- #1
snoopies622
- 840
- 28
A question from a beginner:
If the position operator changes a state vector into a wave function, how does one change a wave function back into a state vector? I have read that the position operator for a one-dimensional vector is simply multiplying by x. Does this mean that what amounts to the inverse position operator (or position inverse operator, perhaps) is putting a
[tex] \frac {1}{x} [/tex]
in front of the wave function? For example, a particle in a one-dimensional "box" of length L has a wave function
[tex]
\psi _n = \sqrt { \frac {2}{L} } sin ( \frac {n \pi x } {L })
[/tex]
Is the corresponding ket
[tex]
| \psi _n \rangle = \frac {1}{x} \sqrt { \frac {2}{L} } sin ( \frac {n \pi x } {L })
[/tex]
This seems logical but something tells me that it is too easy to be correct.
If the position operator changes a state vector into a wave function, how does one change a wave function back into a state vector? I have read that the position operator for a one-dimensional vector is simply multiplying by x. Does this mean that what amounts to the inverse position operator (or position inverse operator, perhaps) is putting a
[tex] \frac {1}{x} [/tex]
in front of the wave function? For example, a particle in a one-dimensional "box" of length L has a wave function
[tex]
\psi _n = \sqrt { \frac {2}{L} } sin ( \frac {n \pi x } {L })
[/tex]
Is the corresponding ket
[tex]
| \psi _n \rangle = \frac {1}{x} \sqrt { \frac {2}{L} } sin ( \frac {n \pi x } {L })
[/tex]
This seems logical but something tells me that it is too easy to be correct.