Result of bra-kets with multiple spins

Question

I'm working on an exercise where I'm calculating the transition probability of a system consisting of two spin-1/2 particles. This system has a Hamiltonian $$ \hat{H} = H_z (\hat{S}_{1x}+\hat{S}_{2x}) $$ where $H_z$ is a constant magnetic field in the z-direction and $\hat{S}$ is the spin operator defined as $$ S=\frac{\hbar}{2} \sigma_x $$ where $\sigma_x$ is the Pauli matrix. After calculating the eigenergies, the eigenstates can be written as \begin{align} | 0,0 \rangle &= \frac{1}{\sqrt{2}}\left( |\uparrow \downarrow + \downarrow \uparrow\rangle \right) \\ | 1,1 \rangle &=| \uparrow \uparrow \,\rangle \\ | 1,0 \rangle &=\frac{1}{\sqrt{2}}\left( |\uparrow \downarrow - \downarrow \uparrow\rangle \right) \\ | 1,-1 \rangle &=| \downarrow \downarrow \,\rangle \end{align} i.e., as combinations of spin up/down. My question is: when calculating some matrix elements (e.g. in perturbation theory), terms like $$ W = \dots = \frac{1}{\sqrt{2}} \langle \, \, (\langle \uparrow \downarrow| + \langle\downarrow \uparrow|) \,\, | V | \, \,(|\uparrow\downarrow + |\downarrow\uparrow\rangle) \, \, \rangle = \frac{2}{\sqrt{2}} V (\text{after reducing)} $$ pop up often. Of course, things like \begin{align} \langle \uparrow \downarrow|\uparrow \downarrow \rangle &= 1 \\ \langle \uparrow \downarrow|\downarrow \uparrow \rangle &= 0 \end{align} are quite straightforward. But what about things like \begin{align} \langle \uparrow \uparrow|\uparrow \downarrow \rangle &= ? \\ \langle \uparrow \uparrow|(\downarrow \uparrow + \uparrow \downarrow)\rangle &= ? \end{align} Or, in general: are there easy, set rules for handling these bras and kets, maybe even with 3 or more spins, or does it need to be evaluated on a more individual basis?

Answer 1

Both of the bases you presented (the individual-spin-direction basis and the $|j,m\rangle$ basis) are orthonormal, so the inner product between two non-identical basis states is 0. The inner product between a bra and a ket is also bilinear, so you can simplify as follows:

$$\langle a|(|b\rangle + |c\rangle)=\langle a|b\rangle+\langle a|c\rangle$$

So, specifically,

$$\langle \uparrow \uparrow|\uparrow\downarrow\rangle=0$$ $$\langle \uparrow \uparrow|(|\downarrow\uparrow\rangle+|\uparrow\downarrow\rangle)=\langle\uparrow\uparrow|\downarrow\uparrow\rangle+\langle\uparrow\uparrow|\uparrow\downarrow\rangle=0$$