- #26

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I might have answered that there is no particle a.It is in the two-particle state, right? I’m not entirely sure what you are asking though.

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- #26

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I might have answered that there is no particle a.It is in the two-particle state, right? I’m not entirely sure what you are asking though.

- #27

Isaac0427

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Right. It’s just a fermion and cannot be identified among other identical fermions in the system. So if you ignore my poor wording, when I talked about the state of particle a I meant ##\psi_a## and when I talked about the state of particle b I meant ##\psi_b##. I am still completely lost as to how these states do not exist.I might have answered that there is no particle a.

- #28

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Yes, but they occur as factors in products, and the state as a whole is a sum of those products. So there are no actual single-particle states, physically, for this system; they are just artifacts of the mathematical way we represent the state. So are the labels ##a## and ##b##. You can't understand what's going on if you focus on these artifacts of the math and don't think about the overall way the state is constructed. You have to actually start from a valid basis for the Hilbert space of the two-particle system, and then see what states you can write down as valid linear combinations of those basis states. So the first question you need to ask is, whatIn this equation ##\psi_a## and ##\psi_b## are single-particle states

- #29

Isaac0427

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I do not see how my example is invalid.Yes, but they occur as factors in products, and the state as a whole is a sum of those products. So there are no actual single-particle states, physically, for this system; they are just artifacts of the mathematical way we represent the state. So are the labels ##a## and ##b##. You can't understand what's going on if you focus on these artifacts of the math and don't think about the overall way the state is constructed. You have to actually start from a valid basis for the Hilbert space of the two-particle system, and then see what states you can write down as valid linear combinations of those basis states. So the first question you need to ask is, whatisa valid basis for the Hilbert space of the two-particle system? (I described how to construct one in a previous post, but you should not take my description for granted; you should construct the basis yourself.)

- #30

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That's because you're not doing what I asked you to do. You are just waving your hands and writing down two one-particle states and assuming that there is some valid state in the Hilbert space of the two-particle system which can be described as one particle being in each of these states. But just waving your hands doesn't show that your assumption is correct. You have to actually show it, by constructing a valid basis for the Hilbert space of the two-particle system and then writing down the state you think you are talking about in that basis. You have not done that. And I think that until you try, you will continue to be confused.I do not see how my example is invalid.

- #31

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That means a tiny and experimentally trivial difference could qualify as "different"? So that you might need a really long run of identical experiments to bring out that tiny difference in state and prove that they are not violating Pauli's principle?The two "one particle states" that make up the two-particle state ... [don't need to be orthogonal]. They only need to be different.

- #32

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If there is a continuous degree of freedom involved, yes. For example, two identical fermions could differ in position by a very small amount, while being in identical spin states, without violating the Pauli exclusion principle. However, if the two identical fermions had exactly the same position, then they could not have spin states that differed by a very small amount, because the spin degree of freedom is not continuous.That means a tiny and experimentally trivial difference could qualify as "different"?

- #33

Isaac0427

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Can you help me with this? I am not exactly sure what constraints you are talking about.If you actually try to write down a valid state for a quantum system consisting of two identical fermions in a single energy level of a bound system, you will find that you cannot write down a state that could be described in words as "one particle is in the state (01)(01)\begin{pmatrix} 0\\ 1\\ \end{pmatrix} and one particle is in the state (.1√.99)(.1.99)\begin{pmatrix} .1\\ \sqrt{.99}\\ \end{pmatrix}". There is no such state that satisifies the constraints. By contrast, if you try to write down a valid state for a quantum system consisting of two identical fermions in free space that both have the same energy, you will find that youcanwrite down such a state. But you actually have to go through the exercise of trying to write down actual states of the actual quantum system under discussion; just naming one-particle spin states won't do it.

- #34

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Have you tried to answer the questions I have been asking? Start with this one: what is a valid basis of the Hilbert space of the quantum system consisting of two identical fermions?I am not exactly sure what constraints you are talking about.

