# I Two particles: Entangled State vs Non interacting State

1. Feb 12, 2017

### fog37

Hello Forum,

Could anyone offer some insight on this topic?

In the case of a physical system composed of multiple particles, there is only one single wavefunction that depends on the three position variables of each particle and the parameter time t. For example, in the case of 2 particles, we have
$$\Psi(r_1, r_2, t)$$
where $r_1=(x_1,y_1,z_1)$ and $r_2=(x_2,y_2,z_2)$.
• For the hydrogen atom, there is only electron which can be in one particular state (among the many) or in a superposition of multiple states. A state that hosts a single electron is called orbital, correct?
• For a system with two electrons, if we apply the approximation that the electrons are not interacting, we can express the two-particle wavefunction as a product of single electron wavefunctions, i.e. orbitals. These orbitals can be eigenstates of a certain observable. The non-interaction seems to be a strong approximation. Because of indistinguishability, we don't know which particle is in which mono-electronic (orbital) state.
• In the case of entanglement between two particles, we say that total state describing the two particle can actually not be expressed as a product of orbital states, i.e. the wavefunction cannot be factorized, correct? In general, even without entanglement, we cannot tell in which state each particle may be (that is due to the fact that the particles are identical and indistinguishable).
• Entanglement is about the existence of a correlation between the observables associated to each one of the two particles. Correlation means that if we know the value of the observable for one particle we automatically know the value of the observable for the other particle. For instance, if particle 1 is in state $|+>$, particle 2 will be for sure in state $|->$. If particle 1 is in state$|->$ , particle 2 will be in state $|+>$. What if the entangled observables have more than two possible values?
• For two possible values of the observables, the two particle can be either in the state $|+ - >$ or $|- - >$ or even a superposition state |Phi>= c_1* $|+ -> + c_2*|- +>$. When the two particle are in the superposition state c_1* $|+ -> + c_2*|- +>$, we say they are entangled because the physical state of one of the particles on its own does not exist. Only the ‘pair state’ exists. But isn't that the case also for a multiple particles system having just a single total wavefunction (if we don't apply the non-interaction approximation) even without entanglement? Entangled particles and interacting particles are not the same thing but their total wavefunctions seem to not be deconstructed in a product of single particle wavefunctions...
Anyhow, I am still clearly struggling to grasp the essence of entanglement...

Thanks,
Fog37

2. Feb 12, 2017

### jfizzix

The quantum state of a single particle or a group of particles in a pure state is defined by its wavefunction (up to a constant term of magnitude 1).

It is still under debate whether a single particle must be in a pure state, but since experiments aren't perfect, we use density matrices to describe the state of a distribution of particles, each of which may be in a slightly different state.

To be particular, an orbital is an eigenstate of energy, the magnitude of orbital angular momentum, and (depending on convention) the z-compinent of orbital angular momentum as well for an electron bound to an atom. For example, the 1s orbital is the quantum state of both minimal energy, and zero angular momentum. The state of the electron may be a superposition of multiple orbitals, and is not guaranteed to be in a single orbital.

If we consider two non-interacting electrons bound to the same proton, it would be a valid approximation to say that their joint wavefunction is the product of their individual wavefunctions, if they were distinguishable particles. However, because they are identical fermions, the joint wavefunction must be rendered anti-symmetric, and both electrons cannot be in precisely the same orbital and spin state at the same time (this is the Pauli exclusion principle).

A pair of particles in a joint pure state is entangled if and only if their joint wavefunction cannot be separated into a product of individual wavefunctions. Alternatively, a pair of particles in an arbitrary state is entangled if and only if their joint state cannot be expressed as a distribution of product states.
A pair of particles can become entangled only through interaction (direct or indirect), or the circumstances of their creation.

Entanglement is more than correlation, though you can't have entanglement without it.
It's possible to have a distribution of product states give the correlations in the z-component of spin as you describe, but only an entangled state can exhibit correlations in the other components of spin as well.

