# I Two particle system, products and entanglement

1. Dec 31, 2017

### fog37

Hello everyone and happy new year. I have a quick dilemma on the concept of entanglement for a two-particle system and I am looking for some clarity:

Say we have particle $A$ and particle $B$. Both particles can be in either of the two states that we call $|\Phi>$ and $|\beta>$ or in a superposition of them.

Particle $A$'s most general state is the superposition $$\Psi_1 = a_1 |\Phi_1> +b_1 |\beta_1>$$
Is it possible for particle $A$ to be in a product state like $|\Phi_1> |\beta_1>$ which is a product of the two states $|\Phi_1>$ and $|\Phi_2>$ ? If so, what would that state physically mean?

Particle $B$'s most general state is $$\Psi_2 = a_2 |\Phi_2> +b_2 |\beta_2>$$. Using the idea of tensor product, we can algebraically multiply $\Psi_1$ by $\Psi_2$. What we get is the following: $$a_1 a_2 |\Phi_1> |\Phi_2> +a_1 b_2 |\Phi_1> |\beta_2> + b_1 a_2 |\beta_1> |\Phi_2> +b_1 b_2 |\beta_1> |\beta_2>$$

This summation is the most general joint state for the two particles together. Each one of the four addends is also a joint state, i.e. it is a product between the state of particle $A$ times the state of particle $B$...

For no entanglement to exist, all terms except one must be zero. Is that correct? If so, why? I am still confused on this simple matter...

thank you

Last edited: Dec 31, 2017
2. Dec 31, 2017

### Staff: Mentor

It is not possible, so the question of what it means does not arise. The state of particle one is represented by a vector in the Hilbert space that is spanned by sums of $|\Phi_1\rangle$ and $|\beta_1\rangle$, not products of those vectors. The products form a completely different Hilbert space, one whose elements are not possible states of particle one, nor of anything else.
You are using the word "joint" to describe two altogether different things, and that is likely contributing to your confusion here. Given two vectors $A$ and $B$, the vectors $\alpha{A}+\beta{B}$ formed by linearly combining them are completely different from the vectors formed by taking their product.

The possible states of particle one are all the possible sums (superpositions) of the vectors describing the state of particle one, and likewise for particle two. The possible state of the combined system is the product of all the possible states of particle one and all the possible states of particle two.
That's pretty much the definition of entanglement: A state that cannot be factorized.

3. Jan 1, 2018

### Zafa Pi

I already find this confusing. You start with both particles A and B in superpositions of states |Φ> and |β>, then |Φ1> and |β1> appear out of no where.
Let me try something simple minded to see if it helps.

Let |0> = [1,0] (a unit vector in the 2D Hilbert Space), and |1> = [0,1]. (quantum computing notation)
Let particle A have state |A> = a1|0> + a2|1> (= [a1,a2]) and particle B have state |B> = b1|0> + b2|1>.
Now |A>|B> = |A>⊗|B> = a1b1|0>|0> + a1b2|0>|1> + a2b1|1>|0> +a2b2|1>|1> = [a1b1,a1b2,a2b1,a2b2]
is a 4D vector in the 4D tensor product space. It factors into |A> and |B> so is not entangled. But notice there could be 4 nonzero terms.
The vector (state) [1,0,0,1] = |0>|0> + |1>|1> does not factor and is thus entangled. It is a state in the tensor product space, but is not a tensor product.
Also note that [0,√½,0,√½] (with 2 nonzeros) is not entangled. If the 4D vector has only 1 zero term it must be entangled.

Last edited: Jan 1, 2018
4. Jan 1, 2018

### fog37

Thank you.

Zafa Pi: my notation was not good. This was my reasoning: I thought I could use the two states $| \Phi>$ and $| \beta>$ for particle $A$ and two different states $| \gamma>$ and $| \eta>$ or particle $B$ thinking that the two particles don't need to be superposition of the same pair of vectors since the particles' states could be in different vector spaces. In your example, the two particles use the same basis vectors, i.e. $|0>$ and $|1>$. Is that necessary?
• Let's go back to your example: the two particles are both in superposition states but the coefficients for particle $A$ are different from the coefficients for particle $B$ implying that their states are different.
• If we perform the tensor product between the two states, we get a joint state which is a superposition of four joint states, each one having a different coefficient: $$[a_1 b_1 , a_1 b_2 , a_2 b_1, a_2 b_2]$$
• If one or two or three of those four coefficients are nonzero, the joint state is entangled. If the four coefficients are all nonzero, it automatically mean that the joint state is not entangled, correct?

