How Does the Dirac Spin Exchange Operator Work in Quantum Mechanics?

[MisterX]

The spin exchange operator would have the property
$$\begin{align*}P\mid \chi_{\uparrow\downarrow} \rangle = \mid\chi_{\downarrow\uparrow} \rangle & &P\mid \chi_{\downarrow\uparrow} \rangle =\mid \chi_{\uparrow\downarrow} \rangle \end{align*}$$
This also implies $P\mid \chi_{\text{sym.}} \rangle = \mid \chi_{\text{sym.}} \rangle$ and $P\mid \chi_{\text{asym.}} \rangle = -\mid \chi_{\text{asym.}} \rangle$.
According to sources this is equivalent to
$$P = \frac{1}{2}\Big(1+ \sum_{i=1}^3 \sigma_{1i}\sigma_{2i} \Big)\,,$$
where $\sigma_{ij}$ applies $j$-th pauli matrix to spin $i$. I am seeking some intuition of this. If I wanted to verify it, maybe I could check its action on a particular vector and then make an argument involving rotational invariance of the form of $P$. I still would like to have some intuition on this and also understand the note from the wikipedia page.

thanks

 

[Ravi Mohan]

The spin exchange operator would have the property
$$\begin{align*}P\mid \chi_{\uparrow\downarrow} \rangle = \mid\chi_{\downarrow\uparrow} \rangle & &P\mid \chi_{\downarrow\uparrow} \rangle =\mid \chi_{\uparrow\downarrow} \rangle \end{align*}$$
This also implies $P\mid \chi_{\text{sym.}} \rangle = \mid \chi_{\text{sym.}} \rangle$ and $P\mid \chi_{\text{asym.}} \rangle = -\mid \chi_{\text{asym.}} \rangle$.
According to sources this is equivalent to
$$P = \frac{1}{2}\Big(1+ \sum_{i=1}^3 \sigma_{1i}\sigma_{2i} \Big)\,,$$
where $\sigma_{ij}$ applies $j$-th pauli matrix to spin $i$. I am seeking some intuition of this. If I wanted to verify it, maybe I could check its action on a particular vector and then make an argument involving rotational invariance of the form of $P$. I still would like to have some intuition on this and also understand the note from the wikipedia page.

thanks

It is little unclear what you are asking. To verify this you can follow what you said. For a formal derivation you can see below. The method I am using can be used to find the explicit form of any operator if you know its action on the basis vectors.

Now as per the property of the operator $\hat{P}$ the spin is exchanged. Here we are dealing with the tensor product of two Hilbert spaces ($\mathbb{C}^2\otimes\mathbb{C}^2$). Let us restrict ourselves to two dimensional space spanned by $|i\rangle\in (|\uparrow \rangle, |\downarrow \rangle )$ for simplicity. Thus the ansatz operator takes the form $\hat{P}_1\otimes\hat{P}_2$.

The property says
$$\hat{P}_1\otimes\hat{P}_2 |i \rangle \otimes|j \rangle = |j \rangle\otimes |i \rangle$$

Now take the combination $\langle i|\otimes\langle j|$ and hit it from the right on the LHS and RHS of above equation and sum over $i,j$. Symbolically (I am abusing the notation to make it clear. Irony!)

$$\sum_{i,j}\left(\hat{P}_1\otimes\hat{P}_2 |i \rangle \otimes |j \rangle\right)\left( \langle i|\otimes\langle j|\right)= \sum_{i,j}\left(|j \rangle\otimes |i \rangle\right)\left( \langle i|\otimes\langle j|\right) \\ \hat{P}_1\sum_{i} |i \rangle\langle i|\otimes \hat{P}_{2}\sum_{j}|j \rangle\langle j|= \sum_{i,j}|j \rangle\langle i|\otimes |i \rangle\langle j|$$

Using the completeness property the last line becomes
$$\hat{P}_1\otimes P_{2} = \sum_{i,j}|j \rangle\langle i|\otimes |i \rangle\langle j|$$

And now it becomes simple (via matrix algebra or vector algebra) to show that the RHS of last equation is equal to $$ \frac{1}{2}\Big(1+ \sum_{i=1}^3 \sigma_{i}\otimes\sigma_{i} \Big)\,,$$

If you use matrix algebra, you might get a clear intuition of what is happening.