Bipartite states whose coefficients are entries of a unitary matrix

3

I've been trying to solve this question

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It seems that in order to show it has unit length, we must show that $$ \frac{1}{d} \sum_{m, n=0}^{d=1} \lvert U_{m, n}\rvert ^2 = 1 $$

I've tried searching online for this type of relation on unitary matrices but haven't succeeded in finding anything. Would appreciate any help.

samlu1999

Posted 2020-06-03T21:16:51.567

Reputation: 33

Answers

2

It may be more helpful to think about the relationship you need to show in terms of the Hermitian form $\sum \limits_{m,n=0}^{d-1} U_{m,n} U_{m,n}^\ast = d$, which is a necessary condition for the defining unitary relationship $UU^\dagger = I$.

To see this explicitly, consider the four equations implicit in the $d=2$ case: $$UU^\dagger=\begin{bmatrix} U_{1,1} & U_{1,2} \\ U_{2,1} & U_{2,2} \end{bmatrix} \begin{bmatrix} U_{1,1}^\ast & U_{2,1}^\ast \\ U_{1,2}^\ast & U_{2,2}^\ast \end{bmatrix} = \begin{bmatrix} 1 & 0 \\ 0 & 1 \end{bmatrix}.$$ The diagonal terms are what you need for the desired relationship: $$(U_{1,1} U_{1,1}^\ast + U_{1,2} U_{1,2}^\ast) + (U_{2,1} U_{2,1}^\ast + U_{2,2} U_{2,2}^\ast) = 1 + 1 = 2.$$ It should not be hard to convince yourself that this generalizes to any $d$.

Jonathan Trousdale

Posted 2020-06-03T21:16:51.567

Reputation: 2 714

2

A matrix is unitary iff its columns (equivalently, rows) are orthonormal (see e.g. Wikipedia).

The sum $\sum_{nm}|U_{nm}|^2$ can be thought of as the sum of the squared norms of the rows of $U$ (equivalently, of the columns of $U$). As each such row (column) has unit norm, the sum of these norms must equal $d$.

Being $U$ unitary, its SVD has the form $U=\sum_k |u_k\rangle\!\langle k|$ for some orthonormal basis $(|u_k\rangle)_k$. The corresponding state $|\Psi\rangle$ thus reads $$|\Psi\rangle = \frac{1}{\sqrt d}\sum_{k=1}^d |u_k\rangle\otimes|k\rangle,$$ which is the maximally entangled state.

glS

Posted 2020-06-03T21:16:51.567

Reputation: 12 247