QM vs SR: The Paradox of Simultaneity in Particle States

[tom.stoer]

You need not consider interactions; formulated in terms of momentum space creation operators it becomes a purely algebraic exercise as you can see from my "calculation".

 

[RUTA]

You need not consider interactions; formulated in terms of momentum space creation operators it becomes a purely algebraic exercise as you can see from my "calculation".

But, if you want to get the new state for the two electrons, which previously were described independently of each other by one in the same algebraic system (you don't have a unique description for every fermion in the universe), something has to be changed. Right? So, how close to the atoms have to be to necessitate a new mathematical description for the pair as opposed to using the same description for the individuals?

 

[tom.stoer]

No, nothing changes. It's only the formulation in position space that is not very helpful as to create the |100+,R> state one has to use a complicated operator which is essentially an infinite product of elementary momentum space creation operators.

Think about a one-dimensional compact space (a circle) which results in discrete momentum space. As momentum is discrete all allowed states can be described using the following notation

$$|n_1^+ n_1^-, n_2^+ n_2^-, n_3^+ n_3^-, \ldots>$$

Here any n can take the values 0 or 1 and + and - refer to the spin quantum number.

Any allowed state can be written as

$$|10, 00, 11, \ldots>$$

You cannot identify one individual electron. You cannot describe any two electrons independently. They always know each other due to the Pauli principle; this principle does not care about there separation in position space.

$$a^\dagger_Y a^\dagger_X |0> = -a^\dagger_X a^\dagger_Y |0>$$

This guarantuees that whenever you try to construct a state where two electrons will be put in the same state you get something like

$$|\ldots, 2, \ldots> = 0$$

 

[RUTA]

No, nothing changes. It's only the formulation in position space that is not very helpful as to create the |100+,R> state one has to use a complicated operator which is essentially an infinite product of elementary momentum space creation operators.

Think about a one-dimensional compact space (a circle) which results in discrete momentum space. As momentum is discrete all allowed states can be described using the following notation

$$|n_1^+ n_1^-, n_2^+ n_2^-, n_3^+ n_3^-, \ldots>$$

Here any n can take the values 0 or 1 and + and - refer to the spin quantum number.

Any allowed state can be written as

$$|10, 00, 11, \ldots>$$

You cannot identify one individual electron. You cannot describe any two electrons independently. They always know each other due to the Pauli principle; this principle does not care about there separation in position space.

$$a^\dagger_Y a^\dagger_X |0> = -a^\dagger_X a^\dagger_Y |0>$$

This guarantuees that whenever you try to construct a state where two electrons will be put in the same state you get something like

$$|\ldots, 2, \ldots> = 0$$

What does this have to do with the energy levels of the electron in a hydrogen atom?

 

[Fwiffo]

But, if you want to get the new state for the two electrons, which previously were described independently of each other by one in the same algebraic system (you don't have a unique description for every fermion in the universe), something has to be changed. Right? So, how close to the atoms have to be to necessitate a new mathematical description for the pair as opposed to using the same description for the individuals?

I don't actually know the answer to this , but, the state includes all degrees of freedom, i.e space, momentum, spin etc... As long as you can tell your electrons aprart (the one on the right side and the one on the left side) there is no pauli principle. The pauli prinicple is in effect only when the electrons cannot be distinguished, i.e have some overlap, in phase space as well as all other degrees of freedom.
If we have one electron in a state |A> and another in a state |B> the "real" state would be |AB>-|BA>, We can ofcourse just re-lable our electorns as the electron in state B and the electron in state A, now we only need to be able to distinguish between them (i know the electron on the moon is on the moon and the one on Earth is on earth, but I have no way of knowing which one is electron number 1 and which is electron number 2).

In other words "good" fermion states are |AB>-|BA> or |AC>-|CA> but never |AA> or |BB>.

 

[Fwiffo]

Neutoron stars are the perfect example of how this whole thing works.

 

[tom.stoer]

What does this have to do with the energy levels of the electron in a hydrogen atom?

It is hard to explain how this works if you don't know the basics of qm.

What I have written down is a general state for some electron configuration in the universe. A state having on single "1" like |0,...,0,1,0,...> is a single electron state, a plane wave with fixed momentum k. Of course this does not apply directly to the state of an electron in a hydron atom. But using more complicated states you can construct all possible states including the |100> you are interested in.

The point is that the Pauli principle does not care about the specific details of the state. All what matters is that the algebraic construction takes care about never allowing two electrons being in the same state.

