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I'm a bit confused about the use of Tensor products in Quantum mechanics. For instance if two electrons are in the state $\frac{1}{\sqrt{2}}(a(r_1)b(r_2)-b(r_1)a(r_2))\otimes \lvert \downarrow \rangle_1 \otimes \lvert \uparrow \rangle_2$, where $r_i$ is the position. Does this simply mean the first electron is in the spin down state and the second in in the spin up state? And would that expression be equivalent to $\frac{1}{\sqrt{2}}(a(r_1)b(r_2)-b(r_1)a(r_2))\otimes \lvert \uparrow \rangle_2 \otimes \lvert \downarrow \rangle_1$?

Thanks in advance!

jx9845
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    Are your electrons distinguishable, or undistinguishable? How do you know which one is electron 1 and which one is electron 2? – Sofia Feb 23 '15 at 11:33
  • I think they are supposed to be indistinguishable. Is such a state possible? – jx9845 Feb 23 '15 at 11:34
  • I am answering you. – Sofia Feb 23 '15 at 11:42
  • Where did you see that expression? What is the meaning of the subscripts $1$ and $2$ for the kets? – Norbert Schuch Feb 23 '15 at 12:20
  • I was trying to figure out wether this state was a realizable state of two electrons. The subscripts confused me as I didn't know if they stand for the electrons. – jx9845 Feb 23 '15 at 17:04
  • I see that you're learning about symmetric and antisymmetric wave functions. This would be a great time to start learning about the real way physicists handle identical particles. I recommend reading this post to get started. – DanielSank Feb 23 '15 at 17:23
  • Well, if $1$ and $2$ stood for the electrons in state $a$ and $b$, respectively, this would make perfect sense. – Norbert Schuch Feb 23 '15 at 18:24

1 Answers1

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Two undistinguishable fermions, electrons in your case, are in an anti-symmetrical state. So, the state you wrote has to be corrected, see below how. Also if you use the notation of the tensor product $\otimes$, it is desirable to use it consistently.

$$\frac {\langle r_1|a\rangle \otimes \langle r_2|b\rangle - \langle r_1|b\rangle \otimes \langle r_2|a\rangle}{\sqrt{2}} \otimes \frac {|\downarrow \rangle_1 \otimes |\uparrow \rangle_2 + |\uparrow \rangle_1 \otimes |\downarrow \rangle_2}{\sqrt{2}}, \tag{i}$$

where the notation $\langle r_1|a\rangle \otimes \langle r_2|b\rangle$ indicates the function $a$ in the representation $r_1$ and the function $b$ in the representation $r_2$.

The meaning of the formula $\text {(i)}$ is that the state of the undistinguishable fermions is anti-symmetrical, i.e. if it is anti-symmetrical in the ordinary space, it is symmetrical in the spin-space, and vice-versa, as shown below

$$\frac {\langle r_1|a\rangle \otimes \langle r_2|b\rangle + \langle r_1|b\rangle \otimes \langle r_2|a\rangle}{\sqrt{2}} \otimes \frac {|\downarrow \rangle_1 \otimes |\uparrow \rangle_2 - |\uparrow \rangle_1 \otimes |\downarrow \rangle_2}{\sqrt{2}}. \tag{ii}$$

All the four spaces (Hilbert, or vector, according to how you construct them), are different spaces, and this is why you use the notation of the tensor product.

Sofia
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  • Notations like $|a;r_1\rangle$ are highly misleading. A "ket" denotes an element of the Hilbert-space, such as $|a\rangle$, whereas $a(r_1)$ are just the components of that Hilbert-vector: $\langle r_1|a\rangle\equiv a(r_1)$ – Nephente Feb 23 '15 at 12:52
  • @nephente I tried a better notation, as you can see. The notation $\langle r_1|a\rangle$ also seemed to me not so good (but thank you for suggesting), because it denotes the vector $|a\rangle$ in the representation $r_1$, while as far as I understand, the user's notation $a(r_1)$ meant the function $a(r)$ at the point $r_1$. – Sofia Feb 23 '15 at 13:26
  • @nephente The truth is that I don't know what the user meant by the notation $r_1$ and $r_2$. Could it be a different notation of the ordinary space for the two electrons? – Sofia Feb 23 '15 at 13:38
  • Thank you very much, this clarifies it a lot for me. Basically what I meant by $a(r_1)$ is one wavefunction or state depending on the coordinates of particle 1. So all of the states in quantum mechanics live in a Hilbert space and whenever I combine the two, I get a tensor product between these two spaces, right? I guess I was confused because I thought the $\otimes \lvert \uparrow \rangle_2 \otimes \lvert \downarrow \rangle_1$ term simply meant particle 1 had spin down and particle 2 spin up. But I think this is just a basis element of the new vector space, correct? Cheers – jx9845 Feb 23 '15 at 16:55
  • @jx9845 more than that! You saw that spin behaves in an independent way, i.e. symmetrizes or anti-symmetrizes separately. So, when there is no interaction between the space coordinates of a particle and the spin variables, the latter arrange themselves independently, as I showed you in the formulas (i) and (ii). And attention! If particles are distinguishable, their joint wave-function doesn't have to be symmetrical (for bosons) or anti (for fermions) anymore. Well, but if you have doubts, you know our site. – Sofia Feb 23 '15 at 17:03
  • OK, so do the space coordinates and the spin coordinates simply have to be arranged in such a way that the total wave function is antisymmetrical (symmetrical for bosons)? Did you mean that by saying they arrange themselves independently? – jx9845 Feb 23 '15 at 17:12
  • Oh wait, I think I get it now. We combine the space coordinates into one Hilbert space and then combine the spin coordinates into another. And then the total description lives in the combined space. – jx9845 Feb 23 '15 at 17:18
  • @jx9845 no! no! Only if the two particles are undistinguishable. If they are distinguishable, then, each particle is described by its own wave-function, e.g. the i-th particle by $f_i(\vec r_i; \sigma_i)$, where $\sigma_i$ are the respective spin variables. – Sofia Feb 23 '15 at 17:19
  • @jx9845 also, pay attention to entanglements. Such states also may arrange the particles in a particular way. But, better ask us when you have such a situation and have doubts. – Sofia Feb 23 '15 at 17:21