Tensor product of operators and ladder operators

[Heidi]

Hi Pfs
i have 2 matrix representations of SU(2) . each of them uses a up> and down basis (d> and u>
If i take their tensor product i will get 4*4 matrices with this basis:
d>d>,d>u>,u>d>,u>u>
these representation is the sum equal to the sum of the 0-representation , a singlet represertation with
m= 0 and a 1-representation of 3*3 matris with m= -1,0,1
i have two states with m= 0 corresponding to u>d> and d>u> in their vector space.
if i start with m = 1 a ladder operator will decrease m to 0 that will be in the triplet .
How to write this action of the ladder operator in the 4 vector basis?

 

[Heidi]

As the total angular mometum of 2 particles is their sum, it seems natural to take a symmetric result to the action on u>u> of the down ladder
and to get u>d> + d>u> (up to a normalization factor.
repeating its action on this vector would give d>d>
for the singulet u>d> - d>u> it would give the nul vector.
Is this correct?
So changin the basis uu>, ud> , du>, dd> into uu>, (ud>+du>),dd,(ud>-du>), would diagonalize by bloc the tensor product of two 2*2 SU(2) representations.

 

[vanhees71]

Consider the addition of two spins $\vec{S}=\vec{s}_1+\vec{s}_2$ and consider that $\vec{s}_1$ and $\vec{s}_2$ are realized in the irreps. with spins $s_1$ and $s_2$, respectively, and we look for the reduction of $\vec{S}$ into irreducible parts of the induced "product representation".

This is achieved by noting that we have the product basis $|s_1,s_2;\sigma_1,\sigma_2 \rangle$, which is the common basis of the compatible operators $\vec{s}_1^2$, $\vec{s}_2^2$, $s_{1z}$, and $s_{2z}$ with (setting $\hbar=1$ for simplicity)
$$\vec{s}_1^2 |s_1,s_2;\sigma_1,\sigma_2 \rangle = s_1 (s_1+1) |s_1,s_2;\sigma_1,\sigma_2 \rangle,$$
$$\vec{s}_2^2 |s_1,s_2;\sigma_1,\sigma_2 \rangle = s_2 (s_2+1) |s_1,s_2;\sigma_1,\sigma_2 \rangle,$$
$$s_{13} |s_1,s_2;\sigma_1,\sigma_2 \rangle =\sigma_{13} |s_1,s_2;\sigma_1,\sigma_2 \rangle,$$
$$s_{23} |s_1,s_2;\sigma_1,\sigma_2 \rangle =\sigma_{23} |s_1,s_2;\sigma_1,\sigma_2 \rangle,$$
where $s_1, s_2 \in \{0,1/2,1,3/2,\ldots \}$ and $\sigma_{13} \in \{-s_1,-s_1+1,\ldots s_1-1,s_1 \}$, $\sigma_{23} \in \{-s_2,-s_2+1,\ldots,s_2-1,s_2 \}$.
Formally these basis vectors are given by the "product basis"
$$|s_1, s_2;\sigma_1,\sigma_2 \rangle = |s_1,\sigma_1 \rangle \otimes |s_2,\sigma_2 \rangle.$$
On the other hand we have $\vec{S}^2$, $S_3$, $\vec{S}_1^2$, $\vec{S}_2^2$ as another compatible set of spin observables, i.e., we can build a common basis $|S,s_1,s_2,\Sigma \rangle$ with
$$\vec{S}^2 |S,s_1,s_2,\Sigma \rangle=S(S+1) |S,s_1,s_2,\Sigma \rangle,$$
$$\vec{s}_1^2 |S,s_1,s_2,\Sigma \rangle=s_1 (s_1+1) |S,s_1,s_2,\Sigma \rangle,$$
$$\vec{s}_2^2 |S,s_1,s_2,\Sigma \rangle=s_2 (s_2+1) |S,s_1,s_2,\Sigma \rangle,$$
$$S_3 |S,s_1,s_2,\Sigma \rangle = S |S,s_1,s_2,\Sigma \rangle.$$
Now the product basis is also an eigenbasis of $S_3$ and
$$S_3 |s_1,s_2;\sigma_1,\sigma_2 \rangle=(\sigma_1 + \sigma_2) |s_1,s_2;\sigma_1,\sigma_2 \rangle.$$
That means that the possible $\Sigma \in \{(s_1+s_2),(s_1+s_2)-1,\ldots,-(s_1+s_2)+1,-(s_1+s_2) \}$ and
$$\langle S,s_1,s_2,\Sigma|s_1,s_2;\sigma_1,\sigma_2 \rangle \propto \delta_{\Sigma,\sigma_1+\sigma_2}.$$
This implies that we must have $S \geq s_1+s_2$, because otherwise there couldn't be the largest eigenvalue $s_1+s_2$ for $S_3=s_{13}+s_{23}$. We can not have $S>s_1+s_2$, because then via the ladder operator $S_+$ there'd be an eigenvector of $S_3$ with eigenvalue $s_1+s_2+1$, but this doesn't exist within our product representation and so cannot exist in the other basis either.

