1,514 reputation
426
bio website
location
age
visits member for 3 years, 7 months
seen 2 days ago

delete me


Jul
21
asked Representation theory and the Nekrasov partition function
Jul
18
awarded  Nice Question
Jul
12
comment Why does a measurement on one qubit force another one into a given state in Simon's algorithm?
I think the issue is as to what do you mean by "measuring in the computational basis". What is this? This is what is forcing the collapse into the $\psi_b$ states. This would make sense only if the measurement is by an operator for whom the $\psi_b$ are the eigenstaes.
Jul
12
comment Why does a measurement on one qubit force another one into a given state in Simon's algorithm?
That $\psi_a$ is a linear combination of the $\psi_b$s can't be the reason why the wave-function collapses into only the $\psi_b$ states! One could have as well chosen a different basis to expand the $\psi_a$s in and gotten a different result!
Jul
12
comment Why does a measurement on one qubit force another one into a given state in Simon's algorithm?
CLARIFICATION : But the probabilities don't add up rightly either $\sum_b ( \bar{\psi_b} \psi_a )^2 = 2$. Wonder is the interpretation of the fact that the probabilites over all these possibilities seem to add up to $2$!
Jul
12
revised Why does a measurement on one qubit force another one into a given state in Simon's algorithm?
added 12 characters in body
Jul
12
comment Why does a measurement on one qubit force another one into a given state in Simon's algorithm?
Also These $\psi_b$s do NOT seem to have this property that they are able to span the entire Hilbert space. They are not a basis. The probabilities don't add up either $\sum_b ( \bar{\psi_b} \psi_a )^2 = 1/2$ Then it becomes more unclear as to why you think something which is not of the $\psi_b$ will not be observed.
Jul
12
comment Why does a measurement on one qubit force another one into a given state in Simon's algorithm?
(1) So is this measurement measuring a certain observable/(Hermitian operator) whose eigenstates are theses $\psi_b$s? Then that makes perfect sense that the observation is one of these $\psi_b$s. (to give an alaogy given a spin 1/2 if one measures its spin in the x direction then its no surprise that one would get either the $|1/2,(1/2)_x>$ or $|1/2,(-1/2)_x>$ state) (2) Can you explain what is a "computational basis"? What do you mean by "measurement on the second register is made in the computational basis" ?
Jul
12
comment Why does a measurement on one qubit force another one into a given state in Simon's algorithm?
Clarification : I mean that the observation could have collapsed it to may be any of the vectors in this $2^{2n}$ dimensional vector space. But then why are these $\psi_b$s special? Is this observation by an operator such that $\psi_b$s are its eigenstates? Then that would make sense.
Jul
11
comment Why does a measurement on one qubit force another one into a given state in Simon's algorithm?
Seems my question isn't clear. This particular $\psi_b$ in my question is not the only state in the Hilbert space of these two qubits. If I understand the notation correctly then the first qubit can be in any of the $|i>$ states of which there are $2^n$ of them and the second qubit can be in any of the $|x_i>$ states of which there are $2^n$ of them. So there are $2^{2n}$ states available in this Hilbert space of these 2 qubits. Then why do we think that states of only this $\psi_b$ type will be measured?
Jul
11
revised Why does a measurement on one qubit force another one into a given state in Simon's algorithm?
deleted 108 characters in body
Jul
11
asked Why does a measurement on one qubit force another one into a given state in Simon's algorithm?
May
12
awarded  Popular Question
Apr
22
awarded  Popular Question
Mar
1
awarded  Autobiographer
Dec
22
awarded  Yearling
Dec
10
awarded  Tumbleweed
Dec
3
asked A question about Ising model
Oct
12
awarded  Nice Question
Oct
2
asked What is the future of complexity theory in black-hole physics and string theory?