# 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 $\frac{1}{2} \,\left( |l_1=1,m_1=1 \rangle |l_2=1,m_2=-1 \rangle \pm |l_1=1,m_1=-1 \rangle |l_2=1,m_2=1 \rangle \right) \otimes \left( |+- \rangle \mp |-+ \rangle \right) \, .$ Commented Jan 22, 2018 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. Commented Jan 22, 2018 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. Commented Jan 22, 2018 at 22:37
• I guess I can be more precise back there: $L^2_i$ is $1(1+1)\hbar^2$ for both states. Commented Jan 22, 2018 at 22:45

## 1 Answer

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.

This way: \begin{align} P_{12}\vert\phi\rangle &= \alpha \left[P_{12}\vert\psi_+\rangle\right]\left[P_{12}\vert\chi_-\rangle\right] + \beta \left[P_{12}\vert\psi_-\rangle\right]\left[P_{12}\vert\chi_+\rangle\right]\, ,\\ &=- \alpha\vert\psi_+\rangle\vert\chi_-\rangle -\beta \vert\psi_-\rangle \vert\chi_+\rangle=-\vert\phi\rangle\, . \end{align} Basically, if you pick the symmetric combination for the spatial part, it must be multiplied by the antisymmetric combinations for the spin part so the product is overall antisymmetric, and vice versa if you pick the antisymmetric combination for the spatial part you have to multiply it by the symmetric one for the spin part.

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.

• Very interesting and thanks @ZeroTheHero! I missed the legitimate possibility of taking a linear combination of the two antisymmetric states. Commented Jan 23, 2018 at 9:14