- #35

Isaac0427

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I really don’t know what you are asking me right now. Can I have some help arriving at the answer? I understand what a Hilbert space is, but I always thought that a valid basis would be a spin up state and a spin down state, or if we are talking about energy, the ground state, the first excited state, the second excited state, etc.Have you tried to answer the questions I have been asking? Start with this one: what is a valid basis of the Hilbert space of the quantum system consisting of two identical fermions?

- #36

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Those states form a basis for one particular Hilbert space, the Hilbert space of spin states for a single spin-1/2 particle.I understand what a Hilbert space is, but I always thought that a valid basis would be a spin up state and a spin down state

Those states would form a basis for another particular Hilbert space, the Hilbert space of a single particle in a potential well. Strictly speaking, this particle would have to have zero spin, since you are only including energy in the specification of the basis states.or if we are talking about energy, the ground state, the first excited state, the second excited state, etc

The Hilbert space you have been implicitly referring to in this discussion is neither of the above. In fact it's not entirely clear which one it is, since we have talked about several different possible scenarios, but we'll start with the simplest possible one that is relevant: the Hilbert space of a quantum system consisting of two identical spin-1/2 particles, which are both in the same energy level of some potential well. At this point we are ignoring any other characteristics of the particle or the potential well: no electric charge, no nucleus of the atom, etc. We are also ignoring any possibility of either particle jumping to some other energy level (for example, by absorbing a quantum of radiation).

This Hilbert space is an antisymmetric tensor product of two copies of the Hilbert space of spin states for a single spin-1/2 particle. Unfortunately, we don't have any way of writing down what that means without labeling the particles, but as I've said before, it's important to realize that these labels are artifacts of our math and do not correspond to anything in the actual system. If we label the particles as "particle 1" and "particle 2", then a simple tensor product (which is not quite what we're looking for, but will give us a start at getting there) of the Hilbert spaces for each particle would be a Hilbert space with a basis consisting of four mutually orthogonal vectors which we could write as ##\vert \uparrow \rangle_1 \vert \uparrow \rangle_2##, ##\vert \uparrow \rangle_1 \vert \downarrow \rangle_2##, ##\vert \downarrow \rangle_1 \vert \uparrow \rangle_2##, ##\vert \downarrow \rangle_1 \vert \downarrow \rangle_2##, where the arrows denote the "up" and "down" basis states for the single-particle Hilbert space and the subscripts denote which particle (1 or 2). In other words, the simple tensor product Hilbert spaces consists of all possible states you can form by multiplying together two single-particle states.

If you want some more information about tensor products of Hilbert spaces, see here:

https://en.wikipedia.org/wiki/Tensor_product_of_Hilbert_spaces

But the Hilbert space we actually want is the antisymmetric tensor product; "antisymmetric" because we are dealing with fermions. This Hilbert space can be constructed in a number of ways, but the simplest is to take all possible states you can form from the following formula:

$$

\vert \psi \rangle_A = \frac{1}{\sqrt{2}} \left( \vert \psi_a \rangle_1 \vert \psi_b \rangle_2 - \vert \psi_b \rangle_1 \vert \psi_a \rangle_2 \right)

$$

where ##\vert \psi_a \rangle## and ##\vert \psi_a \rangle## are states belonging to the single-particle Hilbert space of spin states. So what we need is a valid set of basis states for this Hilbert space. One way to find them is to expand ##\vert \psi_a \rangle## and ##\vert \psi_a \rangle## in terms of the single-particle basis states and substitute into the above formula:

$$

\vert \psi \rangle_A = \frac{1}{\sqrt{2}} \left[ \left( a_{\uparrow} \vert \uparrow \rangle_1 + a_{\downarrow} \vert \downarrow \rangle_1 \right) \left( b_{\uparrow} \vert \uparrow \rangle_2 + b_{\downarrow} \vert \downarrow \rangle_2 \right) - \left( b_{\uparrow} \vert \uparrow \rangle_1 + b_{\downarrow} \vert \downarrow \rangle_1 \right) \left( a_{\uparrow} \vert \uparrow \rangle_2 +a _{\downarrow} \vert \downarrow \rangle_2 \right) \right]

$$

Then we can just expand out the products and collect terms:

