# A proton is trapped in an infinitely deep well of L meters. The proton is in the first excited state. How does the excited state change the question.

The real question was: A proton is trapped in an infinitely deep well of 1*10^-14m. I suppose that is unimportant as that should only help us decided the limits of our integration. What I'm worried about is the second part of the question. "The proton is in the first excited state." Does "The proton being in the first excited state" have some effect on the wavefunction or the question? Or for any probability required should I just find the integral of the wavefunction of a free particle over what ever limits they ask for? For examples. Calculate the probability of finding the proton within 0.25*10^-14m of the left hand wall. Which would give us an integral between say (0... and 0.25*...) or to be more specific (0.25*... and 1*10...). The question also asked for the wavelength of the photon emitted if the proton returns to the ground state. [these are past paper questions I'm working on in preparation for a final so I have no solutions or help *:(*]

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There is something missing with your units and description: what is the potential? gravitational? electric? the 10^-14 m is a width? square well? –  anna v Nov 28 '11 at 5:46

Does "The proton being in the first excited state" have some effect on the wavefunction or the question?

Yes, it does. The wavefunction of the first-excited state is different from the ground state and all the other, higher, excited states. (See Ron Maimon's answer or search the web by "particle in a box".) The difference of the wavefunctions makes difference of properties between the states, such as the probability density of finding the proton at a given position in the box.

For example, if the proton is in the ground state, the probability density to find it at the center of the box is non-zero (actually it is maximum over all the possible position in the box), whereas if the proton is in the first-excited state, the probability density to find it at the center of the box is exactly zero.

As another point, the energy eigenvalue of the first-excited state is different from the ground state and all the other, higher, excited states. The energy of the emitted or absorbed photon depends on the energy difference of the initial and final states of the optical transition. Therefore, whether the proton is in the first-excited state or another higher-excited state makes difference in the photon energy when a photon is emitted by a transition to the ground state.

By the way, given two states, the transition between these two states may be either allowed or prohibited depending on the number of photons emitted/absorbed and depending on the transition is by the dipole/polarizability/hyperpolarizability/.. interaction or quadrupole interaction, etc., with the optical field. A rule whether a transition is allowed or prohibited is generally called a selection rule. This has to do with the symmetries of the interaction as well as the states involved.

In summary, it is not enough to know what system (e.g., a proton in an infinite well) is at hand, but necessary to know what state (e.g., the first-excited state) or a superposition of states (i.e., a wavepacket) the system is in, in order to make any prediction. (Let me try to make a (vague) analogy in classical mechanics. It is not enough to know that a stone of mass 1 kg is flying in a free space, but I would like to know in which direction, at what speed, and at what position it is flying in order to calculate if it is hitting me and to decide if I should call my mama.)

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The eigenstates of Schrodinger's equation in an infinite potential well going from x=0 to x=L are

$$\psi_n(x) = \sin(\pi n x/L)$$

The ground state is n=1, the first excited state is n=2. If you draw the n=2 sine wave, it makes one full wave in the region 0 to L, and you are asked what is the probability of being found in the first 1/4 of the full width.

You can integrate $\psi^2$ to get the answer, but there is a shortcut: the square of the sine wave makes two full waves between 0 and L, so it is fully symmetric between the four intervals (0,L/4)(L/4,L/2)(L/2,3L/4)(3L/4,L), using translations and reflections. So the probability of being in any one of the four intervals is equal, so that the probability is 1/4.

This is not a good problem, because you might have guessed this answer for bad reasons, like thinking the probability is uniform.

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What possible flaw could there be in this obvious answer? –  Ron Maimon Nov 28 '11 at 16:33
None that I can see. Looks completely correct. –  Colin K Nov 29 '11 at 0:31