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 1. Given that for an infinite square well problem, $\psi(x,0)=\frac{6}{a^3}x(a-x)$, I can show by Fourier transform that the probability of measuring $E_n$  for $n$ even is 0. But is there a physical reason for this? 

 2. The ground state, $E_1$ is most probable, but why is its probability $0.99855...\approx 1$? i.e. why sooo close?


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an even function wont go to zero at your boundary? – Nic Sep 1 '11 at 17:58
The eigenfunction for $E_n$ is $$u_n(x)=\sqrt{2/a} \sin (n\pi x /a)$$. So it vanishes for both even and odd functions. – justcurious Sep 1 '11 at 18:10
up vote 6 down vote accepted

I suppose the walls of your well are located at $x=0$ and $x=a$. It is easier to shift the coordianate to $y=x-a/2$ so that the walls are located at $y=\pm a/2$. Your wave function is then $$ \psi(y)=\frac6{a^3}\Bigl(\frac{a^2}4-y^2\Bigr). $$ Since the wave function is even in $y$, it is clear why the probability is zero for all even $n$, since the corresponding stationary state wave functions are odd. (It is generally true that if the potential is an even function, the stationary states have alternating parity: the ground state is even, the first excited state is odd etc.) This answers your first question.

As to the second question, the probability of measuring a particular value of energy is in general given by the overlap of your $\psi$ with the stationary state. Here it is so close to one simply because you have constructed a very good approximation to the ground state wave function. It is even in $y$, has no zeros apart from those at $y=\pm a/2$ (like the ground state and unlike all the excited states), and even the shapes are similar: a sine versus a parabola.

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Thanks a lot, Tomas. – justcurious Sep 1 '11 at 18:33

To add to Tomas' answer, I will point out another reason the probability of the ground state is very high, which is that the probability depends on the square of the wavefunction.

Suppose a wavefunction is mostly $\mid 0 \rangle$ and a little bit $\mid 2 \rangle$ (these represent the ground state and second excited state respectively), so

$$\mid \Psi \rangle = \alpha \mid 0 \rangle + \epsilon \mid 2 \rangle$$

Then the probabilities to measure the energies $E_0$ and $E_2$ are

$$P(E_0) = \alpha^2$$

$$P(E_2) = \epsilon^2$$

If $\epsilon$ is small, then $\epsilon^2$ is much smaller. In your example, we have $\alpha^2 = .9985$, so $\epsilon^2 = .0015$ and $\epsilon = .04$. So the probability to be in the second excited state is very small, even though the contribution of the second excited state to the wavefunction itself is more considerable.

Note: there are actually more states involved than just these two, but I chose them alone for simplicity.

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Thanks for completing my answer :) If one wanted to be really rigorous, one could as well just do the calculation. The probability of finding the k-th even excited state (so k=0 for the ground state) is given by 960/[(2k+1)pi]^6 which drops really fast with k. In particular, for k=1 one gets 0.001370 so the contribution of the higher excited states is truly negligible. – Tomáš Brauner Sep 2 '11 at 7:56
Thanks guys! :) – justcurious Sep 2 '11 at 8:03

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