# Bloch sphere representation of an eigenvector

I'm trying to work through a problem that wants me to determine the Bloch sphere representation of the eigenvectors of $$\sigma_{z}$$.

I'm working in bra-ket notation so these would be $$\ v_{+} = |0\rangle\langle0|$$ and $$v_{-} = -|1\rangle\langle1|$$ but I'm a little uncertain as to whether I've represented these correctly.

My issue is I'm not sure what I'm working towards. I imagine I need to find $$r$$ for each of these vectors. I would appreciate if someone could point me in the right direction.

• WP. Commented Apr 18, 2019 at 18:00
• Related : Understanding the Bloch sphere. Commented Sep 17, 2019 at 12:55
• One correction: Your second eigenvector cannot be $-|1\rangle \langle 1|$. It has to be $+|1\rangle \langle 1|$. Density matrices must have a trace of 1, so the second is not a valid density matrix and doesn't represent a physical state. Commented Sep 17, 2019 at 13:10

What you have identified ($$|0\rangle\langle0|$$ and $$-|1\rangle\langle1|$$) are the matrix elements of the Pauli matrix $$\sigma_z$$. The eigenvectors are simply $$|0\rangle$$ and $$|1\rangle$$.
The Bloch sphere represents possible quantum states of a two-state system. It actually represents states using density matrices rather than kets, so the eigenvectors for your question are $$|0\rangle\langle0|$$ and $$|1\rangle\langle1|$$.
The Bloch sphere representation of a state is the following. The identity matrix and three Pauli matrices form a basis $$\{I, \sigma_x, \sigma_y, \sigma_z\}$$ for the space of 2x2 matrices. The coefficients of the linear combination required to describe an arbitrary state form the Bloch sphere representation of the state. Valid physical states must have trace 1, so the coefficient of $$I$$ is guaranteed to be $$1/2$$. So three numbers suffice to describe a state: $$\frac{1}{2}(I+n_x\sigma_x+n_y\sigma_y+n_z\sigma_z) = |\psi\rangle\langle\psi|$$, and $$(n_x,n_y,n_z)$$ is the Bloch sphere representation of the state $$|\psi\rangle$$ (equivalently $$|\psi\rangle\langle\psi|$$).
So in your case, $$|0\rangle\langle0|$$ is represented by $$(0,0,1)$$ and $$|1\rangle\langle1|$$ represented by $$(0,0,-1)$$.
• Thank you! This makes a lot more sense now. I appreciate the help :) I'm now working on a more complicated example where an operator is given by $v_{x}\sigma_{x}+v_{y}\sigma_{y}+v_{z}\sigma_{z}$ - so applying the same principles the eigenvectors for this mixed state would be $|0><1|, |1><0|,|1><1|, |0><0|.$ such that the Bloch sphere representation would be for |0><1|: v=$(v_{x},-iv_y,0)$, |1><0|: v=$(v_{x},iv_y,0)$ and |1><1|, |0><0| will be the same as the previous example. Although I'm doubting using $v$ in the final form. Commented Jan 29, 2019 at 11:48
• Those eigenvectors cannot be right. You have a 2x2 matrix, so you will have at most two eigenvectors. To find the eigenspaces for an operator $O$, you should solve the equation $O|u\rangle = \lambda|u\rangle$ where $\lambda$ is any constant and $|u\rangle$ is what you are solving for (express it as a linear combination of basis vectors).