How to think about partially entangled states

[zonde]

I am trying to understand how to think about a state of the form:
$|\psi \rangle=\alpha|H_A H_B\rangle+\beta|V_A V_B\rangle$ where $|\alpha|\neq |\beta|$

It is between pure entangled state and a classical state like $\psi \rangle=|H_A H_B\rangle$, but it is not mixture of the two either. So it seems that it can not be represented by vector in some state space and it can not be represented by some combination (mixture) of vectors.

I don't know how to think about it using vectors. Any help?

 

[Nugatory]

So it seems that it can not be represented by vector in some state space and it can not be represented by some combination (mixture) of vectors.

Why not? You just did.

 

[zonde]

Why not? You just did.

Not sure I didn't. It is certainly not mixture of individual vectors in their separate state spaces as there is phase relationship between them.
As I see, when there is fixed phase relationship between vectors they are represented by single vector in some combined state space. This does not seem to work here.

 

[kith]

It is between pure entangled state and a classical state like $\psi \rangle=|H_A H_B\rangle$, but it is not mixture of the two either. So it seems that it can not be represented by vector in some state space and it can not be represented by some combination (mixture) of vectors.

I'm curious: what do you think what kind of mathematical object is your $|\psi \rangle=\alpha|H_A H_B\rangle+\beta|V_A V_B\rangle$ (where $|\alpha|\neq |\beta|$)?

As I see, when there is fixed phase relationship between vectors they are represented by single vector in some combined state space. This does not seem to work here.

Your usage of terminology is a bit nonstandard. Let me suggest a translation:
pure entangled state -> maximally entangled state
classical state -> separable state or product state
the states you are asking about -> partially entangled pure states

All of these are state vectors in the combined state vector space. States which can't be described by state vectors are forced upon us if we go the other direction: only the separable state vectors can be expressed in terms of state vectors of the individual state vector spaces. As soon as there's a nonzero degree of entanglement, the individual states need to be expressed as reduced density matrices.

 

[zonde]

I'm curious: what do you think what kind of mathematical object is your $|\psi \rangle=\alpha|H_A H_B\rangle+\beta|V_A V_B\rangle$ (where $|\alpha|\neq |\beta|$)?

I think it's subspace with specific property.

Your usage of terminology is a bit nonstandard. Let me suggest a translation:
pure entangled state -> maximally entangled state
classical state -> separable state or product state
the states you are asking about -> partially entangled pure states

Thanks for correcting my terminology.

All of these are state vectors in the combined state vector space.

Product state is certainly not a complex vector. Components of product state don't have certain phase relationship between them so that $|\psi \rangle \otimes |\phi \rangle$ I can write as $|\psi \rangle + e^{ix} |\phi \rangle$ where $x$ has no fixed value.

States which can't be described by state vectors are forced upon us if we go the other direction: only the separable state vectors can be expressed in terms of state vectors of the individual state vector spaces. As soon as there's a nonzero degree of entanglement, the individual states need to be expressed as reduced density matrices.

Yes, and that's one of my problems. How do you know the object is a vector when it's projection onto subspace is not a vector but rather a sub-subspace?
Well I might think that vector is okay but there is some problem with subspace. But no, I can have a state that is described by a vector in that subspace, so the subspace should be perfectly fine.

 

[stevendaryl]

Product state is certainly not a complex vector.

I'm not sure what you mean by that it. I would say that it is. In forming entangled states of two spin-1/2 particles, we have a two-particle spin basis:

$|u\rangle |u\rangle$
$|u\rangle |d\rangle$
$|d\rangle |u\rangle$
$|d\rangle |d\rangle$

Those are the basis vectors for a complex vector space: The state $\frac{1}{\sqrt 2} (|u\rangle |d\rangle + |d\rangle |u\rangle)$ is a different state than $\frac{1}{\sqrt 2} (|u\rangle |d\rangle + i |d\rangle |u\rangle)$

 

[zonde]

$|u\rangle |u\rangle$
$|u\rangle |d\rangle$
$|d\rangle |u\rangle$
$|d\rangle |d\rangle$

Those are the basis vectors for a complex vector space ...

The question I'm struggling with is: why do you say that your basis states are complex vectors?
Lets say that $|u_A\rangle$ is a pure state and $|u_B\rangle$ too is a pure state. Is $|u_A\rangle\otimes|u_B\rangle$ a pure state?

 

[stevendaryl]

The question I'm struggling with is: why do you say that your basis states are complex vectors?

I wouldn't say that they are complex vectors, I would say that they are vectors from a complex vector space, which just means that

If $|A\rangle$ is a vector and $|B\rangle$ is a vector, then $\alpha |A\rangle + \beta |B\rangle$ is a vector, for any complex numbers $\alpha$ and $\beta$.

Lets say that $|u_A\rangle$ is a pure state and $|u_B\rangle$ too is a pure state. Is $|u_A\rangle\otimes|u_B\rangle$ a pure state?

Yes, it is.