5. Jan 1, 2018

### Zafa Pi

But you said,
i.e. the states of A and B are both superpositions |Φ> and |β>.
No. I am trying to answer your concern in the simplest fashion.
The particles are not in superposition, the states are. The coefficients for |A> and |B> could be the same. Can a = b?
The coefficients are not necessarily different, but could be different.
False. I did say if exactly three are nonzero then the state is entangled.
No. For example [½,½,½,-½] cannot be factored.

I think you may need to bone up on your linear algebra.

6. Jan 1, 2018

### Staff: Mentor

You are confusing several different concepts:

(1) The Hilbert space of a single particle;

(2) The Hilbert space of a two-particle system;

(3) The choice of basis vectors on a Hilbert space.

It appears that both of your particles, $A$ and $B$, have only two possible states; that means they are both qubits, and the Hilbert space of states of each particle, taken in isolation, is the same--the Hilbert space of a qubit. So your states $\vert \Phi \rangle$ and $\vert \beta \rangle$ are one possible pair of basis states on this Hilbert space, and your states $\vert \gamma \rangle$ and $\vert \eta \rangle$ are another possible pair of basis states on the same Hilbert space. There is nothing stopping you from picking a different basis for each particle, but it's not a good idea in a practical sense, as we'll see below.

Given the above, the Hilbert space of the two-particle system consisting of particles $A$ and $B$ is the Hilbert space of two qubits, which is the tensor product of two copies of the one-qubit Hilbert space. If you use the same pair of basis vectors for both copies of the one-qubit Hilbert space, then it is easy to write down a set of basis vectors for the two-qubit Hilbert space: they are just $\vert \Phi \rangle \vert \Phi \rangle$, $\vert \Phi \rangle \vert \beta \rangle$, $\vert \beta \rangle \vert \Phi \rangle$, and $\vert \beta \rangle \vert \beta \rangle$.

However, if you want to use a different basis for the two copies of the one-qubit Hilbert space, you're going to have a much tougher time writing down a basis for the two-qubit Hilbert space, because all of the vectors in a set of basis vectors must be orthogonal and linearly independent. Your vectors $\vert \gamma \rangle$ and $\vert \eta \rangle$ are not orthogonal to or linearly independent of your vectors $\vert \Phi \rangle$ and $\vert \beta \rangle$. So you can't just write a basis of the two-qubit Hilbert space using vectors that look like $\vert \Phi \rangle \vert \gamma \rangle$ and so forth. That's why it's not a good idea to pick different basis vectors for different copies of the one-qubit Hilbert space in a multi-qubit Hilbert space.

So if we pick the basis vectors in the way that is a good idea, per the above, then the most general state in the two-qubit Hilbert space can be written:

$$\vert \Psi \rangle = a_{00} \vert \Phi \rangle \vert \Phi \rangle + a_{01} \vert \Phi \rangle \vert \beta \rangle + a_{10} \vert \beta \rangle \vert \Phi \rangle + a_{11} \vert \beta \rangle \vert \beta \rangle$$

where the $a$'s are complex numbers whose squared moduli must add up to $1$.

Now, how can we tell whether such a state is entangled? As @Nugatory said, the general rule is that a state is entangled if it cannot be factorized; i.e., if there are no one-qubit states $\vert \Psi_1 \rangle$ and $\vert \Psi_2 \rangle$ such that the state $\vert \Psi \rangle$ can be written:

$$\vert \Psi \rangle = \vert \Psi_1 \rangle \vert \Psi_2 \rangle$$

It is obvious that states in which only one of the $a$'s is nonzero are not entangled, since they can obviously be factorized. However, if you look at the product you wrote down in your OP, you will see that the state will also not be entangled if you can find four complex numbers $a_1$, $b_1$, $a_2$, $b_2$ such that

$$a_{00} = a_1 a_2$$

$$a_{01} = a_1 b_2$$

$$a_{10} = b_1 a_2$$

$$a_{11} = b_1 b_2$$

In other words, all of the states you were writing down were not entangled; you failed to consider that there will be states for which you cannot find numbers $a_1$, $b_1$, $a_2$, $b_2$ that will satisfy the above equations--or, to put it the other way around, there will be sets of coefficients $a_{00}$, $a_{01}$, $a_{10}$, $a_{11}$ which describe valid two-particle states but cannot be factorized into numbers $a_1$, $b_1$, $a_2$, $b_2$ according to the above. The states described by those sets of coefficients which cannot be factorized that way are the entangled states.

