3
votes
Accepted
Why is an operator unitary if and only if its matrix representation is unitary?
The adjoint operator goes in the reverse direction $f^\dagger : W \rightarrow V$. In your notations the defining equality should be
$$
(f^\dagger(|w\rangle ), |v\rangle ) = (|w\rangle , f(|v\rangle )),...
- 7,787
3
votes
Why is an operator unitary if and only if its matrix representation is unitary?
The statement "an operator is unitary if and only if each of its matrix representations is unitary", when taken at face value is wrong. To see this take, for example, the operator $f:\mathbb ...
- 3,474
2
votes
Showing $H(B)_\omega \leq H(B)_\rho$ via concavity of von Neumann entropy
Consider the pure state $|\phi\rangle$. We can extend it to an orthonormal basis of the space $\{|\phi_i\rangle\}_{i=1}^d$ with $|\phi_1\rangle = |\phi\rangle$ and $d$ being the dimension of the space....
- 6,623
2
votes
Computing the expected value of a spin - 1 particle component given density matrix
This reads like it is some kind of homework assignment, so I won't compute the full answer.
For a density matrix, the expectation value of any operator $\hat{O}$ is given by $\langle \hat{O} \rangle = ...
- 536
1
vote
Prove that $U|0\rangle\otimes |0\rangle+U|1\rangle\otimes|1\rangle=|0\rangle\otimes U^T|0\rangle+|1\rangle\otimes U^T|1\rangle$
This is valid for any dimension and any matrix $U$:
\begin{align*}(U \otimes I) |\phi^+\rangle &= \frac1{\sqrt d}\sum_i U|i\rangle |i\rangle \\
&=\frac1{\sqrt d} \sum_{ij} |j\rangle \langle j|...
- 3,073
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