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Suppose we have n electrons and want to construct the wavefunction corresponding to the spins of the electrons. Can we construct this wavefunction (in the $(s_1,s_2,s_3 ... s_n)$ representation, so not the $(S^2,S_z)$ representation; the C.G. coefficients relate the two representations) in terms of Dirac's kets (every arrow in these kets corresponds to a Dirac spinor $\left(\begin{array}{c} 1\\0\\\end{array}\right)$ for one spin up, and $\left(\begin{array}{c} 0\\1\\\end{array}\right)$ for one spin down) by using the Pascal Triangle (PT)?

Let me explain. The 1 in row 0 of the PT corresponds to a single electron in either a spin up or a spin down ket:

$$|\uparrow\rangle {or} |\downarrow\rangle$$

Obviously, here is no superposition invovled.

The superposition of the two spins of one electron spin is composed of 1 spin-up electron and 1 spin-down electron, corresponding to the two 1's in row 1 in the PT. Because the wavefunction has to be normalized and be antisymmetric we'll get the following result:

$$+\frac{1}{\sqrt{2}}|\uparrow\rangle-\frac{1}{\sqrt{2}}|\downarrow\rangle$$

For two electron spins, we have 1 time a spin-up,spin-up ket, 2 times a spin-up, spin-down ket, and one time a spin-down, spin-down ket (corresponding to the 1,2,1 in the second row in the PT). Again, taking normalization and antisymmetry into account [which simply can be done by putting a factor $\frac{1}{\sqrt{n}}$ in front of the wavefunction and distribute the +'s and -'s in a symmetric way: interchanging opposite kets of the total wavefunction, which is obvious in the case of the wavefunction for one electron spin in a superposition state involving a ket for spin up and a ket for spin down (the one above) will yield the negative of the original] we'll get:

$$\frac{1}{\sqrt{4}}(+|\uparrow\uparrow\rangle+|\uparrow\downarrow\rangle-|\downarrow\uparrow\rangle-|\downarrow\downarrow\rangle)$$

So for a superposition of three electron spins, we'll get 1 ket with three spins up, 3 with two up and one down, 3 with one up and two down, and 1 with three spins down, the 1,3,3, and 1 corresponding to the 1,3,3,1, in row 3 in the PT. We get:

$$\frac 1 {\sqrt{8}}(+|\uparrow\uparrow\uparrow\rangle+|\uparrow\uparrow\downarrow\rangle+|\uparrow\downarrow\uparrow\rangle+|\downarrow\uparrow\uparrow\rangle-|\uparrow\downarrow\downarrow\rangle-|\downarrow\uparrow\downarrow\rangle-|\downarrow\downarrow\uparrow\rangle-|\downarrow\downarrow\downarrow\rangle)$$

For a superposition of four electron spins we'll get the number of four spins up (or down) by looking to the left and right of the PT, which is always 1, so there is 1 ket with all spins up and one with four spins down (and a minus sign). There are 4 kets with three spins up, one spin down (and a plus sign), 6 with two up, two down (of which 3 have a plus sign and 3 a minus sign; see the two electron spin superposition above), and 4 with one up, three down (and a minus sign). The normalization factor is $\frac{1}{\sqrt{16}}$. I won't elaborate to write the complete function. The 1,4,6,4,and 1 correspond to the 1,4,6,4,1 in row 4 of the PT.

For a superposition of five electron spins we can use the numbers of row five in the PT to know the distributions of the different kets: 1 up,up,up,up,up, 5 up,up,up,up,down, 10 up,up,up,down,down, 10 up,up,down,down,down, 5 up,down, down,down,down, 1 down,down,down,down,down, after which we add them antisymmetrically, and put the normalization factor in front.

For small numbers of electron spins this is rather trivial, but it comes in handy when the number increases. Can it be done like this?

