How Do You Convert a Density Operator to Matrix Form?

[PatsyTy]

Homework Statement

Write the density operator

$$\rho=\frac{1}{3}|u><u|+\frac{2}{3}|v><v|+\frac{\sqrt{2}}{3}(|u><v|+|v><u|, \quad where <u|v>=0$$

In matrix form

Homework Equations

$$\rho=\sum_i p_i |\psi><\psi|$$

The Attempt at a Solution

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The two first factors $\frac{1}{3}|u><u|$ and $\frac{2}{3}|v><v|$ pertain to the diagonal elements of the 2x2 matrix, I am unsure how to find the off diagonal elements for this matrix and the $|u><v|$ and $|v><u|$ I don't think are included in the sum I listed in the relevant equations.

In matrix form I have

$$\rho=
\left( \begin{array}{cc}
1 & ? \\
? & 2 \end{array} \right)$$

I have the suspicion that $$\frac{\sqrt{2}}{3}$$ is the offdiagonal elements, however, I don't want to put this down just off of my suspicion. Any tips, or things to think of for me would be appreciated.

 

[ShayanJ]

There is nothing deep to be appreciated here. Its just that u and v are somehow serving as the matrix indices. So uu and vv are the diagonal elements and uv and vu are the off-diagonal elements.

 

[PatsyTy]

There is nothing deep to be appreciated here. Its just that u and v are somehow serving as the matrix indices. So uu and vv are the diagonal elements and uv and vu are the off-diagonal elements.

Thank you, that is what I suspected. How do these cross terms arise from the equation

$$\rho=\sum_i p_i|\psi_i><\psi_i$$

It seems that these cross terms do not come from this summation as it is over only one index.

 

[ShayanJ]

The states in that formula are not the base states. That definition is for when you want to assign classical probabilities for the system to be in different states that are not necessarily orthogonal. The off-diagonal terms arise when you expand those states in terms of orthonormal base states.

 

[PatsyTy]

The states in that formula are not the base states. That definition is for when you want to assign classical probabilities for the system to be in different states that are not necessarily orthogonal. The off-diagonal terms arise when you expand those states in terms of orthonormal base states.

Thank you, that makes sense!