- #1
amjad-sh
- 240
- 13
Hello
suppose that we are dealing with a particle moving in an infinite potential well(a box of length L).
Let the wavefunction of the particle be \psi(x,t)=c1\psi_{1}(x,t)+...+cn\psi_{n}(x,t)
suppose that after measurement we find the particle at the energy eigenstate \psi_{1}(x,t).
Now let's change the size of the box to 2L. Let's find the probability of the particle being in state \phi_{1}(x) which is the ground state of the new box.The answer is |\int\phi^{*}_{1}(x)\psi_{1}(x)dx|^{2},which may in many cases be not equal to zero.
My confusion is here: what if we didn't change the box and we computed the same integral above, which is the probability of the particle to be in state \phi_{1}(x) and it is a non allowed state, the probability of course will not be zero because it is the same integral above.
How the probability of the particle in being in a non allowed state can be not equal to zero ?
suppose that we are dealing with a particle moving in an infinite potential well(a box of length L).
Let the wavefunction of the particle be \psi(x,t)=c1\psi_{1}(x,t)+...+cn\psi_{n}(x,t)
suppose that after measurement we find the particle at the energy eigenstate \psi_{1}(x,t).
Now let's change the size of the box to 2L. Let's find the probability of the particle being in state \phi_{1}(x) which is the ground state of the new box.The answer is |\int\phi^{*}_{1}(x)\psi_{1}(x)dx|^{2},which may in many cases be not equal to zero.
My confusion is here: what if we didn't change the box and we computed the same integral above, which is the probability of the particle to be in state \phi_{1}(x) and it is a non allowed state, the probability of course will not be zero because it is the same integral above.
How the probability of the particle in being in a non allowed state can be not equal to zero ?