One example of entanglement between observables with more than two possible values is position-momentum entanglement.
if a particle A decays into a pair of particles B and C, then conservation of momentum requires that the momenta of B and C are highly anti-correlated, so they add up to the momentum of A every time. In addition, the initial positions of B and C are highly correlated because they are right on top of each other, having come from A.
This means it's possible to predict the position of B by knowing the position of A, and it's possible to predict the momentum of B by knowing the momentum of A. The only way to have strong correlations/anti-correlations in both position and momentum is for there to be entanglement between A and B in that degree of freedom. The original EPR paradox is a fine example utilizing position-momentum entanglement.

3. Feb 12, 2017

### Staff: Mentor

No. An orbital is a term for a state with a particular definite set of quantum numbers (n, l, m), where n is the energy level, l is the orbital angular momentum, and m is the z component of orbital angular momentum. But an electron also has a spin quantum number, so each orbital can hold two electrons, in states of opposite spin.

No. You are leaving out electron spin. See above.

This is correct if you say "single particle states" instead of "orbital states". See above.

Yes, but the correlation does not have to be 100%. In other words, measuring one particle does not necessarily give you 100% accurate information about measurement results on the other particle. This can be true even if you measure the same observable (for example, spin in the same direction) on both particles; but if you measure different observables (such as spin in two different directions), things get even more complicated.

This assumes a particular entangled state of the two particles (roughly that total spin is zero), and it also assumes that you are measuring the same spin observable on both. If you were to measure, say, spin in the z direction on one and spin in the x direction on the other, a + result on one would only give a 50-50 chance of a - result on the other.

No. Assuming we are talking about fermions (which electrons are), the wave function has to be antisymmetric, i.e., it has to change sign when you swap particles. So a valid two-fermion state would be $|+-> - |-+>$. But $|+->$ itself would not be valid, since by itself it is not antisymmetric (it has to be combined with $|-+>$ as I just described), and $|-->$ can't even occur as a term in a valid two-fermion state at all (because there's no other term you could possibly combine it with to make the wave function as a whole antisymmetric).

Now a question for you: is the two-fermion state I wrote down just now, $|+-> - |-+>$, entangled? Why or why not?

4. Feb 12, 2017

### fog37

thanks.

Just a quick comment about the orbital before we move forward.

Two electrons cannot have the same set of four quantum numbers by the Pauli exclusion principle. So each electron has to live on a distinct and separate state since each state is identified by the four quantum numbers (Of course, talking about electrons living in different states implies that the electrons are not interacting. If the electrons were strongly interacting, they would both live in the same state).

Assuming the two electrons are not interacting, the term orbital implicitly indicates two states. The wavefunction for each of the two electrons is the product of a spatial part and a spin part (which is not a function but a 2D vector):

$$\Psi(x,y,z) \phi(s)$$

Each electrons is in this state but the spin part is different: $s$ can be either $\frac {1}{ 2} \hbar$ or $\frac {-1}{2} \hbar$. An orbital like $1s$ that can host two electrons is in reality two states that differ from each other only in the spin part of the wavefunction and have the same spatial part $\Psi(x,y,z)$. Right??

5. Feb 12, 2017

### Staff: Mentor

Why do you think this? It's not correct. The Pauli exclusion principle always applies. It doesn't somehow get overridden when the electrons (or any fermions) are strongly interacting.

Yes, which is why I objected when you said it described "a state that hosts a single electron".

Only if the spatial part and the spin part are not entangled. They can be.

Yes. But the term "orbital" is only used for the spatial part (or, if you view it in terms of quantum numbers, it's only used for sets of unique values of the first three quantum numbers (n, l, m), and doesn't include the fourth quantum number s).

Also, there are electron wave functions in which the spatial part is not a single orbital but a superposition of them. And there are also electron wave functions where the spatial part and the spin part are entangled, for example, something like $|1s, +> + \ |2s, ->$. Such a wave function (note that it is for a single electron, not two electrons with different orbitals) does not factorize into functions $\Psi(x, y, z) \phi(s)$.

6. Feb 12, 2017

### fog37

I guess I get hang up on the fact that if we left all possible approximations aside there would always be one single wavefunction, i.e. one single state, describing the two or multi-electron system. The application of the approximations (nuclei are very slow, electrons don't interact, etc.) lead us to talk about multiple state for each separate electron....

