0
$\begingroup$

Consider $E_1,E_2,E_3,E_4$ whose eigenfunctions are $\phi_1,\phi_2,\phi_3,\phi_4$ respectively and given $\Psi = \phi_1 - 2\phi_3 +\phi_4$, calculate the expected value of energy in this state.

I first normalise the wavefunction which is $$\Psi_N = \frac{1}{\sqrt{6}}\phi_1 - \frac{2}{\sqrt{6}}\phi_3 +\frac{1}{\sqrt{6}}\phi_4.$$

The expected value is given by: $$\langle E\rangle=\frac{1}{6}E_1+\frac{4}{6}E_3+\frac{1}{6}E_4.$$

However This is where I'm stuck as I unconfidently want to say: $$\langle E \rangle =\frac{1}{6}+3\,\frac{4}{6}+4\, \frac{1}{6} = \frac{17}{6}.$$

But this doesn't sit right with me as energy should be quantised and I therefore expected an integer back. If it isn't already clear I have only just started the QM module so please be blunt if I've got the entire method wrong!

$\endgroup$
5
  • 3
    $\begingroup$ Try to use \langle and \rangle instead of $<$ and $>$. Just to clarify: Is $E_1 =1$, $E_3 = 3$ and $E_4=4$? If so, the expectation value looks correct. Note that as energy should be quantised does not mean that the energy must be an integer. Take the hydrogen for example; there, the ground state energy is not an integer (in SI units). Indeed, the energy can also be a continuous quantity. Last but not least, even for a system with integer (and discrete) energy levels, for example a two state system, the expectation value of the energy is not expected to be an integer as well. $\endgroup$ Nov 4, 2021 at 18:09
  • $\begingroup$ @Jakob No we are not told $E_n = n$, this is probably a naive assumption I have made. The only information given is that which I have written at the top. I have a feeling I need to utilise the Quantum Harmonic or the Hamiltonian operator but I feel like thats overthinking it? $\endgroup$ Nov 4, 2021 at 18:16
  • 2
    $\begingroup$ If all you're told is that the energies are $E_1$ through $E_4$, then you stop at your second-to-last expression. That's the answer. You can only proceed farther if you are either told what they are, or you know that you have a specific system in mind (like an infinite square well, a harmonic oscillator, a hydrogen atom, etc.) $\endgroup$
    – march
    Nov 4, 2021 at 18:18
  • $\begingroup$ I of course do not know your assignment, but perhaps you just have to write your second equation without specifying the $E_n$; I mean, these could literally be anything without specifying the Hamiltonian / the system. $\endgroup$ Nov 4, 2021 at 18:19
  • 1
    $\begingroup$ @Jakob Thank you for explaining the quantisation it appears I was getting myself confused and made the silly assumption that my output will be an energy level ($n$) and so that is why I was expecting an integer. But you are correct in the fact they do not state what $E_n$ is, mearly that it's eigenfunction is $\phi_n$. And so I think the answer Is likely, as you said, the $\langle E \rangle = \frac{1}{6}E_1...$ equation. Many thanks for the help $\endgroup$ Nov 4, 2021 at 18:39

2 Answers 2

1
$\begingroup$

The energy of the individual eigenstates states is often an integer multiple or fraction of some basic quantity, but the average need not be. Quantization means the energies of eigenstates can only takes discrete values, not that they need be whole numbers.

For instance, if $E_2=2E_1$, and the states with energy $E_2$ and $E_1$ are equally probable, then surely the average is $(E_1+E_2)/2= 3E_1/2$.

As another example, spin is quantized so that $S_z$ only takes value $\pm \hbar$ for spin-1/2, but if spin-up and spin-down are equally probable, then average is $0$.

Here’s yet another example from the classical world. The numbers of the face of a die are all integers (and thus “quantized”), but the average roll will produce $21/6=7/2$, which not only isn’t an integer but not even a possible outcome.

$\endgroup$
0
$\begingroup$

You've calculated the expected value of the energy properly. Assuming $E_1=1$, $E_3=3$ and $E_4=4$ the last equation should be right.

However, it seems to me like you just assumed $E_n=n$, which is not necessarily right. For example look at the energy levels of the hydrogen atom: $E_n=\frac{E_0}{n^2}$.

The energy being quantized does not mean it has and integer value. It just means that the energy does not form a continuum. It takes up discrete values. Again look at the Hydrogen levels: \begin{equation} E_1=E_0=-13.6 eV\; , \;\; E_2=\frac{E_0}{4} \; , \;\;E_3=\frac{E_0}{9}, ... \end{equation} the energy takes discrete values but is not integer.

$\endgroup$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

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