Off Diagonal Zero Entries in Pure State Density Matrix

Question

Let $$\rvert\psi\rangle = a_u\rvert u\rangle + a_d\rvert d\rangle = \langle u\rvert\psi\rangle | u\rangle + \langle d\rvert \psi\rangle |d\rangle,$$ where $$\rvert u\rangle = \begin{pmatrix} 1 \\ 0 \\ \end{pmatrix}\quad\text{and}\quad\rvert d\rangle = \begin{pmatrix}0\\1\\\end{pmatrix}.$$

Then $$\rvert\psi\rangle = \begin{pmatrix} a_u \\ a_d \\ \end{pmatrix} = \begin{pmatrix} \langle u\rvert\psi\rangle \\ \langle d\rvert\psi\rangle \\ \end{pmatrix}, \quad\text{and}\quad \langle\psi\rvert = \begin{pmatrix} \bar a_u & \bar a_d \\ \end{pmatrix} = \begin{pmatrix} \langle\psi\rvert u\rangle & \langle\psi\rvert d\rangle \\ \end{pmatrix},$$ so the density matrix corresponding to $|\psi\rangle$ is given by \begin{align} \rvert\psi\rangle\langle\psi\rvert &= \begin{pmatrix} \langle u\rvert\psi\rangle \\ \langle d\rvert\psi\rangle \\ \end{pmatrix} \begin{pmatrix} \langle\psi\rvert u\rangle & \langle\psi\rvert d\rangle \\ \end{pmatrix} \\& = \begin{pmatrix} \langle u\rvert\psi\rangle\langle\psi\rvert u\rangle & \langle u\rvert\psi\rangle\langle\psi\rvert d\rangle \\ \langle d\rvert\psi\rangle\langle\psi\rvert u\rangle & \langle d\rvert\psi\rangle\langle\psi\rvert d\rangle \end{pmatrix} \\ & = \begin{pmatrix} a_u\bar a_u & \color{blue}{a_u \bar a_d} \\ \color{blue}{a_d \bar a_u} & a_d \bar a_d \end{pmatrix} \\& = \begin{pmatrix} \rvert a_u\rvert^2 & 0 \\ 0 & \rvert a_d\rvert^2\end{pmatrix}. \end{align}

How do you show that the off diagonal terms $\color{blue}{a_u\bar a_d}$ and $\color{blue}{a_d\bar a_u}$ come out to 0 ?

Answer 1

Those terms are only zero if one of $a_u$ or $a_d$ is zero, i.e. if $|\psi\rangle$ coincides with one of the basis vectors. The off-diagonal terms of a density matrix are known as its coherences and they are one of the most important parts of the matrix. (The diagonal terms, on the other hand, are known as populations.)

For a pure state that's not a basis state, in general, the coherences will not vanish. You can make them vanish by considering a mixed state, which has the form $$ \rho = \sum_i p_i |\psi_i\rangle \langle \psi_i|, $$ where the $p_i$ are nonnegative weights that sum to one. In that case, the coherences will typically be smaller than for a pure state with the same populations, and in some special cases they can vanish completely.

For a pure state, though, this won't happen.