# Projection operators and positive operators

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.

• 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. – AHusain Nov 20 '18 at 21:27
• I tried with eigen values but could not figure out the final steps...better if you can help me @AHusain – Adex Nov 21 '18 at 7:15

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: $$|\vec{n}|\leq 1$$.