Spin of two identical particles

Question

I read that when I have two identical particles with spin 1/2 there are 4 possibilities:

|↓↓⟩,|↑↑⟩,|↑↓⟩,|↓↑⟩.

Then since there is the symmetrization requirement I can take as eigenvalues the following 4 states:

|↓↓⟩; |↑↑⟩; 1/2(|↑↓⟩-|↓↑⟩); 1/2(|↑↓⟩+|↓↑⟩)

My question is: if the particle are indistinguishable what's the difference between these 2 states:

|↑↓⟩,|↓↑⟩

Aren't they the same state that tells there is a particle with spin up and another with spin down? Why is it so important to symmetrize the state respect to them?

Answer 1

My feeling is that you are confused by notation.

I read that when I have two identical particles with spin 1/2 there are 4 possibilities

These are not four possibilities, but these are four basis states. They form a complete orthonormal set of states for two spins. These basis states span the Hilbert space of all possible two-spin states.

The notation, for example $\left|\left.\uparrow\downarrow\right>\right.$, is a shorthand for $$ \left|\left.\uparrow\downarrow\right>\right. \equiv \left|\left.\uparrow\right>\right._1\otimes\left|\left.\downarrow\right>\right._2,$$ meaning that particle 1 is in state $\left|\left.\uparrow \right>\right.$ and particle 2 is in state $\left|\left.\downarrow\right>\right.$ (note, there is a logical and).

what's the difference between these 2 states $$ \left|\left.\uparrow\downarrow\right>\right., \left|\left.\downarrow\uparrow\right>\right.$$

Given the above, you can now tell the difference: in the first case, particle 1 is in state $\left|\left.\uparrow\right>\right.$ and particle 2 is in state $\left|\left.\downarrow\right>\right.$, while in the second case particle 1 is in state $\left|\left.\downarrow\right>\right.$ and particle 2 is in state $\left|\left.\uparrow\right>\right.$.

So far, we made no use of the indistinguishability of the two particles, we even labeled them as particle 1 and particle 2, so we treated them as distinguishable particles.

Then since there is the symmetrization requirement I can take as eigenvalues the following 4 states: $$ \left|\left.\uparrow\uparrow\right>\right., \left|\left.\downarrow\downarrow\right>\right.,\frac{1}{\sqrt{2}}\left(\left|\left.\uparrow\downarrow\right>\right.+\left|\left.\downarrow\uparrow\right>\right.\right), \frac{1}{\sqrt{2}}\left(\left|\left.\uparrow\downarrow\right>\right.-\left|\left.\downarrow\uparrow\right>\right.\right).$$

Note first that I have put factors of $1/\sqrt{2}$ instead of your factors $1/2$. This is, because we want the states to remain normalized. Second, please be aware that these four states are in no way eigenvalues in this context, they are also not eigenstates, because you did not specify any hamiltonian, of which these states could be the eigenstates. These four new states simply form an alternative complete and orthonormal set of basis states for two spins.

Now let me try to explain, why these four states are well suited to describe indistinguishable particles. To have indistinguishable particles should mean that it should not matter in our description of the two particles, which of them is in which state. In other words, when we exchange the labels 'particle 1' and 'particle 2', we should get the same observable properties of the system. This is indeed the case for our four new basis states. If you remember our notation convention from above, then, for example, $$ \left|\left.\psi\right>\right. = \frac{1}{\sqrt{2}}\left(\left|\left.\uparrow\downarrow\right>\right.-\left|\left.\downarrow\uparrow\right>\right.\right) = \frac{1}{\sqrt{2}}\left(\left|\left.\uparrow\right>\right._1\otimes\left|\left.\downarrow\right>\right._2-\left|\left.\downarrow\right>\right._1\otimes\left|\left.\uparrow\right>\right._2 \right).$$ Now exchanging the labels for particle 1 and 2, we have $$ \frac{1}{\sqrt{2}}\left(\left|\left.\uparrow\right>\right._2\otimes\left|\left.\downarrow\right>\right._1-\left|\left.\downarrow\right>\right._2\otimes\left|\left.\uparrow\right>\right._1 \right) = -\frac{1}{\sqrt{2}}\left(\left|\left.\uparrow\downarrow\right>\right.-\left|\left.\downarrow\uparrow\right>\right.\right). = -\left|\left.\psi\right>\right.$$ The exchange of the particle labels gave the same wave function with a minus sign!