The antisymmetrization Fwiffo mentions is already build into this so called Fock space construction. You can use it to check how it works with wave functions. I will elaborate on that in a couple of minutes ...

 

[tom.stoer]

... back again.

From relativistic quantum field theory one can derive the so-called spin-statistics theorem; it says that wave functions of fermions (= spin 1/2) particles are antismmetric under exchange of any two particles.

In a two-particle state this will look as follows. Let the two particles with two different coordinates be described by two wave functions $$u_X(r_1)$$ and $$v_Y(r_1-R)$$ u and v determine the shape of the wave function, X and Y determine all other quantum numbers, in our case this is just spin.

Now let's assume that both electrons are in the 100 ground state but in different atoms lokated at different positions. That means u=v and X=Y. The antisymmetrized wave function of the two-electron system looks like

$$U(r_1, r_2) = u(r_1) u(r_2-R) - u(r_2) u(r_1-R)$$

Both vectors r₁ and r₂ are independent; they label what we would call the position of the individual particles. But in qm there are no individual particles; both particles are entangled, you are no longer allowed to say "electron A is here and electron B is there". All what you can say is "one electron is here the other one is there". There are no additional labels A and B. You see the difference?

Now if try to locate both particles at the same position, meaning you set R=0 you automatically get

$$U(r_1, r_2) = u(r_1) u(r_2) - u(r_2) u(r_1) = 0$$

That's basically my argument from above, now expressed in terms of wave functions instead of Focks states

 

[RUTA]

It is hard to explain how this works if you don't know the basics of qm.

What I have written down is a general state for some electron configuration in the universe. A state having on single "1" like |0,...,0,1,0,...> is a single electron state, a plane wave with fixed momentum k. Of course this does not apply directly to the state of an electron in a hydron atom. But using more complicated states you can construct all possible states including the |100> you are interested in.

The point is that the Pauli principle does not care about the specific details of the state. All what matters is that the algebraic construction takes care about never allowing two electrons being in the same state.

The antisymmetrization Fwiffo mentions is already build into this so called Fock space construction. You can use it to check how it works with wave functions. I will elaborate on that in a couple of minutes ...

I've taken and taught many courses on QM. I also just wrote a paper on the path integral interpretation of QFT, so I can probably keep up :-)

Your next post answers the question very well. I'll respond there, thanks.

 

[RUTA]

... back again.

From relativistic quantum field theory one can derive the so-called spin-statistics theorem; it says that wave functions of fermions (= spin 1/2) particles are antismmetric under exchange of any two particles.

In a two-particle state this will look as follows. Let the two particles with two different coordinates be described by two wave functions $$u_X(r_1)$$ and $$v_Y(r_1-R)$$ u and v determine the shape of the wave function, X and Y determine all other quantum numbers, in our case this is just spin.

Now let's assume that both electrons are in the 100 ground state but in different atoms lokated at different positions. That means u=v and X=Y. The antisymmetrized wave function of the two-electron system looks like

$$U(r_1, r_2) = u(r_1) u(r_2-R) - u(r_2) u(r_1-R)$$

Both vectors r₁ and r₂ are independent; they label what we would call the position of the individual particles. But in qm there are no individual particles; both particles are entangled, you are no longer allowed to say "electron A is here and electron B is there". All what you can say is "one electron is here the other one is there". There are no additional labels A and B. You see the difference?

Now if try to locate both particles at the same position, meaning you set R=0 you automatically get

$$U(r_1, r_2) = u(r_1) u(r_2) - u(r_2) u(r_1) = 0$$

That's basically my argument from above, now expressed in terms of wave functions instead of Focks states

Very nice, I can appreciate that you're not going to construct the Fock space representation for electrons in two hydrogen atoms -- your general approach suffices :-)

So, the question remains: When do I have to use the two particle entangled state as opposed to two single particle states? The answer can't be "always," because you'd have to put every fermion in the universe into every calculation.

 

[vanhees71]

So, the question remains: When do I have to use the two particle entangled state as opposed to two single particle states? The answer can't be "always," because you'd have to put every fermion in the universe into every calculation.

That's a good question. The answer is that fortunately, the particles are welldescribed by a local QFT which implies the linked-cluster theorem, according to which any experiment localized here and now is unaffected from unrelated other events in outer space. A very detailed analysis about this important feature of local QFT can be found in Weinberg, Quantum Theory of Fields, Vol. I.