7. Feb 12, 2017

### Staff: Mentor

That's true. But I don't see how it supports what you said that I objected to. If you want to talk about one single state describing, say, a two-electron system, then that state has to be antisymmetric under electron exchange. That is true whether the electrons are strongly interacting or not.

8. Feb 13, 2017

### fog37

Thanks.
• Yes, if the atom/molecule has only a single electron then the orbital is just a single state but if there are many multiple electrons, an orbital becomes a set of two states occupied by two different electrons.
• As you mention, there is always a single wavefunction for two or more electrons. That single total wavefunction must be antisymmetric in the sense that i under particle interchange (which is equivalent to interchange the particles' position), the total wavefunction must be:
$$\Psi (r_1, r_2) =- \Psi(r_2, r_1)$$
I guess the total wavefunction $\Psi(r_1, r_2)$ also and implicitly includes the spin part $\eta(s)$ as well. So a function multiplied by a 2D spin vector seems to produce a vector having two wavefunctions as components:
$$\Psi(r_1, r_2)= \psi(r_1, r_2) \eta(s)= \psi(r_1, r_2) \begin{pmatrix} s_1 \\ s_2 \end{pmatrix}= \begin{pmatrix} \psi(r_1, r_2) s_1 \\ \psi(r_1, r_2) s_2 \end{pmatrix}$$
For some reason, this does not help me yet to see how fermions must obey the exclusion principle and be in different states since for the function $\Psi(r_1, r_2)$ still contemplates both electrons in one state.
• When we assume that two electrons are not interacting, then the total wavefunction $\Psi(r_1, r_2)$ can be expressed in terms of single electron wavefunctions $\psi_a (r_1)$ and $\psi_b (r_2)$ which are two different states, the first being the state for particle at location $r_1$ and the other being the state for the electron located at $r_2$. We are also implicitly referring to $\psi_a (r_1) = psi_a (r_1) \eta(s_1)$ and $\psi_a (r_1) \eta(s_2)$
$$\Psi(r_1, r_2)= \psi_a (r_1) \psi_b (r_2)$$
But since the particles are indistinguishable, we could write
$$\Psi(r_1, r_2)= \psi_a (r_2) \psi_b (r_1)$$
For fermions like electrons,
$$\Psi(r_1, r_2)= \frac {1}{\sqrt {2}} [\psi_a (r_1) \psi_b (r_2)- \psi_a (r_2) \psi_b (r_1)]$$

Then $\Psi(r_1, r_2)$ becomes zero when either electron occupies both states $\psi_a (r_1) = psi_a (r_1) \eta(s_1)$ and $\psi_a (r_1) \eta(s_2)$

9. Feb 13, 2017

### DrChinese

That's no problem. Usually there is some kind of conservation rule at work. Total momentum conserved, total energy, etc. So the values can be a range as long as the total does not change.

10. Feb 13, 2017

### Staff: Mentor

No, it isn't. You're confusing the orbital with the state of the electron. They're not the same. The orbital is a set of two possible states of the electron. The single electron can be in one of those states, but that doesn't make the orbital a single state.

No, it isn't. You have to take into account the entire wave function, not just the spatial part.

No, the total wave function is not $\Psi(r_1, r_2)$. It's $\Psi(r_1, r_2, s_1, s_2)$, i.e., it's a function of two positions and two spins. In some cases you can factor this into separable position and spin parts, i.e., $\Psi(r_1, r_2, s_1, s_2) = \Phi(r_1, r_2) \eta(s_1, s_2)$. But in some cases you can't.

No, it doesn't. It describes one two-electron state, but that is not the same as two electrons both being in the same single-electron state. (Also, as above, the arguments to $\Psi$ have to include the spins, not just the positions.)