 

[Demystifier]

$|\psi \rangle \otimes |\phi \rangle$ I can write as $|\psi \rangle + e^{ix} |\phi \rangle$

No, you can't write the former as the latter. I'm not sure where exactly your confusion stems from, but it sure stems from some very basic misconception.

 

[zonde]

I have to think it over.

 

[Nugatory]

Lets say that $|u_A\rangle$ is a pure state and $|u_B\rangle$ too is a pure state. Is $|u_A\rangle\otimes|u_B\rangle$ a pure state?

Yes. All kets and linear combinations of kets represent pure states (as opposed to mixed states, which must be represented with density matrices).

You do have to be aware of which Hilbert space your kets belong to. $|u_a\rangle$, $|u_b\rangle$, and $|u_a\rangle\otimes|u_b\rangle$ are all pure states, but they belong to different Hilbert spaces so are pure states of different things.

In your original post we have three Hilbert spaces to consider:
1) One for the states of particle A, considered as an isolated quantum system. This one contains vectors $|H_A \rangle$, $|V_A \rangle$, and all of their linear combinations. It is a two-dimensional space because it can be spanned with two basis vectors.
2) Likewise for the states of particle B.
3) The Hilbert space that is the tensor product of the first two; the kets in this Hilbert space represent the pure states of the (potentially entangled) two-particle system. It is a four-dimensional space containing the vectors $|H_A H_B\rangle$, $|V_A H_B\rangle$, $|H_A V_B\rangle$, and $|V_A V_B\rangle$ and all their linear combinations.

A linear combination of kets (like the $|\psi\rangle+e^{ix}|\phi\rangle$ you mention above) only makes sense if all the kets are vectors in the same Hilbert space, and the result is necessarily also a ket in that Hilbert space.

 

[zonde]

3) The Hilbert space that is the tensor product of the first two; the kets in this Hilbert space represent the pure states of the (potentially entangled) two-particle system. It is a four-dimensional space containing the vectors $|H_A H_B\rangle$, $|V_A H_B\rangle$, $|H_A V_B\rangle$, and $|V_A V_B\rangle$ and all their linear combinations.

Okay, I might have found the source of my confusion. I am trying to keep track of phase factor. So when we speak about dimension $|H_A H_B\rangle$ there appears a question. Complex dimensions $|H_A\rangle$ and $|H_B\rangle$ have additional degree of freedom - a phase factor. So when we view combine together these two dimensions how do we combine together phase factors?
Alternatively I could think about single complex dimension as two real dimensions. So if we combine together two complex dimensions how the four real dimensions fit together?

 

[Demystifier]

So when we view combine together these two dimensions how do we combine together phase factors?

Suppose that $|\psi'_1\rangle =e^{i\varphi_1}|\psi_1\rangle$, $|\psi'_2\rangle =e^{i\varphi_2}|\psi_2\rangle$. Then
$$|\psi'_1\rangle \otimes |\psi'_2\rangle =e^{i\varphi_1}e^{i\varphi_2}|\psi_1\rangle \otimes |\psi_2\rangle$$
$$|\psi'_1\rangle + |\psi'_2\rangle =e^{i\varphi_1}|\psi_1\rangle + e^{i\varphi_2} |\psi_2\rangle$$
Does it answer your question?

 

[vsv86]

Okay, I might have found the source of my confusion. I am trying to keep track of phase factor. So when we speak about dimension $|H_A H_B\rangle$ there appears a question. Complex dimensions $|H_A\rangle$ and $|H_B\rangle$ have additional degree of freedom - a phase factor. So when we view combine together these two dimensions how do we combine together phase factors?
Alternatively I could think about single complex dimension as two real dimensions. So if we combine together two complex dimensions how the four real dimensions fit together?

What helped me with these sorts of things, is getting a book on linear and multi-linear algebra and reading about tensor product (direct product). You have 2 two-dimensional vector spaces ($V,W$), both over complex numbers. You then create another vector space $V\otimes W$ out of these two vector spaces. Using the definition of the tensor product (dual space to the space of bilinear maps to complex numbers, i.e. $V \otimes W = Hom\left(V \times W,\mathbb{C}\right)^{*}$) you can work out its properties. Such as number of dimensions, what happens with factors etc.

However! For finite-dimensional vector spaces you don't really need this, since finite-dimensional spaces are isomorphic to trivial ones. Thus to understand the tensor product of a 2d vector $v=(a, b) \epsilon \mathbb{C}^2$ and a 2d vector $w=(c,d) \epsilon \mathbb{C}^2$, simply look at the properties of $v\otimes w=(ac, ad, bc, bd)$

 

[Nugatory]

Complex dimensions $|H_A\rangle$ and $|H_B\rangle$ have additional degree of freedom - a phase factor. So when we view combine together these two dimensions how do we combine together phase factors?

