I recently came across the concepts of operators. However with current my knowledge I am unable to solve the following problem.Given an operator $$\vec{A}=\frac{1}{2}(I+\vec{n}.\vec{\sigma})$$ where $\vec{n}=n_x\hat{x}+n_y\hat{y}+n_z\hat{z}$ is a unit vector and $\vec{\sigma}=\sigma_x\hat{x}+\sigma_y\hat{y}+\sigma_z\hat{z}$ in usual Pauli matrices notation, under what condition on $\vec{n}$, is $\vec{A}$ is a positive operator and in other condition $\vec{A}$ is a projection operator.

  • 3
    $\begingroup$ What have you tried already? Have you tried computing the eigenvalues as functions of $n_x$, $n_y$ and $n_z$ yet? It's only 2 by 2, you can write out the characteristic polynomial without issue. $\endgroup$
    – AHusain
    Nov 20, 2018 at 21:27
  • $\begingroup$ I tried with eigen values but could not figure out the final steps...better if you can help me @AHusain $\endgroup$
    – Aleph
    Nov 21, 2018 at 7:15

1 Answer 1


This is really a question about eigenvalues:

  • A projector has eigenvalues 1 and 0. So, for a qubit, that could be eigenvalues $\{1,1\}$ or eigenvalues $\{1,0\}$.

  • A positive operator is one for which all eigenvalues $\lambda$ satisfy $\lambda>0$.

One could calculate the eigenvalues by brute force, but there are a couple of tricks that will help you. Firstly, the trace is equal to the sum of the eigenvalues $$ \text{Tr}(A)=\sum_{i=1}^2\lambda_i. $$

Reader exercise: justify that $\text{Tr}(A)=\frac12$.

Hence, it is impossible that the projector has eigenvalues $\{1,1\}$ (this should be obvious, as that can only give you $\mathbb{I}$). We're looking for the projector to have eigenvalues $\{1,0\}$. Note also that this must be the edge case for positivity.

Now, to progress with the calculation of the eigenvalues, evaluate $$ \text{Tr}(A\cdot A)=\sum_{i=1}^2\lambda_i^2=\lambda_1^2+\left(\frac12-\lambda_1\right)^2. $$ To answer the specific question, we don't need a full calculation of the eigenvalues. Instead, we just need to know when $\text{Tr}(A\cdot A)=1$. \begin{align} \text{Tr}(A\cdot A)&=\frac{1}{4}\text{Tr}\left((\mathbb{I}+\vec{n}\cdot\vec{\sigma})(\mathbb{I}+\vec{n}\cdot\vec{\sigma})\right) \\ &=\frac{1}{4}\text{Tr}\left(\mathbb{I}+2\vec{n}\cdot\vec{\sigma}+(\vec{n}\cdot\vec{\sigma})(\vec{n}\cdot\vec{\sigma})\right) \\ &=\frac12+\frac14\text{Tr}\left((\vec{n}\cdot\vec{\sigma})(\vec{n}\cdot\vec{\sigma})\right) \end{align} Reader Exercise: Justify that $\text{Tr}\left((\vec{n}\cdot\vec{\sigma})(\vec{n}\cdot\vec{\sigma})\right)=2\vec{n}\cdot\vec{n}$.

Hence, the condition for being a projector is $|\vec{n}|=1$, and the condition for being positive is $|\vec{n}|<1$, although I would suggest that of more relevance for density matrices is the positive semi-definite condition: $|\vec{n}|\leq 1$.


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