Consider a generic $2\times 2$ Hermitian matrix written as $$\rho =\alpha\sigma_0+\beta\hat{\vec n}\cdot\vec\sigma\quad ,$$ where $\hat{\vec n}$ is a unit vector and the coefficients are real numbers.

My question is this; how do I figure out what the coefficients have to be for a pure state and a mixed state. I know theoretically what I'm supposed to do, focusing on pure state I should find $$Tr[\rho^2]=1$$ and use the identity $(\vec a\cdot\vec\sigma)(\vec b\cdot\vec\sigma)=(\vec a\cdot\vec b)I+i(\vec a\times\vec b)\vec\sigma$. However somewhere along these lines I fail and can't get the right answer. I know an easy method would be to just calculate the matrix itself by expanding the pauli vector into its components but I am trying to avoid that if possible, any suggestions?

EDIT: Here is my full attempted solution

I start with the condition that $\mathrm{Tr}[\rho]=\mathrm{Tr}[\rho^2]=1$ for pure states. Starting with $\mathrm{Tr}[\rho]$ I can solve it according as $$\mathrm{Tr}[\alpha\sigma_0]+\mathrm{Tr}[\beta\hat{\vec n}\cdot\vec\sigma]=1\Longleftrightarrow \mathrm{Tr}[\alpha\sigma_0]=1\Longleftrightarrow \alpha=\frac{1}{2}\quad .$$ Where I used the fact that the Pauli vector is traceless. Now I try to find $\rho^2$. $$\rho^2=(\alpha\sigma_0+\beta\hat{\vec n}\cdot\vec\sigma)(\alpha\sigma_0+\beta\hat{\vec n}\cdot\vec\sigma)=|\alpha|^2\sigma_o+2\alpha\beta\hat{\vec n}\cdot\vec\sigma+|\beta|^2(\hat{\vec n}\cdot\hat{\vec n})\sigma_0+i|\beta|^2(\hat{\vec n}\times\hat{\vec n})\vec\sigma\\ =|\alpha|^2+2\alpha\beta\hat{\vec n}\vec\sigma+|\beta|^2\sigma_0.\\ \Longrightarrow \mathrm{Tr}[\rho^2]=2|\alpha|^2+2|\beta|^2=1 \\ \Longleftrightarrow |\beta|^2=\frac{1}{4} \Longleftrightarrow\beta=\frac{1}{2}$$

Is this train of thought correct in any manner or have I made some baseless assumptions along the way?

  • 1
    $\begingroup$ I've edited some of your equations. Feel free to undo if you want. $\endgroup$ Mar 15, 2022 at 20:58
  • $\begingroup$ I've deleted my answer as I've seen to misunderstood your question. $\endgroup$ Mar 15, 2022 at 21:10

1 Answer 1


That seems to be right to me, but don't forget that $\beta = -\frac{1}{2}$ because you had absolute value in your equation and $\beta$ can be any real number. Also, another way to think about pure states is that they are points on the edge of the unit ball; that is to say, if when the density matrix is written in the form $$\rho = \frac{1}{2}(I +r_x\sigma_x + r_y\sigma_y+r_z\sigma_z),$$ and if $r_1^2+r_2^2+r_3^2=1$, then the density matrix represents a pure state. If $r_1^2+r_2^2+r_3^2<1$, it is a mixed state. The form we're given is $$\rho = \alpha\sigma_0 + \beta \hat{\vec{n}} \cdot \vec{\sigma}= \alpha I + n_x\beta\sigma_x + n_y\beta\sigma_y + n_z\beta\sigma_z = \frac{1}{2} ( 2\alpha I + 2n_x\beta\sigma_x + 2n_y\beta\sigma_y + 2n_z\beta\sigma_z)$$ Using this method, it is clear that $\alpha = \frac{1}{2}$, and we have that $r_1^2+r_2^2+r_3^2 = (2\beta n_x)^2 + (2\beta n_y)^2 + (2\beta n_z)^2 = 4\beta^2$. Since we want this to equal $1$, we have $1=4\beta^2 \implies \beta = \frac{1}{2},$ or $\beta= -\frac{1}{2}$.

Hope this helps.

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    $\begingroup$ This helped a lot! Thank you! $\endgroup$ Mar 16, 2022 at 5:41

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