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Suppose I have a density matrix like $\rho = \frac{1}{2}[I + \hat{n}\vec{\sigma}]$.

The claim is that $\rho$ is an orthogonal projector for the state $|+\rangle$ in an arbitrary direction $\hat{n}$.

How do I convince myself of this claim?

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2 Answers 2

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If a matrix is a projector, it squares to itself. So, you just have to verify that $\rho^2=\rho$. You'll need the assumption that the length of the Bloch vector is 1.

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Suppose I have a density matrix like $\rho = \frac{1}{2}[I + \hat{n}\vec{\sigma}]$.

Explicitly, we have $$ \rho = \frac{1}{2}\left(\begin{matrix} 1+n_z & n_x - in_y\\ n_x + in_y & 1 - n_z \end{matrix}\right) $$

The claim is that $\rho$ is an orthogonal projector for the state $|+\rangle$ in an arbitrary direction $\hat{n}$.

How do I convince myself of this claim?

You can convince yourself based on the well known Bloch sphere expression for a ket in the direction $\hat n$: $$ |\psi\rangle = \cos(\theta/2)|0\rangle + e^{i\phi}\sin(\theta/2)|1\rangle\;, $$ where $$ \hat n = (\sin(\theta)\cos(\phi), \sin(\theta)\sin(\phi), \cos(\theta))\;. $$

Then the density operator is: $$ \hat \rho = |\psi\rangle\langle\psi| $$ and, for example the (0,0) matrix element of the above operator is: $$ \rho_{00} = \langle 0|\psi\rangle\langle\psi|0\rangle = \cos(\theta/2)^2 = \frac{1+\cos(\theta)}{2}\;, $$ which is the (0,0) element of your original matrix, since: $$ \begin{align} \rho &= \frac{1}{2}\left(\begin{matrix} 1+n_z & n_x - in_y\\ n_x + in_y & 1 - n_z \end{matrix}\right)\\ &= \frac{1}{2}\left(\begin{matrix} 1+\cos(\theta) & \sin(\theta)(\cos(\phi)) -i\sin(\phi))\\ \sin(\theta)(\cos(\phi)) +i\sin(\phi)) & 1 - \cos(\theta) \end{matrix}\right)\;. \end{align} $$

And so on.

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