# Matrix form of Density Operator

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1. Jun 28, 2015

### Peter Yu

Hi All,
I have spent hours trying to understand the matrix form of Density Operator. But, I fail. Please see page 2 of the attached file. (from the book "Quantum Mechanics - The Theoretical Minimum" page 199).
Most appreciated if someone could enlighten me this.
Many thanks in advance.
Peter Yu

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2. Jun 28, 2015

### ShayanJ

I find the notation in that book really awful so let me change the notation. Imagine I have a set of basis vectors $\{|\varphi_i\rangle\}$ and I prepared my system in the mixed state $\rho=\sum_i p_i |\psi_i\rangle\langle \psi_i |$ where $|\psi_i\rangle=\sum_k c_{ik}|\varphi_k\rangle$.

The matrix element of an operator L in the above basis are defined to be $L_{mn}=\langle \varphi_m|L|\varphi_n\rangle$.

About the second equation, we have $Tr(\rho L)=\sum_i \langle \varphi_i | \rho L |\varphi_i\rangle$. Now if I put the above expression of $\rho$ in the formula for trace, I'll have:

$Tr(\rho L)=\sum_i \langle \varphi_i | (\sum_i p_i |\psi_i\rangle\langle \psi_i |) L |\varphi_i\rangle=\sum_i \sum_j p_j\langle \varphi_i |\psi_j\rangle\langle \psi_j | L |\varphi_i\rangle= \\ \\ \sum_i \sum_j p_jc_{ji}(\sum_k c_{jk}^* \langle \varphi_k |) | L |\varphi_i\rangle=\sum_i \sum_j \sum_k p_jc_{ji}c_{jk}^* \langle \varphi_k | L |\varphi_i\rangle= \\ \\ \sum_i \sum_j \sum_k p_jc_{ji}c_{jk}^* L_{ki}=\sum_i\sum_k (\sum_j p_j c_{ji}c_{jk}^*) L_{ki}=\sum_i\sum_k \rho_{ik} L_{ki}$

3. Jun 28, 2015

### Dyatlov

Hello!
Density matrix is a matrix that describes the expectation value(L) of a of a state vector from a quantum subsystem which is a part of composed entangled system.
If you have two systems, each with it's own space of states HA and HB, the density matrix simply gives you the information of one of the subsystems from it's own point of view, the second system being just a simple observer and not interfering with any measurement done on the first.

For determining <L>A from HA X HB (note the product state of the entangled system): <L>A = <ψ(x’y’)|L|ψ(xy)>.
<L>A = Σxy ψ*(x’y’)ψ(xy)Lx’x. Here <x’|L|x> = Lx’x = are just the matrix elements of the observable. If we apply δy’y for <L>A (this means we just replace y' with y), we get Σxy ψ*(x’y)ψ(xy)Lx’x. Now we sum over the eigenvalues of the eigenvectors of the second system y (the observer) just so the first system won't have a dependency on those values, so we can truly measure LA. Therefore<L>A = Σy ψ*(x’y)ψ(xy) Lx’x. ρx’x = Σy ψ*(x’y)ψ(xy) is the density matrix and the more non-zero eigenvectors you find inside the matrix of ρ the more entangled the system is. Maximum entanglement is when all eigenvalues are non-zero, equal and of the 1/n form (n being the dimentionality of the space).
NOTE: Σy = sum over y, since I don't know how to put the value under the sum.
Hope this helped!

Last edited: Jun 28, 2015
4. Jun 28, 2015

### Peter Yu

Hi Shyan and Dyatloc,
Heartfelt thank for your assistance. I need some time to absorb your explanation.
Many Thanks
Peter