(Note: I have left out one aspect of all this, that the two-qubit Hilbert space is actually the antisymmetric tensor product of two copies of the one-qubit Hilbert space. For a "B" level thread that seemed like too much additional complication. But it's worth mentioning to be sure it's understood that this rabbit hole goes deeper. )

7. Jan 7, 2018

### fog37

Hello everyone. This is great information. I am in the process of absorbing and making sure your help was not in vain. Let me start by addressing my confusion on concept (1), i.e. The Hilbert space of a single particle.

a) A single particle $A$ that has only two possible states (and all possible linear combinations of these two states) is called a qubit. The basis contains only two basis vectors in this vector space (which is a two-dimensional Hilbert space). In this same vector space, there are infinite other possible bases also containing just two vectors. Are all the two vectors in each different and possible basis always eigenvectors of the observable (linear operator) that we are implicitly referring to for that vector space?

b) The observable spin has only two states (spin up and spin down). But it is possible for the same particle $A$, in a particular physical situation, to have only two states (or three...), for, say, the energy observable or some other observable. For a specific physical problem, if we focus on a particular observable, the number of states in the basis depends on the boundary conditions of the problem ,except for the observable spin for which a particle has always and only bases with two states. The Hilbert space defined by spin is always 2D. For other observables, the dimension of the Hilbert space depends on the physical problem.

Is that correct? any correction? Thank you for the patience.

8. Jan 7, 2018

### Staff: Mentor

Ok so far.

This is stated confusingly. Here is how I would state it: each pair of basis vectors in the Hilbert space of a qubit are eigenvectors of a corresponding Hermitian operator: the one that corresponds to measuring the spin of the qubit in a particular direction. The eigenvalues of the two basis vectors correspond to the "up" and "down" results for that measurement. Each different pair of basis vectors corresponds to a spin measurement in a dfiferent direction.

More precisely, the operator corresponding to the observable "spin in direction d" has only two eigenstates, corresponding to the up and down results of measuring the observable. But there are an infinite number of such observables, one for each possible direction in which you can measure the spin.

Energy is a different degree of freedom from spin; the qubit Hilbert space does not represent any other degrees of freedom besides the qubit's spin. If you want to represent its energy as well, you need to use a bigger Hilbert space that includes more degrees of freedom.

More precisely, the full Hilbert space including all relevant degrees of freedom, if the particles in question are qubits, will be a tensor product of the qubit Hilbert space with some other Hilbert space describing the other degrees of freedom. The number of states in a basis of the other Hilbert space will depend on the specifics of the problem. The number of basis states for the full Hilbert space will just be the number of states in the basis of the other Hilbert space, times two--since the number of basis states in a tensor product Hilbert space is just the product of the numbers of basis states in the individual Hilbert spaces.

9. Jan 7, 2018

### Staff: Mentor

No. Any two linearly independent vectors will work as a basis (not necessarily orthonormal). For a trivial example, if $|+\rangle$ and $|-\rangle$ are eigenvectors of your observable $\hat{O}$, then they form a basis. But so do $|\psi_+\rangle=|+\rangle+|-\rangle$ and $|\psi_-=|+\rangle-|-\rangle$ and these are not eigenfunctions of $\hat{O}$.

10. Jan 8, 2018

### vanhees71

Sure, and for any basis $|u_k \rangle$ ($k \in \{1,2 \}$) you can easily construct a self-adjoint operator for which this basis is the eigenbasis:
$$\hat{O}=\sum_{k=1}^2 o_k |u_k \rangle \langle u_k |, \quad o_k \in \mathbb{R}.$$