Thus there's exactly one eigenvector $|S=s_1+s_2,s_1,s_2,\Sigma=s_1+s_2 \rangle$, and we can choose it to be the product-basis vector $|s_1,s_2;\sigma_1=s_1,\sigma_2=s_2 \rangle$. Thus we must have as one part of the new basis the complete basis to the irreducible representation with $S=s_1+s_2$, which we can get from a repeated application of $S_-=S_x-\mathrm{i} S_y$ on this eigenvector with maximum $\Sigma=s_1+s_2$, which finally stops with the eigenvector with $\Sigma=-(s_1+s_2)$.

Now you consider the orthogonal complement of the subspace spanned by these $2(s_1+s_2)+1$ eigenvectors for $S=s_1+s_2$. The largest eigenvalue for $S_3$ in this orthogonal complement can only be $s_1+s_2-1$, and there's only one eigenvector to this eigenvalue (since there are two in the product basis, i.e., the ones where $\sigma_1=s_1-1, \quad \sigma_2=s_2$ and $\sigma_1=s_1, \quad \sigma_2=s_2-1$, i.e., the eigenspace $\text{Eig}(\Sigma,s_1+s_2-1)$ is two-dimensional, and one basis vector we have constructed with $S=s_1+s_2$. Now with the same argument as above for $S$ when restricting $\vec{S}^2$ to the orthonal complement to the $S=s_1+s_2$ irrep. we must have $S=s_1+s_2-1$, and there's a unique (up to a factor) vector orthogonal to $|S=s_1+s_2, s_1,s_2,\Sigma=s_1+s_2-1$, which we have to choose as the eigenvector of $S=s_1+s_2-1,s_1,s_2,\Sigma=s_1+s_2-1$, and again we get all other eigenvectors for the irrep. with $S=s_1+s_2-1$ by acting repeatedly with $S_-$ to this vector.

Then we iterate the argument to the orthogonal complement of both the $S=s_1+s_2$ and $S=s_1+s_2-1$ subspaces etc. It becomes clear that the iteration stops with the lowest possible value $S=|s_1-s_2|$. This you get by simple counting of the resulting dimensions. We know that the entire space is $(2s_1+1)(2s_2+1)$-dimensional (spanned by the product basis) and the dimensions of the $S$ bases of the irred. subspaces just constructed iteratively add indeed up to (make $s_1 \geq s_2$ for simplicity)
$$[2(s_1+s_2)+1]+[2(s_1+s_2-1)+1]+[2 (s_1+s_2-2)+1]+\cdots + [2(s_1-s_2)+1]=(2s_2+1)2(s_1+s_2) - 2(1+2+\cdots 2s_2)+(2s_2+1)= 2 (2s_2+1)(s_1+s_2) -2 \frac{1}{2} (2s_1+1)(2s_2)+2s_2+1=(2s_1+1)(2s_2+1).$$
By explicitly doing this iterative scheme you can derive the Clebsch-Gordan coefficients
$$C(S,s_1,s_2,\Sigma|s_1,s_2;\sigma_1 \sigma_2)=\langle S,s_1,s_2,\Sigma|s_1,s_2;\sigma_1,\sigma_2 \rangle.$$
For your example we have
$$|S=1,s_1=1/2,s_2=1/2,\Sigma=1 \rangle=|s_1=1/2,s_1=1/2,\sigma_1=1/2,\sigma_1=1/2 \rangle,$$
$$|S=1,s_1=1/2,s_2=1/2,\Sigma=0 \rangle = \frac{1}{\sqrt{2}} (|s_1=1/2,s_2=1/2,\sigma_1=-1/2,\sigma_2=1/2 \rangle + |s_1=1/2,s_2=1/2,\sigma_1=1/2,\sigma_2=-1/2 \rangle),$$
$$S=1,s_1=1/2,s_2=1/2,\Sigma=-1 \rangle = |s_1=1/2,s_2=1/2,\sigma_1=-1/2,\sigma_2=-1/2 \rangle.$$
The remaining eigenvector with $\Sigma=0$ must be orthogonal to the above given with $S=1$ and $\Sigma=0$, which uniquely (up to a phase factor) is
$$|S=0,s_1=1/2,s_2=1/2,\Sigma=0 \rangle=\frac{1}{\sqrt{2}} (|s_1=1/2,s_2=1/2,\sigma_1=-1/2,\sigma_2=1/2 \rangle -|s_1=1/2,s_2=1/2,\sigma_1=1/2,\sigma_2=-1/2 \rangle).$$