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  • $\begingroup$ The composition of n identical spins is quantified by the Catalan triangle. $\endgroup$ – Cosmas Zachos Feb 12 at 1:59
  • $\begingroup$ But doesn't a superposition state of three spins consists of eight kets? How would this state look like according to this triangle (in the ($s_1,s_2,s_3$) representation? $\endgroup$ – descheleschilder Feb 12 at 18:18
  • $\begingroup$ Adding 3 spin 1/2s gives one spin 3/2, and two spin 1/2s, so see here. $\endgroup$ – Cosmas Zachos Feb 12 at 19:06
  • $\begingroup$ Further and eqn (22) of this paper . $\endgroup$ – Cosmas Zachos Feb 12 at 19:14
  • $\begingroup$ ... so you recognize the shifted sequences going to the right at 45 degrees in the WP table: 1,1; 1,2; 1,3,2; ... multiplicities. $\endgroup$ – Cosmas Zachos Feb 12 at 19:40
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Antisymmetry refers to permutation in the spin of two different particles, not of the possible states of a single particle. Indeed $\vert \uparrow \rangle - \vert \downarrow\rangle$ is an eigenstate of $\sigma_x$.

The state $$\frac{1}{\sqrt{4}}(+|\uparrow\uparrow\rangle+|\uparrow\downarrow\rangle-|\downarrow\uparrow\rangle-|\downarrow\downarrow\rangle)$$ actually doesn't have well-defined permutation symmetry since interchanging spins $1$ and $2$ gives $$\frac{1}{\sqrt{4}}(+|\uparrow\uparrow\rangle+|\downarrow\uparrow\rangle-|\uparrow\downarrow\rangle-|\downarrow\downarrow\rangle)$$ which is not a multiple of your initial state.

For completeness note that $\vert \uparrow\uparrow \rangle \pm \vert\downarrow\downarrow\rangle$ is symmetric, irrespective of the relative sign, since interchanging the particles will give the same state. An antisymmetric state would be $\vert\uparrow\downarrow \rangle - \vert \downarrow\uparrow\rangle$.

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  • $\begingroup$ Doesn't, in this case, antisymmetrical means that flipping all the spins (rather than interchange the spins in each ket) gives the negative of the original superposition? How do you interchange the spins of a ket with three arrows? In the example with three spins, it would be, according to you, only the part of the total without the left and right ket that's antisymmetric? $\endgroup$ – descheleschilder Feb 10 at 21:30
  • $\begingroup$ I suppose you can decide that antisymmetry refers to states rather than particles but then there’s no connection with spin-statistics, which refers to the (anti)symmetry under particle exchange for fermions, not state exchange. Your second state also is not an eigenstate of $S_z$, and there is a priorI no reason to reject eigenstates of $S_z$ as eigenstates. $\endgroup$ – ZeroTheHero Feb 10 at 21:41
  • $\begingroup$ Sorry for reacting a bit late! I went for a walk with the dog in the rainy, scary, and dark forest (where you can think very well, when not playing with the dog). What does $S_z$ look like when applied to the second state? And ain't it so that the chance is $\frac{1}{4}$ ($\frac {1}{\sqrt{4}}$ squared) of finding two electrons with spin up, $\frac{1}{2}$ of finding both electrons with opposite spins, and again $\frac{1}{4}$ to find the two spins with spin down? $\endgroup$ – descheleschilder Feb 11 at 0:14
  • $\begingroup$ Don't have to exchange all spinors (1,0) with a (0,1) spinor and vice-versa? $\endgroup$ – descheleschilder Feb 11 at 0:25
  • $\begingroup$ Applying $S_z$ doesn’t give a probability: it gives another state. $S_z$ on your second state produces $\frac{1}{2}\left(\vert \uparrow\uparrow\rangle + \vert\downarrow\downarrow\rangle\right)$ as $S_z$ kills the $m=0$ states $\vert\uparrow\downarrow\rangle$ and $\vert \downarrow\uparrow\rangle$. I do not know what is a $(1,0)$ spinor in the context of your question. $\endgroup$ – ZeroTheHero Feb 11 at 4:22

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