# Why my density matrix trace is over 1?

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.

• I've just converted your maths into latex format. Can you check that I have converted it to mean what you intended? Dec 21, 2022 at 7:41

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