Yes, you can attach a constant multiplier to one of these kets - for example, $e^{i\theta}|H_A\rangle$ - but that doesn't represent anything you don't already have. The linear combinations of $|H_A\rangle$ and $|V_A\rangle$ are of the form $\alpha|H_A\rangle+\beta|V_A\rangle$ so it's just the linear combination with $\alpha=e^{i\theta}$ and $\beta=0$. In fact, you could choose to use $e^{i\theta}|H_A\rangle$ as a basis vector instead of $|H_A\rangle$ and all that would happen is that coefficients would change everywhere, as you'd expect from any basis change. @Demystifier has already posted the recipe for handling this change in the tensor product.

Alternatively I could think about single complex dimension as two real dimensions. So if we combine together two complex dimensions how the four real dimensions fit together?

The whole point of the bra-ket notation is that the kets are, as @stevendaryl has already pointed out, vectors in a complex vector space. A real vector space with twice as many dimensions is not an equivalent mathematical construct.

 

[zonde]

Suppose that $|\psi'_1\rangle =e^{i\varphi_1}|\psi_1\rangle$, $|\psi'_2\rangle =e^{i\varphi_2}|\psi_2\rangle$. Then
$$|\psi'_1\rangle \otimes |\psi'_2\rangle =e^{i\varphi_1}e^{i\varphi_2}|\psi_1\rangle \otimes |\psi_2\rangle$$
$$|\psi'_1\rangle + |\psi'_2\rangle =e^{i\varphi_1}|\psi_1\rangle + e^{i\varphi_2} |\psi_2\rangle$$
Does it answer your question?

Thanks, it does.

 

[zonde]

Lets say that $|u_A\rangle$ is a pure state and $|u_B\rangle$ too is a pure state. Is $|u_A\rangle\otimes|u_B\rangle$ a pure state?

Yes, it is.

So $\psi \rangle=|H_A H_B\rangle$ is a vector in product space. But then I can write it as superposition of two different vectors. Let's say that $|C_A C_B\rangle$ and $|D_A D_B\rangle$ are basis vectors when both particles are viewed in $+45^\circ/-45^\circ$ basis.

Naive version similar to the way how superposition would be written for single particle gives entangled state:
$1/\sqrt{2}(|C_A C_B\rangle + |D_A D_B\rangle)\neq|H_A H_B\rangle$

So how can one write $|H_A H_B\rangle$ as a superposition in different basis?

 

[stevendaryl]

So $\psi \rangle=|H_A H_B\rangle$ is a vector in product space. But then I can write it as superposition of two different vectors. Let's say that $|C_A C_B\rangle$ and $|D_A D_B\rangle$ are basis vectors when both particles are viewed in $+45^\circ/-45^\circ$ basis.

Naive version similar to the way how superposition would be written for single particle gives entangled state:
$1/\sqrt{2}(|C_A C_B\rangle + |D_A D_B\rangle)\neq|H_A H_B\rangle$

So how can one write $|H_A H_B\rangle$ as a superposition in different basis?

I don't know what the coefficients for the transformation are, but let's suppose that:

$|C\rangle = \frac{1}{\sqrt{2}} (|H\rangle + |V\rangle)$

$|D\rangle = \frac{1}{\sqrt{2}} (|H\rangle - |V\rangle)$

Then the inverse transformation is

$|H\rangle = \frac{1}{\sqrt{2}} (|C\rangle + |D\rangle)$

$|V\rangle = \frac{1}{\sqrt{2}} (|C\rangle - |D\rangle)$

So $|H_A\rangle |H_B\rangle = \frac{1}{2} (|C_A\rangle |C_B\rangle + |C_A\rangle |D_B\rangle + |D_A\rangle |C_B\rangle + |D_A\rangle |D_B\rangle)$

 

[zonde]

Then the inverse transformation is

$|H\rangle = \frac{1}{\sqrt{2}} (|C\rangle + |D\rangle)$

$|V\rangle = \frac{1}{\sqrt{2}} (|C\rangle - |D\rangle)$

So $|H_A\rangle |H_B\rangle = \frac{1}{2} (|C_A\rangle |C_B\rangle + |C_A\rangle |D_B\rangle + |D_A\rangle |C_B\rangle + |D_A\rangle |D_B\rangle)$

After you wrote it seems sort of obvious. Thanks a lot.

 

[zonde]

So the state from post #1

$|\psi \rangle=\alpha|H_A H_B\rangle+\beta|V_A V_B\rangle$

given

$|H\rangle = \frac{1}{\sqrt{2}} (|C\rangle + |D\rangle)$
$|V\rangle = \frac{1}{\sqrt{2}} (|C\rangle - |D\rangle)$

I can rewrite as superposition of two entangled states:

$|\psi \rangle=\alpha|H_A H_B\rangle+\beta|V_A V_B\rangle=(\alpha + \beta)(|C_A\rangle |C_B\rangle + |D_A\rangle |D_B\rangle) + (\alpha - \beta)(|C_A\rangle |D_B\rangle + |D_A\rangle |C_B\rangle)$

So I think I have got the answer for question in post #1. Without doubt the state transforms as vector.
Thanks to all who participated in the thread.