# Indices of a Density Matrix

• I
I am reading Leonard Susskind's Theoretical Minimum book on Quantum Mechanics. Excercise 7.4 is as follows:

Calculate the density matrix for ##|\Psi\rangle = \alpha|u\rangle + \beta|d\rangle##.

$$\psi(u) = \alpha, \quad \psi^*(u) = \alpha^* \\ \psi(d) = \beta, \quad \psi^*(d) = \beta^*\\ \rho_{a'a} = \begin{pmatrix} \alpha^*\alpha & \alpha^*\beta \\ \beta^*\alpha & \beta^*\beta \end{pmatrix}$$

From my understanding, if we use ##\rho_{a'a} = \psi(a')\psi^*(a)##, then I notice that the matrix is equivalent to:

$$\rho_{a'a} = \begin{pmatrix} \psi^*(u)\psi(u) & \psi^*(u)\psi(d) \\ \psi^*(d)\psi(u) & \psi^*(d)\psi(d) \end{pmatrix} = \begin{pmatrix} \rho_{uu} & \rho_{du} \\ \rho_{ud} & \rho_{dd} \end{pmatrix}$$

I was wondering how the indices affect the position of each ##\rho_{a'a}## in the density matrix.

In other words, why is the matrix not:

$$\rho_{a'a} = \begin{pmatrix} \alpha^*\alpha & \beta^*\alpha \\ \alpha^*\beta & \beta^*\beta \end{pmatrix}$$

Orodruin
Staff Emeritus
Homework Helper
Gold Member
Because it is not defined that way and because it is not symmetric (it is Hermitian).

• vanhees71
Because it is not defined that way and because it is not symmetric (it is Hermitian).

But both matrices are Hermitian, would it be defined such that:

$$\rho_{aa'} = \begin{pmatrix} \rho_{11} & \rho_{12} \\ \rho_{21} & \rho_{22} \end{pmatrix}$$

And u would correlate to the 1 while d correlates to the 2, since ##|u\rangle =
\begin{pmatrix}
1 \\
0
\end{pmatrix}## while ##|d\rangle## is orthonormal?

DrClaude
Mentor
The density operator for state ##| \Psi \rangle## is defined as ##\hat{\rho} = | \Psi \rangle \langle \Psi|##. When you go to a density matrix representation, you can choose the order of the components, e.g., for ##|\Phi \rangle = a | u \rangle + b | d \rangle##,
$$\begin{pmatrix} a \\ b \end{pmatrix} \mbox{ or }\begin{pmatrix} b \\ a \end{pmatrix}.$$
Once that order is chosen (the one on the left here), the position of the elements of the density matrix is fixed, so that the result of ##\hat{\rho} |\Phi \rangle## is the same as the corresponding matrix-vector multiplication.

• doggydan42
The density operator for state ##| \Psi \rangle## is defined as ##\hat{\rho} = | \Psi \rangle \langle \Psi|##. When you go to a density matrix representation, you can choose the order of the components, e.g., for ##|\Phi \rangle = a | u \rangle + b | d \rangle##,
$$\begin{pmatrix} a \\ b \end{pmatrix} \mbox{ or }\begin{pmatrix} b \\ a \end{pmatrix}.$$
Once that order is chosen (the one on the left here), the position of the elements of the density matrix is fixed, so that the result of ##\hat{\rho} |\Phi \rangle## is the same as the corresponding matrix-vector multiplication.

That makes sense, since when you find the eigen values of of the density matrix with the state vector and make it depend on the order of components, you get 1 for both options.

Thank you.

vanhees71
Gold Member
Well, let's do the calculation in Dirac's way, which is much simpler, because you don't need to remember so much. The statistical operator of a pure state, given by one representative state ket is
$$\hat{\rho}=|\psi \rangle \langle \psi|=(\alpha |u \rangle + \beta |d \rangle) (\alpha^* \langle u| + \beta^* \langle d|).$$
Now by definition in a basis representation, it's written as
$$\hat{\rho} = \sum_{ij} |i \rangle \langle j | \rho_{ij}.$$
Now you multiply out the ket-bra products, ans you get
$$\rho_{uu}=\alpha \alpha^*, \quad \rho_{ud}=\alpha \beta^*, \quad \rho_{du}=\beta \alpha^*, \quad \rho_{dd}=\beta \beta^*.$$
There's no need to write this in matrix form, but usually in a basis representation you write the vectors as column vectors and the co-vectors as row vectors. In this sense we map
$$|\psi \rangle \rightarrow \psi=\begin{pmatrix}\alpha \\ \beta \end{pmatrix}.$$
The corresponding co-vector is then
$$\psi^{\dagger}=(\alpha^* \quad \beta^*).$$
Then the corresponding projector is according to the usual rules of matrix multiplication
$$\hat{\rho}=\psi \psi^{\dagger} = \begin{pmatrix} \rho_{uu} & \rho_{ud} \\ \rho_{du} & \rho_{dd} \end{pmatrix}=\begin{pmatrix}\alpha \\ \beta \end{pmatrix} (\alpha^* \quad \beta^*) = \begin{pmatrix} \alpha \alpha^* & \alpha \beta^* \\ \beta \alpha^* & \beta \beta^* \end{pmatrix}.$$

• Terran Thrawn and odietrich