# Calculating Entanglement Spectrum from reduced density matrix

It says:

After a system is partitioned into subsystems {A} and {B}, a state of the entire system may now be expressed as a matrix with rows and columns given by the state vectors from the Hilbert space for {A} and the state vectors from the Hilbert space for {B}.

If we have an initial Hilbert space that contains 4 spins, the basis states can be labelled as such: $$|0000\rangle,|1000\rangle,|0110\rangle$$, etc. When we bipartition the system, each of these basis states can be split as, e.g., $$|00\rangle_{A}|00\rangle_{B}, |10\rangle_{A}|00\rangle_{B}, |01\rangle_{A}|10\rangle_{B}$$, etc. We can then construct any state matrix with the A states on one axis, and the B states on the other. The ground-state matrix takes the following form, for example:

We can construct the reduced density matrix by taking the dot product of this ground state matrix with its hermitian conjugate:

We call the spectrum of the eigenavalues of the reduced density matrix the entanglement spectrum of this state.

I wanna calculate Entanglement Spectrum.

First I wanna calculate reduced density matrix $$\rho_A$$, I get from mutiply matrix

$$\rho_A = \pmatrix{0.91 & 0 & 0 & -0.276 \\ 0 & 0 & 0 & 0 \\ 0 & 0 & 0 & 0 \\ -0.276 & 0 & 0 & 0.0952 }$$

but problem! Trace is not $$1$$ but trace $$Tr(\rho_A) = 1.0052$$.

Questions: What is no $$1$$? Should normalize $$Tr(\rho_A)$$? How to do? How to understand correct way to calculate?

Your $$V_{GS}$$ isn't normalized (the sums of the squared elements don't add up to 1). Also, I'm not sure if the $$|0011 \rangle$$ coefficient should be 0.1 instead of 0.3. The sums of squares then add up to 1.0052, the value you got for the trace. If you start with the matrix for the normalized state $$V_{GS}$$, your trace should sum to 1.