement.

Anyways, we rotate the row qubit by 45 degrees by treating each row as a 2d vector to be rotated. We ignore annoying factors of $\sqrt{2}$:

$\begin{bmatrix}-1 & 1 \\ -1 & -1\end{bmatrix}$

It's harder to see that this is entangled, because there's not a nice clear diagonal in this basis. Anyways, we then rotate the column qubit by -45 degrees, as instructed, by treating the columns as 2d vectors:

$\begin{bmatrix}-1 & 0 \\ 0 & -1\end{bmatrix}$

And the entries being along the diagonal tells us the result is still entangled.

Last edited: Feb 19, 2015
12. Feb 19, 2015

### jk22

I was trying something like (there seems to be some errors in the display) :

$$\left(\begin{array}[cccc] 1&-1&0&0\\1&1&0&0\\0&0&1&1\\0&0&-1&1\end{array}\right)\left(\begin{array}[c] 0\\1\\-1\\0\end{array}\right)=\left(\begin{array}[c] -1\\1\\-1\\1\end{array}\right)$$

The later being simply equals to $$\left(\begin{array}[c] 1\\1\end{array}\right)\otimes\left(\begin{array}[c] -1\\1\end{array}\right)$$

This let me think about the original epr paradox where we consider an entangled initial state delta(x1+x2)

Then we consider measurement at distant points so the particle takes some time to travel and here happens a time evolution so that the wavefunction becomes a gaussian in the form exp(-(x1+x2)^2). Here we see that at the time of the messurement process we can no more predict with certainty the position of the other particle knowing the first one.

Last edited: Feb 19, 2015
13. Feb 19, 2015

### kith

I am not sure if you are still trying to get rid of entanglement by a basis transformation. If yes, I want to emphasize that this is not a basis transformation. Your final state is not the same state in a different basis, but a different state in the same basis. That you can transform a product state into an entangled state (and vice versa) by a unitary operation should be no surprise.

If you want to use your matrix to transform the basis, you have to apply the matrix (resp. its inverse) to the old basis vectors in order to get the new ones. Then you have to write the old basis vectors as a superposition of the new ones and use this to express your vector in the new basis. Since every basis vector is now entangled, you cannot write the resulting component vector as a tensor product like you did in your last line.

Last edited: Feb 19, 2015
14. Feb 19, 2015

### Strilanc

The transformation you are showing is not a transformation that rotates the first qubit by 45 degrees and the second by -45 degrees. Instead, it is rotating the first qubit by 45 degrees conditioned upon the second qubit being false. It rotates the first qubit by -45 degrees if the second qubit is true.

An operation on the first qubit looks like this:

$\begin{bmatrix}A & B & 0 & 0 \\ C & D & 0 & 0 \\ 0 & 0 & A & B \\ 0 & 0 & C & D \end{bmatrix}$

Whereas an operation on the second qubit looks like this:

$\begin{bmatrix}A & 0 & B & 0 \\ 0 & A & 0 & B \\ C & 0 & D & 0 \\ 0 & C & 0 & D \end{bmatrix}$

Try playing around in this quantum circuit simulator for a bit, if you want more of a feel of how the operations relate to the matrices (when your cursor is over a gate, the bottom area shows the matrix for that gate's column of operations as well as the state after that gate's column of operations, the circles with lines each represent a complex number).

15. Feb 20, 2015

### jk22

Yes it is a direct sum it is impossible to separate with a tensor product.

16. Mar 22, 2015

### jk22

The question is : does there exist a unitary transformation that performs the separation ?

17. Mar 23, 2015

### Strilanc

You can't disentangle a state with single qubit transformations (i.e. operations of the form $M \otimes I_2$ or $I_2 \otimes M$). A controlled-not would work, but that's a 2-qubit operation.

If you could disentangle the state with single qubit operations, you could run that process backwards to remotely entangle things without making them interact. Too good to be true.

One easy proof that you can't do it with single-qubit operations involves laying the state out like a matrix... then you pretend it actually *is* the matrix $\begin{bmatrix}0 & 1 \\ -1 & 0 \end{bmatrix}$. What do our single-qubit operations correspond to now?
1. Operations on the first qubit, of the form $A \otimes I_2$, are equivalent to post-multiplying the state matrix by $A^T$.
2. Operations on the second qubit, of the form $I_2 \otimes B$, are equivalent to pre-multiplying the state matrix by $B$.
3. Because the initial state is a unitary matrix, and the desired final state $\begin{bmatrix}1 & 0 \\ 0 & 0 \end{bmatrix}$ is not unitary, and matrix multiplication is closed over unitary matrices, and single-qubit operations are turning into unitary matrix multiplications, you're not going to be able to turn the initial state into the final state with single-qubit operations.

18. Mar 23, 2015

### jk22

But this operation can not be of the form $$A\otimes B$$

19. Mar 23, 2015

### Strilanc

Which operation?

20. Mar 23, 2015

### jk22

An operation that disentangle