amjad-sh
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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 [itex]\psi(x,t)=c1\psi_{1}(x,t)+...+cn\psi_{n}(x,t)[/itex]
suppose that after measurement we find the particle at the energy eigenstate [itex]\psi_{1}(x,t)[/itex].
Now let's change the size of the box to 2L. Let's find the probability of the particle being in state [itex]\phi_{1}(x)[/itex] which is the ground state of the new box.The answer is [itex]|\int\phi^{*}_{1}(x)\psi_{1}(x)dx|^{2}[/itex],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 [itex]\phi_{1}(x)[/itex] 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 [itex]\psi(x,t)=c1\psi_{1}(x,t)+...+cn\psi_{n}(x,t)[/itex]
suppose that after measurement we find the particle at the energy eigenstate [itex]\psi_{1}(x,t)[/itex].
Now let's change the size of the box to 2L. Let's find the probability of the particle being in state [itex]\phi_{1}(x)[/itex] which is the ground state of the new box.The answer is [itex]|\int\phi^{*}_{1}(x)\psi_{1}(x)dx|^{2}[/itex],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 [itex]\phi_{1}(x)[/itex] 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 ?