Ordinary language is a very poor tool for this job; you need to look at the math. For example, consider the following two-electron state:

$$\Psi(r_1, r_2, s_1, s_2) = \vert 1s, + \rangle \vert 2s, - \rangle - \vert 2s, - \rangle \vert 1s, + \rangle$$

This is an allowed two-electron state in which the position and spin parts are entangled. In ordinary language, we might describe it as "one electron in a 1s orbital with spin up, and one electron in a 2s orbital with spin down". So it is not the case that both electrons are "in the same state" in the single-electron sense--we clearly have two electrons in two different single-electron states. But the two electrons are entangled so we can't factor the wave function into separate wave functions for each electron. We can't even factor it into a two-electron wave function for position and a two-electron wave function for spin.

I think you need to take a step back and re-think everything you're saying, using the above two-electron state as an example: if what you're saying can't be true of that state, then it's wrong.

11. Feb 20, 2017

### fog37

Thanks PeterDonis,

I am still processing your answer and get back to that soon. Thank you.

On a different and simpler side note, I have a quick doubt about the action of a linear operators in QM. In linear algebra a linear operator is a transformation that converts one vector in the vector space into a different vector in the same vector space. In general, a linear operator $\hat {F}$ acts on the state $|\Psi>$ of the system and converts that state into a different state $|\Phi>$ (unless the initial state is an eigenstate of the operator itself):
$$\hat {F} |\Psi > = |\Phi >$$

where $|\Psi> \neq | \Phi>$.
In QM, the act of measuring an observable corresponds to the action of the operator $\hat {F}$ on the current state of the system $| \Psi>$. After the measurement, the system always ends up in one of the eigenstates of the operator itself (unless it is already in one of the eigenstates) and the measurement result is the eigenvalue of the corresponding eigenstate. That implies that a linear quantum mechanical operator corresponding to an observable always converts a certain arbitrary state into one of the eigenstates (upon measurement):
$$\hat {F} |\Psi > = |\Phi >$$
where $| \Phi>$ is one of the operator $\hat {F}$ eigenstates. So operators corresponding to observables always and only convert any input state into one of their eigenstates. It is not possible to obtain any other type of output state, correct?

I know there are also linear operators in QM that are not Hermitian or that do not correspond to observables (for example, unitary operators). I guess these operators can then convert some generic input state into a state that is not necessarily one of their eigenstates.

Am I on the right track?

Thanks,
Fog37

12. Feb 20, 2017

### Staff: Mentor

Only on a collapse interpretation. And on a collapse interpretation, something else has to happen to the quantum state over and above the action of the linear operator representing the observable. Wave function collapse is a nonlinear operation.

No, it doesn't; this is impossible for a linear operator, since a linear operator must obey the superposition principle. In other words, if $\vert \Psi \rangle = a \vert \alpha \rangle + b \vert \beta \rangle$, then $F \vert \Psi \rangle = a F \vert \alpha \rangle + b F \vert \beta \rangle$. This is part of the definition of a linear operator. So if $\vert \alpha \rangle$ and $\vert \beta \rangle$ are eigenstates of $F$, then $F \vert \Psi \rangle$ must be a superposition of eigenstates, not an eigenstate itself.

This should make it obvious why wave function collapse cannot be a linear operation. It also might make it clearer where no collapse interpretations like the MWI come from: they just leave out the nonlinear collapse part.

No. See above.

13. Feb 20, 2017

### fog37

Ok, thanks PeterDonis. I see how the wavefunction collapse process is nonlinear and hides certain steps. So linear operators can covert a certain state into a different state other than one of its eigenstates.

So what is a simple example of a linear operator that corresponds to an observable that changes the state of the system into a different state? I struggling to think of a particular situation like. Most of the time we discuss eigenvalues equations and eigenvectors in relations to linear operators...

14. Feb 23, 2017

### Staff: Mentor

Any operator will do that to any state that is not an eigenstate. Linear operators have the additional property of obeying the superposition principle (which I stated in my previous post). That is what makes them important in QM.