11. Jan 8, 2018

### fog37

Thank you!
• @PeterDonis, I was overlooking the important fact that the spin in direction d has only two eigenstates and that each different pair of basis vectors corresponds to a spin measurement in a different direction d. The Hilbert space for spin is always finite and 2-dimensional. The direction $d$ along which the spin measurement is carried out can be any arbitrary direction in space. So, in 3D space, the observable spin comprises three operators which are the three spin operators $s_x$ , $s_y$ and $s_z$. Once we orient the three Cartesian spatial axes in space, a spin measurement along any general spatial direction $d$ between the three Cartesian axes $x,y,z$ must refer to a spin operator that is a (linear) combination of the spin operators of $s_x$ , $s_y$ and $s_z$. Is that correct?
• Just to make sure: as you mention, the energy operator $E$ is a different degree of freedom from spin. I think I there are situations in which the system's energy states can be described using bases with a finite number of energy states. I think finite quantum well problems are an example in which the basis of energy eigenvectors could be two dimensional depending on the well height. So a system in that situation would seem to be a qubit system as well.
• I think the following is correct but please feel free to rip it apart: there are many observables (energy, position, spin, angular momentum, etc.) that can describe a physical system. Each different observable "defines" a Hilbert vector space of possibly different dimension $D$ within which we find all the possible states the system can be in. The "total" overall state $|\Psi_{total}>$of the system is a ket state vector that lives in the "total" Hilbert vector space equal to the tensor product of the Hilbert vector spaces associated to each different observable (Pardon the language. I am not sure it is correct to say that a Hilbert space is associated to an observable...). Using imprecise language, it seems that the system's characteristics and properties exist across different Hilbert spaces associated to different observables....Different observables are also different degrees of freedom. Are all observables independent degrees of freedom or are some observables that are interdependent?
• The wavefunction $\Psi(x)$ is a function of the space variable $x$. The function $\Psi(x)$ represents the projection of the system's state $|\Psi>$ onto the eigenstates $|x>$ of the position operator. Just to make sure, my understanding is that $\Psi(x)$ the projection of the total system's state $|\Psi_{total}>$ that lives in the total Hilbert space onto the eigenstates $|x>$ of the position operator. We could project the total system's state $|\Psi>$ onto the eigenstates $|p_{x}>$ of the momentum operator and obtain the function $\Psi(p_x)$. In principle, we would obtain a function $\Psi(x,y,z,p_{x,}p_{y},p_{z},L_{x},L_{y},L_{z},s_{x},s_{y},s_{z},...)$ if we could project the total state $|\Psi_{total}>$ onto the eigenstates of all Hermitian operators....Is that completely wrong?

12. Jan 10, 2018

### fog37

So sorry if I am overwhelming everyone with my numerous queries. I am trying hard to piece together all your hints and teachings and relate them to the content of the book I am reading (quantum mechanics by Griffiths). I can assure you that your help goes a long way.

Thanks again,
fog37

13. Jan 10, 2018

### Staff: Mentor

Yes; the linear combination is the dot product of a unit vector in the chosen direction with the "vector" whose components are the spin operators in the $x, y, z$ directions. Wikipedia has a brief discussion here:

https://en.wikipedia.org/wiki/Spin_(physics)#Measurement_of_spin_along_an_arbitrary_axis

No, there aren't. Bound quantum systems have discrete energy eigenstates--i.e., the states are labeled by a discrete index (usually called $n$) instead of a continuous parameter. But there are an infinite number of such discrete eigenstates for any bound quantum system. For example, the electron in a hydrogen atom is a bound quantum system, with an infinite number of possible energy levels. It's just that all of them except for a few, corresponding to finite and fairly small values of $n$, are irrelevant in practice because there is such a low probability of the electron occupying them.

No. The system has one Hilbert space. It's just that, for our human convenience, we express that Hilbert space as a tensor product of various smaller Hilbert spaces corresponding to different observables. From the standpoint of the total Hilbert space of the system, operators that measure particular observables are really operators that act non-trivially on only a portion of the total state vector; they just act as the identity operator on the rest.

For example, consider the electron in the hydrogen atom again, and suppose the only observables we are interested in are energy and spin, so we model the system in a Hilbert space that is the tensor product of the qubit Hilbert space (for spin) and a Hilbert space with a countably infinite basis of eigenstates labeled by $n$ (for energy). Write this Hilbert space as $H = H_E \otimes H_s$, where the $E$ subscript stands for energy and the $s$ subscript stands for spin. Then an operator that measures spin in some particular direction would be written as $I_E \otimes S_\hat{n}$, where $I_E$ is the identity operator acting on $H_E$ and $S_\hat{n}$ is the spin operator in the direction of the unit vector $\hat{n}$ acting on $H_s$. Similarly, an operator that measures the electron's energy would be written $E \otimes I_s$, where $E$ acts on $H_E$ and $I_s$ is the identity operator on $H_s$.

Not necessarily. You have to look at the actual representation of the Hilbert space and how it factors. For example, the position and momentum observables both act on the same factor of the system's total Hilbert space, so they are not different degrees of freedom; they are just different representations of the same degrees of freedom.

A better way to say this would be: the wavefunction $\Psi(x)$ that represents the position/momentum degrees of freedom, in the position representation, is a function of the space variable $x$. This wavefunction does not represent the entire Hilbert space unless there are no other observables except position/momentum and observables that are functions of them (like energy). For example, it does not represent the spin degrees of freedom.