 

[tom.stoer]

Very nice, I can appreciate that you're not going to construct the Fock space representation for electrons in two hydrogen atoms -- your general approach suffices :-)

So, the question remains: When do I have to use the two particle entangled state as opposed to two single particle states? The answer can't be "always," because you'd have to put every fermion in the universe into every calculation.

It depends on the states you want to describe. In principle you have to use the entangled state whenever there is "non-zero" overlap. So in principle this means "always".

The difference between Fock states and bound states known from qm (hydrogen atom) is that Fock states are constructed from plane waves which are not localized, whereas bound states are always localized. The question is not when one has to take all infinitly many electrons into account, but that using a Fock space construction requires always an infinite number of Fock states to construct a bound state. But that's not the question.

For bound states non-rel. qm is certainly enough. Then the approach is standard: use antisymmetrized states and start with a perturbation series to check if the cross-terms are relevant for your specific problem.

 

[alxm]

Going to be a bit nitpicky here, but the spin-statistics theorem is not a consequence of special relativity -- it's better to phrase it as: being compatible with special relativity. It's because you also have spin-statistics theorems for ordinary Euclidean space-time -- you don't need the full symmetry of special relativity as an input.

Well AFAIK, you need Lorentz invariance to prove the spin-statistics theorem. After which you're free to take the non-relativistic limit and show that the resulting antisymmetry relation still holds. So that's what I mean by 'consequence'.

It depends on the states you want to describe. In principle you have to use the entangled state whenever there is "non-zero" overlap. So in principle this means "always".

Or never.. (Since I recently wrote up a PF library entry on the Slater determinant ) If you describe your interacting electrons with non-interacting electronic states as a basis (configuration interaction), you can exactly describe your system as an expansion in terms of single-particle functions (with the caveat that a single 'orbital' function no longer represents the state of a single electron) So in that framework, the interaction energy can be related to the overlap integrals without any approximation being made.

In any case, the energy of the interaction between two atoms at a distance is pretty well known to die off as r^-6.

 

[tom.stoer]

In non-rel. qm no one "forces" you to use the antisymmetrized wave functions. It appears as a kind of interaction. Using rel. QFT you learn that it is a deeper principle which is automatically build in the Fock space approach but hidden in non-rel. qm. It is not an interaction.

That's why I started with the Fock space.

 

[alxm]

Well I didn't say it was an interaction, unless you interpreted 'configuration interaction' that way. It's a historical misnomer that stuck. Hartree himself disliked the name and always called it 'superposition of configurations', which is a better description.

 

[Fwiffo]

Very nice, I can appreciate that you're not going to construct the Fock space representation for electrons in two hydrogen atoms -- your general approach suffices :-)

So, the question remains: When do I have to use the two particle entangled state as opposed to two single particle states? The answer can't be "always," because you'd have to put every fermion in the universe into every calculation.

Again the use of language is misleading. seemingly the state |AB>-|BA> is a singlet state, and in some cases it is entangled. But in the most general case there is no "real" entanglement, this is because the particles are identical so if I discover one particle in |A> I have no idea if it was the first one or the second one. Two electrons in different parts of the universe are both in state |1>, the "proper" description if we want to give numbers to the electrons is |electron on Earth in state 1, electron on Andromeda in state 1>-|electron on Andromeda in state 1, electron on Earth in state 1> . This is not an entanglement resource in fact it is very true to say, "the electron on Earth is in state 1". It is never true to say (even after measurement) electron number 1 is the electron on earth. More to come...

 

[Fwiffo]

... sorry about braking this into two posts

so when do we have to anti-symmetrize the state? The answer is, when it becomes meaningful for our purposes. You could say the state of all fermions in the universe is at all times antisymmetric, this should be true (as far as we know). If I look at the two ground state electrons on the He atom I could say there are two electrons one in the spin up state and the other in the spin down state. This would be good in most cases, i.e it is enough to say the state is |01> (first quantized). But this is not the "real" state because the real state is |01>-|10> when we are numbering the electrons as "electron number 1" and "electron number 2" but in most cases this is meaningless. It is uninteresting except for the fact that we cannot write an antisymmetric version of |11> so that is an impossible state.

Another example: two fermions one in NY and the other in Paris, the one in NY is a proton in the Up state (1) the one in Paris is an electron in the Down state (0). It is perfectly good to write the state as |10> although a more precise way to write it would be |p,NY,1;e,P,0> and even better would be to antisymmetrise it, but that is completely unnecessary because a proton is not an electron.

So to answer the question "when dose the wave function become antisymmetric?" It always is , we just don't care about it until the "labels" become meaningful.