$$

\vert \psi \rangle_A = \frac{1}{\sqrt{2}} \left[ \left( a_{\uparrow} b_{\uparrow} - b_{\uparrow} a_{\uparrow} \right) \vert \uparrow \rangle_1 \vert \uparrow \rangle_2 + \left( a_{\uparrow} b_{\downarrow} - b_{\uparrow} a_{\downarrow} \right) \vert \uparrow \rangle_1 \vert \downarrow \rangle_2 + \left( a_{\downarrow} b_{\uparrow} - b_{\downarrow} a_{\uparrow}\right) \vert \downarrow \rangle_1 \vert \uparrow \rangle_2 + \left( a_{\downarrow} b_{\downarrow} - b_{\downarrow} a_{\downarrow} \right) \vert \downarrow \rangle_1 \vert \downarrow \rangle_2 \right]

$$

You can see that two of these four terms vanish, and the other two have the same factor in front, just with a sign change, so we end up with (note that the factor involving the ##a## and ##b## coefficients must be ##1## if the state is normalized):

$$

\vert \psi \rangle_A = \frac{1}{\sqrt{2}} \left( \vert \uparrow \rangle_1 \vert \downarrow \rangle_2 - \vert \downarrow \rangle_1 \vert \uparrow \rangle_2 \right)

$$

In other words, the antisymmetric tensor product Hilbert space has just

What does this mean, physically? It means that, if you want to put two spin-1/2 particles into the same energy level of a potential well,

Note also that, since the two particles must have opposite spins, there is

I'll stop here and let you digest the above. But you should understand it thoroughly before we try to go on to more complicated examples.

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- #37

Isaac0427

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So, I see that the spin up state and the spin down state are orthogonal. It would appear that any other two orthogonal states would provide the same results (i.e. having ##| \uparrow \rangle \longrightarrow \frac{1}{\sqrt{2}}\left ( | \uparrow \rangle + | \downarrow \rangle \right )## and ##| \downarrow \rangle \longrightarrow \frac{1}{\sqrt{2}} \left ( | \uparrow \rangle - |\downarrow \rangle \right )##). Is this correct?

Thank you very much for all the help.

- #38

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Yes, you can pick any two orthogonal single particle spin states and use them as a basis for the single particle Hilbert space, and things will work out the same way.It would appear that any other two orthogonal states would provide the same results

- #39

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Hm, Griffiths seems to confuse students often with his quantum mechanics textbook. It seems not as good as his electrodynamics book then? I don't know his book, so I can't tell.Well, I do see where I got confused. Griffiths did say that you could pretend that the particles are distinguishable for practical matters. I’m assuming then that it is not really the limit of the identical fermion equation, it is just another way look at the situation that yields the same practical result.

First of all one should indeed distinguish two things first: According to all what we know in Nature there are to kinds of indistinguishable particles only, namely bosons and fermions, for which many-body states are completely symmetric for bosons (antisymmetric for fermions) under exchange of two particles. There are esceptions for certain "quasiparticles" in interesting effectively two-dimensional quantum many-body systems, but let's leave out this somewhat exotic topic here. As far as the elementary particles are concerned there are indeed only bosons or fermions (at least as far as we know and as we more than successfully describe by the Standard Model).

Then there is spin, and according to the mathematics of the rotation group SO(3) and its covering group, the SU(2), there are half-integer and integer spin realizations. At a first glance, this seems to have nothing to do with whether a given particle with spin is a boson or fermion, but in relativistic quantum field theory, making some quite intuitive assumptions (locality, microcausality, boundedness of the Hamiltonian from below and the existence of a stable ground state) implies that for the proper realization of the proper orthochronous Poincare group for both massless and massive particles you are forced to quantize integer-spin (half-integer-spin) fields as bosons (fermions) to fulfill all these assymptions, which in turn lead to a physically meaningful particle interpretation (in the sense of asymptotic free states), a unitary S-matrix, and the linked-cluster principle. Indeed, all empirical evidence shows that half-integer-spin particles are always fermions and integer-spin particles are always bosons.

For the details see Weinberg, Quantum Theory of Fields, vol. I, which is a tough book, but it really provides the complete insight why QFT is the way it is!

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