- #1
TimeRip496
- 254
- 5
Let's start with a vacuum, I will write |0000...> as |0>.
$$ \psi^+ (x) |0\rangle = \sum_k e^{-ikx} a^+ (k) |0\rangle $$
Qns: What happens if you superpose/add up state vectors corresponding to a particle of momentum k and you add up with coefficient e-ikx?
$$ \sum_k e^{-ikx} a^+ (k) |0\rangle $$ where a+ is a creation operator
Ans: The ans is it is a position state. It is a state with a definite position. What is a position? x. If you add up momentum states with e-ikx, that gives you a position state located at particle x. This is one particle quantum mechanics.
Source: Starts at 40:21
I am lost at this part whereby how multiplying e-ikx with a+(k) gives the position state?
$$ \psi^+ (x) |0\rangle = \sum_k e^{-ikx} a^+ (k) |0\rangle $$
Qns: What happens if you superpose/add up state vectors corresponding to a particle of momentum k and you add up with coefficient e-ikx?
$$ \sum_k e^{-ikx} a^+ (k) |0\rangle $$ where a+ is a creation operator
Ans: The ans is it is a position state. It is a state with a definite position. What is a position? x. If you add up momentum states with e-ikx, that gives you a position state located at particle x. This is one particle quantum mechanics.
Source: Starts at 40:21
I am lost at this part whereby how multiplying e-ikx with a+(k) gives the position state?