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?


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,

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


Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.