The best simple examples IMO are the spin operators. However, we have to be careful about saying what "changes the state" actually means. Look at the general formula I gave for the action of a linear operator $F$ on a state $\vert \Psi \rangle$ that is a superposition of eigenstates $\vert \alpha \rangle$ and $\vert \beta \rangle$ of $F$. Since $\vert \alpha \rangle$ and $\vert \beta \rangle$ are eigenstates, we must have $F \vert \alpha \rangle = A \vert \alpha \rangle$ and $F \vert \beta \rangle = B \vert \beta \rangle$, where $A$ and $B$ are numbers. So if $\vert \Psi \rangle = a \vert \alpha \rangle + b \vert \beta \rangle$, then from the linear operator formula I gave, we must have $F \vert \Psi \rangle = a A \vert \alpha \rangle + b B \vert \beta \rangle$.

Note that this state $F \vert \Psi \rangle$ looks very similar to $\vert \Psi \rangle$; the only thing that keeps it from being just a multiple of $\vert \Psi \rangle$ (which would have all the same physical properties as $\vert \Psi \rangle$ itself) is if the condition $A \neq B$ is satisfied. In other words, as long as $A \neq B$, then $F$ does indeed change $\vert \Psi \rangle$ to a physically different state.

For the case of spin operators, take for example the $z$ spin operator, usually represented as the Pauli matrix $\sigma_z$. In the simplest case, where this operator acts on a single spin-1/2 particle, it has two eigenstates, which we can represent as $\vert z+ \rangle$ and $\vert z- \rangle$, with eigenvalues $+ 1/2$ and $- 1/2$. So a general spin state $\vert \Psi \rangle = a \vert z+ \rangle + b \vert z- \rangle$ will be acted on as follows: $\sigma_z \vert \Psi \rangle = \frac{1}{2} a \vert z+ \rangle - \frac{1}{2} b \vert z- \rangle$. This is evidently a different state from $\vert \Psi \rangle$.

15. Feb 25, 2017

### fog37

Thanks PeterDonis.

1) I guess your point (if I understood it) is that an operator will always change a state $|\Psi>$ that is not one of operator's eigenvectors (and is therefore a superposition of eigenvectors) into a different state $|\Phi>$ (not an eigenvector either, hence also a superposition (different expansion coefficients) of the operator's eigenvectors;

2) Interesting: if the state $|\Psi>$ is a superposition of two eigenstates that have different eigenvalues $A$ and $B$, then the final state $\hat{F} |\Psi> = | \Phi>$ will be different from the initial state $|\Phi>$. If the two eigenstates had the same eigenvalue (degeneracy) then the final state would be the same physical state as the initial input state (just a scaled version, which in QM means the same physical state. Scaled versions of the same state are the same physical state because once normalized they represent the same state?).

Thanks.

16. Feb 25, 2017

### Staff: Mentor

Correct.

17. Feb 25, 2017

### fog37

Ok, back to observables and entanglement. Let me summarize:
• The observables that are used to describe a system are the momentum, energy, position, angular momentum and spin. Am I forgetting any other? Is the spin observable on a different footing than the other observables? I think so. How is spin, which is an internal property of the system, different from the other types of observable? All observables have eigenstates that admit a position representation with the exception of spin. Is that the difference? Some systems are two-level systems which means that there are only two observable eigenstates for a specific observable (for ex. the energy observable). That is similar to spin which always has two eigenstates (regardless of the system under study). But still, the two state observable and spin seem to be observables on different footing.
• Orbital: jfizzix said that an orbital is an eigenstate.This eigenstate is simultaneously an eigenstate for the energy, magnitude of orbital angular momentum and z-component of orbital angular momentum operators because all these mentioned operators commute in the case of an electron bound to an atom under a central force field. This commutation may not always exist for other potential function $V(r)$.
• An electron may be in a state that is a superposition of multiple orbitals. In the case of two non-interacting electrons (indistinguishable particles like any particle in QM) bound to the same proton, we can approximate their total joint wavefunction as the product of their individual wavefunctions. That joint wavefunction must be anti-symmetric. Both electrons cannot be in precisely the same state of the orbital (Pauli exclusion principle). The orbital is actually and always comprising two different states which are difference because of the spin.
• Entanglement: that same pair of electrons (or any indistinguishable particles) in a joint pure state is entangled iff their joint wavefunction cannot be separated into a product of individual wavefunctions. To go a step further, why does the lack of this product imply entanglement? What does the ability to separate the wavefunction into a product imply entanglement?
• I am still confused about particles interacting and non-interacting. If two particles are non-interacting it means they are independent and can never be entangled, correct? But two interacting particles are not necessarily entangled. The joint wavefunction of interacting particles cannot separated either.
• It is possible to entangle two or more particles and two or more of their observables. jffix provided theposition-momentum entanglement. But that does not imply that one observable is cross-correlated with the other. It just means that identical observables are correlated for the two particles. By cross-correlation, I mean , classically, that given two properties A and B and two systems 1 and 2, we could correlate property A of system 1 with property B of system 2. Can that happen with entanglement?

