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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? Cheers. |
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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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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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