Why my density matrix trace is over 1?

Question

Suppose this operator $$ \rho=\frac{a^2}{\cosh^2(r)}\sum_{n=0}^{\infty}\tanh^{2n}(r)|0,n\rangle\langle 0,n|+\frac{b^2}{\cosh^4(r)}\sum_{n=0}^{\infty}(n+1)\tanh^{2n}(r)|1,n+1\rangle\langle 1,n+1| $$ where $a^2+b^2=1$.

I wanted to obtain the matrix representation of sectors $n$ and $n+1$ on basis $\{|0,n\rangle,|0,n+1\rangle,|1,n\rangle,|1,n+1\rangle\}$

These are my calculations

1. $\langle0,n|\rho|0,n\rangle = a^2$
2. $\langle0,n+1|\rho|0,n+1\rangle = a^2\tanh^2(r)$
3. $\langle1,n|\rho|1,n\rangle=b^2$
4. $\langle1,n+1|\rho|1,n+1\rangle=b^2$

The rest of the elements are 0

Also I used below formulas $$ \sum_{n=0}^{\infty}\tanh^{2n}(r)=\cosh^2(r) $$ $$ \sum_{n=0}^{\infty}(n+1) \tanh^{2n}(r)=\cosh^4(r) $$ As you see according to my calculations the trace of the matrix is over 1.

What did i miss? Plz help me.

Answer 1

I don't think you've calculated any of your diagonal elements correctly. You should have:

1. $\langle 0,n|\rho|0,n\rangle=\frac{a^2}{\cosh^2(r)}\tanh^{2n}(r)$
2. $\langle 0,n+1|\rho|0,n+1\rangle=\frac{a^2}{\cosh^2(r)}\tanh^{2n+2}(r)$
3. $\langle 1,n|\rho|1,n\rangle=\frac{b^2}{\cosh^4(r)}n\tanh^{2n-2}(r)$
4. $\langle 1,n+1|\rho|1,n+1\rangle=\frac{b^2}{\cosh^4(r)}(n+1)\tanh^{2n}(r)$