How would we now interpret such a wave function? We would say: either particle 1 is in state $\left|\left.\uparrow\right>\right.$ and particle 2 is in state $\left|\left.\downarrow\right>\right.$, or particle 1 is in state $\left|\left.\downarrow\right>\right.$ and particle 2 is in state $\left|\left.\uparrow\right>\right.$ (note that you can read the $\otimes$ as a logical and, and the $+$ or $-$ as a logical or). Another way of saying the same thing: one particle is in state $\left|\left.\uparrow\right>\right.$ and the other is in state $\left|\left.\downarrow\right>\right.$. So we have somehow lost the information, which particle is which.

Now the question remains, why can no measurement distinguish two systems, whose wave functions differ only by a minus sign. Here is the answer: suppose you have a measurement operator $\hat{O}$. For the system with wave function $\left|\left.\psi\right>\right.$ its expectation value is $\left<\left.\psi\right|\right.\hat{O}\left|\left.\psi\right>\right.$, and for the system with wave function $-\left|\left.\psi\right>\right.$ it is $$(-1)^2\left<\left.\psi\right|\right.\hat{O}\left|\left.\psi\right>\right.= \left<\left.\psi\right|\right.\hat{O}\left|\left.\psi\right>\right.,$$ so it is the same! This means indeed, the antisymmetrized state describes a system, in which the particles are indistinguishable, because it cannot be observed if their labels are exchanged. We can no longer tell, who is who.

For this explanation, we have used only one of the four new basis states. If you repeat the argument with the other three, you will see that for them, the wave functions do not change sign, when the particle labels are exchanged. So it appears already for two particles, that there are two classes of states: bosonic states (no sign change upon exchange of particle labels) and fermionic states (sign change upon exchange of particle labels). In both cases, any observable $\hat{O}$ has the same expectation value for the original state, and the state with the exchanged particle labels. This is what we mean by saying, the particles cannot be distinguished (by measurement).

Answer 2

Let $\mathcal{H}_1$ be the one-particle Hilbert space and $|\psi\rangle_a$ be a one-particle quantum state (the subscript $a$ referring to the $a^{th}$-particle in case we have more than one).

A physical system of 2 indistinguishable particles lives in the symmetrised (respectively anti-symmetrised) 2-particle Hilbert space $$\hat{S}\Big(\mathcal{H}_2\Big) = \hat{S}\Big(\mathcal{H}_1\otimes\mathcal{H}_1\Big)$$ where each state is of the form $\hat{S}\big(|\psi\rangle_1\otimes|\psi\rangle_2\big)$, the subscript $1,2$ referring to the first and second particle, respectively. The operator $\hat{S}$ denotes the symmetrisation (or anti-symmetrisation) of the combination it acts on.

For the example at hand let us assume we are describing a system of two fermions: nature is so (and we do not discuss it here) that such a state must be taken to be anti-symmetric. For a spin-zero total state one must have $$ |\phi; S=0\rangle = c_1\Big(|+\rangle_1\otimes|-\rangle_2\Big) \pm c_2\Big(|+\rangle_2\otimes|-\rangle_1\Big) $$ where you notice that the second piece exchanges the roles of the particles $1,2$.

My question is: if the particle are indistinguishable what's the difference between these 2 states:

The difference is that in the first state the particle 1 is in the state $|+\rangle$ and the particle 2 in the state $|-\rangle$; the second combination has the opposite. The particles being indistinguishable means that no observable measurement exists such that one can distinguish which particle is in which state - but this does not imply that the two particles are the same particles. There are two possible ways of combine two particles into $|\pm\rangle$ states, corresponding to the different permutations of the number 2. For many more particles the same holds mutatis mutandis.

Answer 3

If you apply the exchange operator to $|\uparrow \downarrow>$, you get

$P|\uparrow \downarrow> = |\downarrow \uparrow> $

So this is not an eigenstate of the exchange operator, so a system of 2 fermions or 2 bosons cannot be in this state as the state is not symmetric.

Answer 4

I think your question is at the heart of quantum mechanics. The particles are not "identical" (that is a classical concept), they are "indistinguishable".

Though you say "indistinguishable" in the question, it's a really a valid question for "identical" particles. For example: classically, if I have 2 identical balls in a bag, and one is red and one is green, does it matter which one is red or green? No. It's a bag with 2 "identical" balls: one red, one green. Each configuration of the bag of balls is the same.

When considering "indistinguishable" particles, there is no classical analog, and it's not a concept that is amenable to classical thinking. We're talking about particle labels, and when you observe (1,2) it's the same as observing (2,1), and when I say "same", I mean it is physical the same existence, and hence they both exist at the same time.

To really answer you question satisfactory, you have to get into "what is a quanta in quantum field theory?" and "why is the spin-statistics theorem true?". The former is a hotly contested topic on this site, and the latter is famously difficult to understand with any level of intuition.