New answers tagged

2

Your expressions for $\langle X_0X_1 \rangle$ and $\langle Y_0Y_1 \rangle$ are correct under the assumption that the two qubits are independently random. In the case that they are correlated, these expressions will not yield the right answer. This is because you have to think of $X_0X_1$, for example, as an operator in its own right, rather than just a ...


2

For your more general question in the title, the answer is yes. This is more of a lesson in linear algebra than in quantum computing. What you found out is that actually the set of eigenvectors of a matrix $A$ corresponding to a given eigenvalue $\lambda$ forms a vector subspace of the space upon which $A$ acts, this is called the eigenspace of $\lambda$, $...


7

The quantum states that differ by a global phase (i.e., by a complex number multiple which has absolute value of 1) are considered the same quantum state, since they can not be distinguished using any operations or measurements. Thus, the eigenstates for hω are $|+\rangle = \frac{1}{\sqrt2}(|0\rangle + |1\rangle)$, $-|+\rangle = \frac{1}{\sqrt2}(-|0\...


3

The Pauli matrices form an orthogonal basis of $\mathcal{M}_2$, this vector space can be endowed with a scalar product called the Hilbert-Schmidt inner product $$ \langle A,B\rangle=\mathrm{Tr}(A^\dagger B)$$ since the Pauli matrices anticommute, their product is traceless, and since they are Hermitian this implies that they are orthogonal with respect to ...


Top 50 recent answers are included