No. It only projects the portion of the Hilbert space that represents the position/momentum degrees of freedom. See above.

14. Feb 18, 2018

### fog37

Hello Zafa Pi and everyone else,

I am honestly still confused on how a composite state with the coefficients like [½,½,½,-½] cannot be factored. In general, I am clearly missing the point of how to determine if a state is entangled or not based on the value of the four coefficients. In some cases, it seems that having even three or four nonzero coefficients (depending on what the coefficients are) can result in either an entangled (not a product state) or non entangled state ( a product state). Given the expression below,
$$\vert \Psi \rangle = a_{00} \vert \Phi \rangle \vert \Phi \rangle + a_{01} \vert \Phi \rangle \vert \beta \rangle + a_{10} \vert \beta \rangle \vert \Phi \rangle + a_{11} \vert \beta \rangle \vert \beta \rangle$$
where the $a$'s are complex numbers whose squared moduli must add up to $1$, is it all about recognizing if there the four product states in the summation above form two basis vectors? In that case, the summation would be a superposition of two basis vectors resulting in a single, non entangled, product state...

I am clearly still not getting this important point. Any hint?

15. Feb 18, 2018

### Staff: Mentor

No. Write the state this way:

$$\vert \Psi \rangle = a_{00} \vert \Phi \rangle_1 \vert \Phi \rangle_2 + a_{01} \vert \Phi \rangle_1 \vert \beta \rangle_2 + a_{10} \vert \beta \rangle_1 \vert \Phi \rangle_2 + a_{11} \vert \beta \rangle_1 \vert \beta \rangle_2$$

The subscripts 1 and 2 make it clear which ket applies to which subsystem (particle 1 or particle 2) of the total system. Now see if you can factor the RHS of the above into an expression of this form:

$$\left( A_0 \vert \Phi \rangle_1 + A_1 \vert \beta \rangle_1 \right) \left( B_0 \vert \Phi \rangle_2 + B_1 \vert \beta \rangle_2 \right)$$

If you cannot do this, then the state is entangled.

So we have $a_{00} = a_{01} = a_{10} = - a_{11} = \frac{1}{2}$ in the first expression above. Now multiply out the factors in the second expression; that gives

$$A_0 B_0 \vert \Phi \rangle_1 \vert \Phi \rangle_2 + A_0 B_1 \vert \Phi \rangle_1 \vert \beta \rangle_2 + A_1 B_0 \vert \beta \rangle_1 \vert \Phi \rangle_2 + A_1 B_1 \vert \beta \rangle_1 \vert \beta \rangle_2$$

This means that, for the composite state with coefficients [½,½,½,-½] to factor, we must have $A_0 B_0 = A_0 B_1 = A_1 B_0 = - A_1 B_1 = \frac{1}{2}$. But that's impossible. The first equality requires $B_0 = B_1$, but the last equality requires $B_0 = - B_1$. The only way for that to be true is for $B_0 = B_1 = 0$, but that amounts to saying particle 2 does not exist and we only have a one-particle system.

16. Feb 18, 2018

### fog37

Thank you PeterDonis. Very grateful.

17. Apr 11, 2018

### fog37

Hello PeterDonis,

I was re-reading your last commment/explanation in this thread. It is now clear how to understand if a composite state is entangled or not given the set of four coefficients $[a_{00}, a_{01}, a_{10}, a_{11}]$. But I am now wondering about what this means. Given the expression $$A_0 B_0 |\Phi>_1 |\Phi>_2 +A_0 B_1 |\Phi>_1 |\beta>_2 + A_1 B_0 |\beta>_1 |\Phi>_2 + A_1 B_1 |\beta>_1 |\beta>_2$$

Entanglement or not (if that expression can be reduced to a single product), after the measurement is carried out, the two particle system will collapse to one of those four joint states in the above superposition. For instance, say the system collapses to $A_1 B_1 |\beta>_1 |\beta>_2$ which means that particle 1 will be in state $\beta>$ and particle 2 in state $\beta>$ as well. This type of end result happens if the particles are entangled and also where the particles are not entangled. For particles that are not entangled, the composite state is surely separable and can be expressed as the product of a state pertaining just to particle 1 (the state can be a superposition or not) and a state pertaining just to particle 2 (this state can be a superposition or not) but we seem to still know even in that case that one particle will end in one state and the other particle will end up in the other since we have a single joint state to worry about. I am clearly missing something fundamental here. The presence of entanglement is all about being able to predict what the other particle's observable value will be while when entanglement is not present that is not possible.

