did i calculate the matrix elements correctly?

Question

Suppose the below operator $$ x\sum_{n=0}^{\infty}\tanh^{2n}(r)|0,n\rangle\langle 0,n| +y\sum_{n=0}^{\infty}\tanh^{2n}(r)|1,n\rangle\langle 1,n| +z\sum_{n=0}^{\infty}\tanh^{2n}(r)(n+1)|1,n\rangle\langle 0,n+1|+z\sum_{n=0}^{\infty}\tanh^{2n}(r)(n+1)|0,n+1\rangle\langle 1,n|+q\sum_{n=0}^{\infty}\tanh^{2n}(r)|0,n+1\rangle\langle 0,n+1| $$

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. $\langle0,n|\rho|0,n\rangle = x\tanh^{2n}+q\tanh^{2n-2}$
2. $\langle1,n|\rho|1,n\rangle = y\tanh^{2n}$
3. $\langle1,n|\rho|0,n+1\rangle=z\tanh^{2n}$
4. $\langle0,n+1|\rho|1,n\rangle=z\tanh^{2n}$
5. $\langle0,n+1|\rho|0,n+1\rangle=x\tanh^{2n+2}+q\tanh^{2n}$
6. $\langle1,n+1|\rho|1,n+1\rangle=y\tanh^{2n+2}$

The rest of the elements are 0

I feel there is something wrong

What did i miss? did i calculate them correctly?

Answer 1

Recall the definition of a trace, for a complete set of orthonormal basis states $|\mu\rangle$, is $$\mathrm{Tr}(X)=\sum_\mu \langle \mu|X|\mu\rangle.$$ Note that I am specifically using some other label $\mu$, not $n$, because the sum for the trace is not the same as the sum for your operator - there will be a double sum in the result.

For example, take $X=\sum_{n=0}^\infty t^{2n}|0,n\rangle\langle 0,n| $ as at the beginning of your operator. It is convenient to use a basis for the trace that has nice inner products with your operator's expansion: we want a basis where $\langle \mu|0,n\rangle$ is easy to compute. Since there are two labels for the states, we want a basis with two labels as well. Then the best choice is to just use $|\mu\rangle=|i,j\rangle$ such that the trace becomes $$\mathrm{Tr}(X)=\sum_{i,j} \langle i,j|X|i,j\rangle.$$ It is now easy to compute $\langle i,j|0,n\rangle=\delta_{i,0}\delta_{j,n}$; the inner product is only nonzero when $i=0$ and $j=n$. We find the sum to collapse: \begin{align} \mathrm{Tr}(X)&=\sum_{i,j} \langle i,j|(\sum_{n=0}^\infty t^{2n}|0,n\rangle\langle 0,n|)|i,j\rangle\\ &=\sum_{i,j} \sum_{n=0}^\infty t^{2n} \delta_{i,0}\delta_{j,n}\\ &=\sum_{n=0}^\infty t^{2n}. \end{align} Note that this is different from your answer because there is still a sum over $n$ (which is a geometric series that can be analytically computed).

As for the other terms, one proceeds in the same way. Note that the trace has the same expression appear on both sides for each term, so off-diagonal elements like $\langle 1,j|\rho|0,j\rangle$, $\langle 1,j|\rho|1,j+1\rangle$, etc. do not appear in the trace.