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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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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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