Last edited by a moderator: Apr 11, 2018
18. Apr 11, 2018

### Staff: Mentor

This expression had a typo in the second term ($\Phi_2$ instead of $\beta_2$). I have edited to fix.

A better way of putting it is that the two particle system gets changed to two separate one particle systems by the measurement. In other words, you can no longer exchange the two particles, because you've measured (at least) one of them and so you now have a record of which one is which. So the question of what happens to the overall wave function when you exchange the particles is now meaningless, since you can't.

No, if the state is separable then you don't have a "joint state", because there is no way to exchange the particles. See above.

No, the presence of entanglement means you do not have two separate particles; you have one quantum system with two "particles" worth of degrees of freedom. When you make a measurement of one "particle", you are changing the system. See above.

19. Apr 16, 2018 at 5:04 PM

### fog37

Hello PeterDonis and everyone else. Thanks agina.
I have been thinking about this thread and tried to summarize what I understood. Hope this is a good summary for everyone else too that is seeking clarity like me:

Let's consider two photons and the orthogonal basis composed by $|H>$ and $|V>$, the horizontal and vertical polarization states. With subscript $1$ indicating photon 1 and subscript $2$ indicating photon 2, photon $1$ is in the superposition state $(a|H>_1+b |V>_1)$ and photon 2 in the state $(c |H>_2+d |V>_2)$. The two-photon joint state is $$|\Psi>_{12} = \alpha |H>_1|H>_2+ \beta |H>_1|V>_2+ \gamma |V>_1|H>_2+ \delta |V>_1|V>_2$$
with $\alpha = ac$, $\beta = ad$, $\gamma = bc$ and $\delta = bd$.
• The probability of getting photon 1 with horizontal polarization and photon 2 with vertical polarization is $\beta^2$ if we carried out the first measurement on photon 1 and a second measurement on photon 2. When we measure horizontal polarization for photon 1 and vertical polarization for photon 2, the two-photon joint state becomes exactly $|H>_1|V>_2$.
• The probability of getting horizontal polarization for photon 1 without worrying about photon 2 (i.e. without executing the 2nd measurement on photon 2) is equal to $\alpha^2 + \beta^2$. If a measurement has been carried out only on photon 1 and obtained horizontal polarization, the state of the composite system becomes $c |H>_1|H>_2+ d |H>_1|V>_2$ after the measurement. The coefficient $a$ has become 1 and the terms in the superposition that involve $|V>_1$ are removed.
After measuring the polarization of photon 1, if the states of the two photons are not entangled, the probability of obtaining horizontal polarization for photon 2 is $c^2$ and vertical polarization $d^2$. These are the same probabilities that existed even before we measured the polarization of photon 1. The fact that we measured the polarization of the first photon does not influence the results ( and associated probabilities) for photon 2. This is true if the joint state is separable.

But if the joint state is not separable and the states of the two photons are entangled, then the situation is different. For example, for the joint state $$\frac {1}{\sqrt 2} |H>_1|H>_2 + \frac {1}{\sqrt 2} |V>_1|V>_2$$, if we measured horizontal polarization for photon 1, the polarization of photon 2 will be determined with certainty and vice versa.

The state $$\frac {1}{\sqrt 2} [|H>_1|H>_2 - |H>_1|V>_2 - |H>_1|V>_2 + |V>_1|V>_2]$$ is instead separable: if we measured horizontal polarization for photon 1, the (unnormalized) system state would become $$\frac {1}{\sqrt2} [|H>_1|H>_2 - |H>_1|V>_2 - |H>_1|V>_2 ]$$. The polarization of photon 2 is not automatically determined in this case. It can be either horizontal with probability $\frac {1}{2}$ or vertical with probability $\frac {1}{2}$. By the way, I am not sure if the $\frac {1}{\sqrt 2}$ term is correct and if it should be just $\frac {1}{ 2}$ since the sum of the squared coefficients must equal 1.

Thanks for any correction and comment.

20. Apr 16, 2018 at 7:42 PM

### Staff: Mentor

If you are saying this, it means you don't have a two-photon system; you have two one-photon systems.

But if you are saying this, then you don't have two one-photon systems, you have a two-photon system.

So which case do you want to discuss?

If this is the case, then the state factors into two one-photon states, so it's not entangled; as above, you really have two one-photon systems, not a two-photon system. So again, which case do you want to discuss?

I can't comment on the rest of your post because of the above.