4 votes

Prove that $\text{Tr}(M|ψ\rangle\langleϕ|)=\langleϕ|M|ψ\rangle$

Your proof is not general, it assumes implicitly that the operator $M|\psi\rangle\langle \phi|$ is diagonalizable (there is a basis of eigenvectors). You can instead just use basic fact of Trace, ...
Pierre-Paul T.'s user avatar
3 votes

conditions for two hermitians operators same up to unitary

It holds precisely if they have the same spectrum: You already argue for necessity. For sufficiency: $A=VDV^\dagger$, $B=WDW^\dagger$, then $D=W^\dagger BW$ and thus $A=VW^\dagger DV^\dagger W$. Thus, ...
Norbert Schuch's user avatar
2 votes

What Hamiltonians generate Hadamard and CNOT matrices?

To add to the other answer: multiple such Hamiltonians are possible, in general. A simple way to see it is to notice that you are looking for Hermitians $H$ such that $e^{iH}=U$ for a given unitary $U$...
glS's user avatar
  • 23.9k
1 vote

Prove that $\text{Tr}(M|ψ\rangle\langleϕ|)=\langleϕ|M|ψ\rangle$

$\newcommand{\bra}[1]{\left<#1\right|}\newcommand{\ket}[1]{\left|#1\right>}\newcommand{\bk}[2]{\left<#1\middle|#2\right>}\newcommand{\bke}[3]{\left<#1\middle|#2\middle|#3\right>}$ If ...
FDGod's user avatar
  • 1,739
1 vote

Performing a projective measurement, is the resulting expectation value $\langle \Psi|M|\Psi\rangle$ bounded between $+1$ and $-1$?

The measurement operators $\{M_i\}$ obey two conditions, firstly they are positive operators, $M_i\geq 0$, which is $\forall |\psi \rangle,\, \langle \psi | M_i |\psi \rangle \geq 0$. Secondly, they ...
Ethan's user avatar
  • 51
1 vote

help understanding gate to hamiltonian and representation

There are many ways that you could go about doing this. (You might also want to take a look at what you"know" about exponentiating a matrix, because you've exponentiated $A$ not $iA$.) One ...
DaftWullie's user avatar
  • 56.9k
1 vote

Quantum teleportation of unknown qubit when the entangled state is not a Bell state

I am assuming that $|j|=1$, i.e. $j$ is just a phase factor. Otherwise, the state would not be normalized (as already mentioned in a comment above) and the state would also not be maximally entangled (...
Tobias95's user avatar
1 vote

How to prove that the mutual information is subadditive?

Unlike entropy, mutual information can be either subadditive or superadditive. If $A$, $B$, $C$ are bits with equal probabilities and $A=B=C$, then $$ I(A,B:C) = 1,\\ I(A:C) + I(B:C) = 2, $$ ...
Adam's user avatar
  • 11

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