I am very new to quantum mechanics and I don't know how to tackle this problem :

A particle of mass $m$ is in an infinite, square quantum well of length $L$.

The particle is in the state described by $$\psi= \frac{1}{\sqrt{14}}(\psi_1 + 2\psi_2 + 3\psi_3)$$

With $$\psi_n(x) = \sqrt{\frac{2}{L}}\sin\left(\frac{n\pi x}{L}\right)$$

and I am to figure out the energy $E$ of $\psi$ as a multiple of $E_1$ among 2, 5, 9, 14 and 16.

I know that $$E_n = n²\frac{h²}{8mL²}$$

but I don't understand how to find the energy of the superposition state $\psi$

  • 1
    $\begingroup$ When you measure the energy of a superposition state you get different values, each with a certain frequency. Perchance were you asked for the expected value of the energy? That would be a weighted average, each valued weighted by the frequency it occurs. $\endgroup$ – Timaeus Sep 29 '15 at 18:37
  • $\begingroup$ No, actually the exact question is "what is a possible result for measuring the energy of state $\psi$" $\endgroup$ – mwa1 Sep 29 '15 at 18:49
  • $\begingroup$ What techniques has your text already covered? Projections onto eigenspaces? Probabilities? The Born rule? $\endgroup$ – Timaeus Sep 29 '15 at 18:51
  • 1
    $\begingroup$ I think I get it. The measured energy can be alternatively either E1 E2 or E3. since E3 is 9 times E1 that must be it, right ? $\endgroup$ – mwa1 Sep 29 '15 at 18:54

In the rules of quantum mechanics, every state $|\psi\rangle$ is a "vector" which has a "dual", which is usually a complex conjugate $\langle \psi|$ and every measurement in some state is described by an average $\langle A\rangle$ and an operator $\hat A$ which is its own conjugate transpose: together these say that in state $|\psi\rangle$ the average measurement is $\langle A\rangle = \langle \psi | \hat A |\psi\rangle.$

Here in general $\langle\phi|\psi\rangle = \int_{-\infty}^\infty dx~\phi^*(x)~\psi(x),$ if that helps you understand what we mean when we take the "transpose" of a function.

Linearity/superposition happens at this "expectation level" too, so if you have $$|\psi\rangle = a_1 |\psi_1\rangle + a_2 |\psi_2\rangle + a_3 |\psi_3\rangle$$ then the expected value of $\hat A$ is going to be:$$\begin{align} \langle \psi | \hat A |\psi\rangle =&~a_1^* a_1 \langle\psi_1|\hat A|\psi_1\rangle + a_1^* a_2 \langle\psi_1|\hat A|\psi_2\rangle + a_1^* a_3 \langle\psi_1|\hat A|\psi_3\rangle + \\ &~a_2^* a_1 \langle\psi_2|\hat A|\psi_1\rangle + a_2^* a_2 \langle\psi_2|\hat A|\psi_2\rangle + a_2^* a_3\langle\psi_2|\hat A|\psi_3\rangle + \\ &~a_3^* a_1\langle\psi_3|\hat A|\psi_1\rangle + a_3^* a_2\langle\psi_3|\hat A|\psi_2\rangle + a_3^* a_3\langle\psi_3|\hat A|\psi_3\rangle \end{align}$$ I hear you saying, "goodness gracious, do I really need to do nine integrals?!" And the answer in general is "no, because the operator is its own conjugate transpose, $\langle a|\hat A|b\rangle = (\langle b|\hat A|a\rangle)^*,$ so you only need to do 6 integrals," but the answer for you in particular is "No, you can get away with doing no integrals."

The reason for this is really particular, it is that you are looking for the average energy which is given by the Hamiltonian operator $\hat H$, but actually your basis states are normalized eigenvectors of the Hamiltonian, $\langle \psi_m|\psi_n\rangle = \{1 \text{ if }m = n\text{, else } 0\},$ with $\hat H|\psi_n\rangle = E_n |\psi_n\rangle.$

Using those properties tells you that the average energy in the state is simply $\langle E\rangle = \sum_n E_n |a_n|^2.$

  • $\begingroup$ That gives me : $E_1+4E_2+9E_3 = E_1 + 16E_1 + 81E_1 = 98E_1$. But 98 is not in my list... i'm confused $\endgroup$ – mwa1 Sep 29 '15 at 19:20
  • $\begingroup$ @mwa1 You need to divide by 14, check the normalization of the wavefunction. Also, it's better to focus on the physics and the math instead of some hints that don't seem to work. You're better off moving on to another problem once you've understood this one. There is no need to worry about "98". $\endgroup$ – Count Iblis Sep 29 '15 at 19:27
  • $\begingroup$ If I divide by 14 I get 7. Closer but still not in the list :) $\endgroup$ – mwa1 Sep 29 '15 at 19:31
  • 1
    $\begingroup$ @mwa1: That's the answer for the average energy, indeed. However, you mentioned in a comment that I had not seen that this was not the question, which was "what is a possible measurement for the energy", and the possible measurements for the energy are $n^2 E_1$ for $n=1,2,3,\dots.$ $\endgroup$ – CR Drost Sep 29 '15 at 19:41

Not the answer you're looking for? Browse other questions tagged or ask your own question.