1
$\begingroup$

During course of quantum mechanics we dealt with addition of angular momenta. If we have two particles with spin $j_1$ and $j_2$ we can introduce total spin operator:

$$\mathbf{J} = \mathbf{j}^{(1)} + \mathbf{j}^{(2)},$$ and new basis: $$\mathbf{J}^{2}|J,M\rangle = J(J+1)|J,M\rangle,$$ $$\mathbf{J}_{z}|J,M\rangle = M|J,M\rangle,$$ where $J = |j_1 - j_2|, \ldots, j_1 + j_2$ and $M = -J, \ldots, J$.

The same thing can be done for $N$ particles with e.q. spin $1/2$. One can introduce collective spin operator:

$$\mathbf{J} = \sum\limits_{i=1}^{N}\mathbf{j}^{(i)},$$ and new basis in the Hilbert space $\mathcal{H} = \mathcal{H}_{1} \otimes \ldots \otimes \mathcal{H}_{N}$, $$\mathbf{J}^{2}|J,M\rangle = J(J+1)|J,M\rangle,$$ $$\mathbf{J}_{z}|J,M\rangle = M|J,M\rangle.$$ States with highest momentum $J = N/2$ are known as the Dicke states: $$|N/2, M\rangle, \ \ M = -N/2, \ldots, N/2$$ I always thought that these states are not degenerate with respect to a quantum number $M$, but here they claim that only symmetric Dicke states are uniquely defined. I can't see this cause I construct states with $M$ less than $N/2$ by applying the lowering operator to $|N/2, N/2\rangle$ state (this seems to give me only symmetric Dicke states).

Another issue is the Hilbert space itself. I think it does not take into account that particles are indistinguishable. We simply deal with tensor product of single particle spaces. No symmetrization or antisymmetrization. Am I right?

$\endgroup$
  • $\begingroup$ See Sec.II in lptms.u-psud.fr/ressources/publis/2010/… $\endgroup$ – udrv Apr 25 '16 at 5:25
  • $\begingroup$ @udrv I think they mention only about symmetric Dicke states. I wonder if this is actually some subclass of all Dicke states that are degenerate with respect to quantum number $M$? $\endgroup$ – WoofDoggy Apr 25 '16 at 10:14
  • $\begingroup$ Correct, the construction starts as you already noticed with the direct product of N spin-1/2 spaces (distinguishable qubits) and the corresponding highly degenerate basis. Then the symmetric states are constructed as the basis for the symmetric subspace. I'm having trouble finding better refs. fast, but these 2 give good enough overviews: see.IIA of arxiv.org/pdf/1511.03281v1.pdf, and rsta.royalsocietypublishing.org/content/369/1939/1137. $\endgroup$ – udrv Apr 25 '16 at 14:52
  • $\begingroup$ @udrv I think I finally understand. As the second paper states: dimension of the full $N$-particle Hilbert space is equal to $2^N$, but the space of spin states grows much slower, that is why it needs to be degenerate with respect to $M$ quantum number. The procedure of lowering the highest spin state produces symmetric states and they live in Hilbert space of indistinguishable particles. For 3 particles of spin 1/2 the product Hilbert space has dimension $8$, but the total spin seems to have dimension $6$ ($s=3/2, 1/2$). What would be the states from degenerate space (non-symmetric ones)? $\endgroup$ – WoofDoggy Apr 25 '16 at 22:33
  • $\begingroup$ I guess if $|D^{(N)}_M\rangle$ is the normalized symmetric state for given M and $|P\rangle = |P\lbrace\underbrace{00\dots 0}_{\frac{N-M}{2}}\underbrace{11\dots 1}_{\frac{N+M}{2}}\rbrace\rangle$ are the corresponding tensor product states for permutations $P$ of $(N+M)/2$ states $|1\rangle$, then one may start with $|u_P\rangle = |P\rangle - \langle D^{(N)}_ M| P\rangle |D^{(N)}_M\rangle$, $\langle D^{(N)}_ M| u_P\rangle = 0$, and build an orthonormal basis of non-symmetric states thereof. But I am not aware of any more explicit form. $\endgroup$ – udrv Apr 26 '16 at 5:34

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Browse other questions tagged or ask your own question.