We can write a Singlet state of two $\frac{1}{2}$ spin particles like this:

$$|S\rangle = \frac{1}{\sqrt{2}}\left( |+ \rangle ⊗ |-\rangle - |-\rangle ⊗|+\rangle \right) $$

is this the same as writing this:

$$|S\rangle = \frac{1}{\sqrt{2}}\left( |+ \rangle |-\rangle - |-\rangle |+\rangle \right) $$ ?

Another example(from the Clebsch-Gordan Coeffiecients):

We have two particles one with spin $\frac{3}{2}$ and the other with spin $\frac{1}{2}$, we can then write the state $|2,1 \rangle$ the following way:

$$|2,1 \rangle = \frac{1}{2}\left|\frac{3}{2},\frac{3}{2} \right\rangle\left|\frac{3}{2},-\frac{1}{2} \right\rangle +\sqrt{\frac{3}{4}}\left|\frac{3}{2},\frac{1}{2} \right\rangle\left|\frac{3}{2},\frac{1}{2} \right\rangle$$

Is this as well to be meant as a tensor product? otherwise I dont see how to mulitply those kets.

  • 1
    $\begingroup$ Yes; the $\otimes$ is implied. $\endgroup$ – ZeroTheHero Jun 21 at 18:14

Yes, all those notations mean the same thing—tensor products. $$|A\rangle|B\rangle\equiv|A\rangle\otimes|B\rangle$$


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