18. Feb 25, 2017

### DrChinese

1. Time is often thrown in the list of basic observables. When it comes to entanglement, spin entanglement is not on a different footing than other entanglement.

2. Yes (depending on where you draw the line). There are many combinations of entanglement possible, and composite observables can be created. You can entangle a photon and an electron, for example. In fact, there are many weird things that can be entangled, and these lead to some odd looking Bell inequalities. Here are a couple that happen to involve Bose-Einstein Condensates (BEC).

https://arxiv.org/abs/1604.06419
Characterizing many-body systems through the quantum correlations between their constituent particles is a major goal of quantum physics. Although entanglement is routinely observed in many systems, we report here the detection of stronger correlations - Bell correlations - between the spins of about 480 atoms in a Bose-Einstein condensate. We derive a Bell correlation witness from a many-particle Bell inequality involving only one- and two-body correlation functions. Our measurement on a spin-squeezed state exceeds the threshold for Bell correlations by 3.8 standard deviations. Our work shows that the strongest possible non-classical correlations are experimentally accessible in many-body systems, and that they can be revealed by collective measurements.

https://arxiv.org/abs/1201.0248
We propose a robust scheme to prepare three-dimensional entanglement states between a single atom and a Bose-Einstein Condensate (BEC) via stimulated Raman adiabatic passage (STIRAP) techniques. The atomic spontaneous radiation, the cavity decay, and the fiber loss are efficiently suppressed by the engineering adiabatic passage. Our scheme is also robust to the variation of atom number in the BEC.

19. Mar 2, 2017

### fog37

Thank you.

I am winding things back to the expression of an orbital for moment.

For the hydrogen atom, the stationary states are wavefunctions $\psi_{n \ell m_\ell}$ where $n$, $\ell$ and $m_\ell$ are the primary quantum number, the secondary (or orbital) quantum number, and the magnetic quantum number respectively. The state $\phi_{n \ell m_\ell}$ is a complicated function with a radial and angular component:
$$\psi_{n \ell m_\ell} = R_{n \ell} Y_{\ell}^{m_\ell}(\theta, \phi)$$
When there are two electrons involved, we are forced to introduced the 4th quantum number $m_s$, i.e. the spin quantum number. How does the total state with the four quantum numbers
$$\psi_{n \ell m_\ell m_s}$$
look like? Is it a function? How do we incorporate the label $m_s$ into the wavefunction? By itself, $\psi_{n \ell m_\ell}$ is a function with 3 labels and three spatial coordinates $(r, \theta, \phi)$.

For instance, the two states $\psi_{000, \frac{1}{2}}$ and $\psi_{000,\frac {-1}{2}}$ have identical wavefunctions (same probability distribution density) except for the spin quantum numbers. Do we simply distinguish these two states from the label $m_s$ but their wavefunction is not affected and is exactly the same?

I don't think the single electron in the hydrogen atom can be in a superposition of two the states $\psi_{000, \frac{1}{2}}$ and $\psi_{310, \frac{1}{2}}$ with the same spin quantum number $m_s = \frac {1}{2}$, correct?

What about the superposition of the following states $\psi_{320, \frac{1}{2}}$, $\psi_{321, \frac {-1}{2}}$ and $\psi_{32,-2, \frac{1}{2}}$? Is that an allowed superposition?

Thanks!