# Two identical particles

A system made up of two spin-$1/2$ identical particles is prepared such that:

1. a measurement of $\mathbf{L}_{1}^2$ and $\mathbf{L}_{2}^2$ gives $2\hbar^2$ with certainty;
2. a measurement of $L_{1z}$ and $L_{2z}$ gives $+\hbar$ for one particle and $-\hbar$ for the other one;
3. a measurement of $S_{1z}$ and $S_{2z}$ gives $+\hbar/2$ for one particle and $-\hbar/2$ for the other one.

Write the most general quantum state, $\left|\alpha\right\rangle$, compatible with these measurements.

Since there are two fermions, I know the state must be antisymmetric.

From 1., I know $\ell_1=\ell_2=1$, so the total orbital angular momentum can be $\ell=0,1,2$.

From 2. and 3., $m_{\ell_1}=+1$ and $m_{\ell_2}=-1$, $m_{s_1}=+1/2$ and $m_{s_2}=-1/2$; so, $m_\ell=m_s=0$.

How can I write the state of this system?

• I'd say $\frac{1}{2} \,\left( |l_1=1,m_1=1 \rangle |l_2=1,m_2=-1 \rangle \pm |l_1=1,m_1=-1 \rangle |l_2=1,m_2=1 \rangle \right) \otimes \left( |+- \rangle \mp |-+ \rangle \right) \, .$ – secavara Jan 22 '18 at 22:28
• @secavara, thank you for your answer. But how would you justify this? I can't understand if this is the most general state it can be written. – Vincenzo Ventriglia Jan 22 '18 at 22:36
• It's an antisymmetric state under permutation (that's the reason for the $\pm,\mp$) which is needed since they are two identical fermions. And the rest of the properties mentioned are satisfied: $L^2_i$ is 1 for both states, we can only have $m_i$ equal to 1 for one and -1 for the other in each measurement and similarly for the spin. – secavara Jan 22 '18 at 22:37
• I guess I can be more precise back there: $L^2_i$ is $1(1+1)\hbar^2$ for both states. – secavara Jan 22 '18 at 22:45

You know that you have $\ell_1=\ell_2=1$, and the $m_{\ell_1}=+1$ while $m_{\ell_2}=-1$. Hence you must have (up to a factor of $1/\sqrt{2}$): $$\vert\psi_\pm\rangle = \vert 1 1\rangle_1\vert 1,-1\rangle_2 \pm \vert 1,-1\rangle_1\vert 1 1\rangle_2$$ Now for the spin part, you can conclude likewise that $$\vert\chi_\pm\rangle = \vert 1/2,1/2\rangle_1\vert 1/2,-1/2\rangle_2\pm \vert 1/2,-1/2\rangle_1\vert 1/2,1/2\rangle_2\, .$$ Note that, under the interchange of particle labels: $$P_{12}\vert\psi_\pm\rangle =\pm \vert\psi_\pm\rangle\, , \qquad P_{12}\vert\chi_\pm \rangle= \pm \vert\chi_\pm\rangle\, .$$ To have overall antisymmetry you should therefore have $$\vert\phi\rangle=\alpha \vert\psi_+\rangle\vert\chi_-\rangle + \beta \vert \psi_-\rangle \vert\chi_+\rangle$$ with $\alpha$ and $\beta$ chosen so your states are properly normalized.
Note that the situation gets considerably more complicated with $3$ particles as the permutation group $S_3$ of three objects have a 2-dimensional representation of mixed symmetry which do not necessarily transform back to a $\pm 1$ multiple of themselves under permutation. The combination of such types of functions is a little more delicate than in the 2-particle case.