Can a Qubit's Mixed State Density Operator Be Expressed Using Pauli Matrices?

mtak0114
Messages
46
Reaction score
0
1. Show that an arbitrary density operator for a mixed state qubit may be written as


2. \rho = \frac{I+r^i\sigma_i}{2}, where ||r||<1
(Nielsen and Chuang pg 105)

3. So my attempt was as follows
Given that a \rho is hermitian it may be written as a linear combination of the pauli matrices and the identity.
\rho = aI+r^i\sigma_i where r^i is arbitrary
the trace condition tr(\rho)=1 implies a=1/2 and
the positivity condition \langle\varphi|\rho|\varphi\rangle \geq 0
implies that 2cos\theta |r||\langle\varphi|\sigma^i|\varphi\rangle| \geq -1 which implies that
|r|\leq 1/2, fnally redefing r above gives the result QED
is this correct or am i assuming too much?

cheers

Mark
 
Last edited:
Physics news on Phys.org
Everything fine!
 
Thread 'Need help understanding this figure on energy levels'
This figure is from "Introduction to Quantum Mechanics" by Griffiths (3rd edition). It is available to download. It is from page 142. I am hoping the usual people on this site will give me a hand understanding what is going on in the figure. After the equation (4.50) it says "It is customary to introduce the principal quantum number, ##n##, which simply orders the allowed energies, starting with 1 for the ground state. (see the figure)" I still don't understand the figure :( Here is...
Thread 'Understanding how to "tack on" the time wiggle factor'
The last problem I posted on QM made it into advanced homework help, that is why I am putting it here. I am sorry for any hassle imposed on the moderators by myself. Part (a) is quite easy. We get $$\sigma_1 = 2\lambda, \mathbf{v}_1 = \begin{pmatrix} 0 \\ 0 \\ 1 \end{pmatrix} \sigma_2 = \lambda, \mathbf{v}_2 = \begin{pmatrix} 1/\sqrt{2} \\ 1/\sqrt{2} \\ 0 \end{pmatrix} \sigma_3 = -\lambda, \mathbf{v}_3 = \begin{pmatrix} 1/\sqrt{2} \\ -1/\sqrt{2} \\ 0 \end{pmatrix} $$ There are two ways...
Back
Top