I believe my Lecturer and the textbook have contradicted one another. My lecturer gave the example that if the spatial part of the wavefunction of a particle is given by

$\psi(x) = c_1\psi_1(x) + c_2\psi_2(x)$

for the infinite square well potential (where $\psi_1$ and $\psi_2$ are the ground and first excited energy eigenfunctions). He stated that if we were to measure an observable, for example the energy of the particle, the wave function will collapse to one of the two energy eigenstates.

Where as in Griffiths the following is stated:

enter image description here

Which one is correct? I have no reasoning for which one I think could be correct, so I can't figure it out...


2 Answers 2


Both are correct, actually. If you measure an observable for that wave function you'll either find the eigenvalue corresponding to state 1 with probability $|c_1|^2$ (similarly for state 2), subject to the condition $|c_1|^2 + |c_2|^2 = 1$.

Edit: What Griffiths is saying is that before you perform the measurement, the particle is neither in state 1 or 2, but in a quantum superposition. Only the act of measurement forces the wavefunction to collapse to a particular state (at least according to the orthodox interpretation of quantum physics).

  • $\begingroup$ I'm still not convinced, Griffiths states "the particle is in state $\Psi$, not $\Psi_n$". Contradicting what my lecturer stated, that it collapses to $\Psi_n$ $\endgroup$
    – charl1e
    Apr 21, 2016 at 8:56
  • 1
    $\begingroup$ The particle is in state Psi before the measurement. The collapse to Psi_n (e.g., Psi_1) happens after a measurement takes place. $\endgroup$
    – Zelador
    Apr 21, 2016 at 8:58
  • $\begingroup$ The word collapse invariably means a measurement has taken place, therefore the original quantum superposition has been lost. $\endgroup$
    – Zelador
    Apr 21, 2016 at 9:01
  • $\begingroup$ The confusion comes about, possibly, since the probability of finding Psi_n is |c_n|^2 $\endgroup$
    – Zelador
    Apr 21, 2016 at 9:02
  • $\begingroup$ Yep I understand now, thanks so much for taking the time to answer! :) $\endgroup$
    – charl1e
    Apr 21, 2016 at 11:18

It sounds like both Griffiths and your lecturer are trying to come up with a mechanical rule for what you should say about states and measurement results instead of an explanation. I could be wrong about that since I have only your report to go on, but the appropriate explanation follows.

Say you have a state $\psi = a\phi_1 + b\phi_2$, where $\phi_1,\phi_2$ are eigenstates of an observable $\hat{A}$. A measurement of $\hat{A}$ transfers information about $\hat{A}$ from the system being measured to a measuring instrument in some ready state $\alpha_0$. An interaction that does this would do the following $\phi_j\alpha_0 \to \phi_j\alpha_j$. The joint state of the measuring apparatus and system would then be $$a\phi_1\alpha_1 + b\phi_2\alpha_2.$$

After the measurement the measurement device is present in two different versions: one for each possible outcome. These two versions will be unable to undergo interference as a result of decoherence:


As a result, each version of the measurement instrument only sees one result. We can say that the relative states of the measurement instrument and particle are $\phi_1\alpha_1$ for one version and $\phi_2\alpha_2$ for the other. We can also say that in each of those relative states the observable has the corresponding eigenvalue. This is commonly taken to mean you can then ignore quantum mechanics, but decoherent systems can carry quantum information, so this assumption is wrong, see




In a particular relative state, you can say the state of the system is $\phi_1$ say and that the measurement of the observable had a particular outcome. Saying that the measurement had a particular outcome or the system has a particular state outside that context is wrong. The lecturer and the textbook author are both wrong if they told you anything else.

  • $\begingroup$ "After the measurement the measurement device is present in two different versions" sounds like the Many Worlds Interpretation $\endgroup$ Oct 23, 2020 at 21:43
  • $\begingroup$ The MWI is what you get when you take QM equations of motion seriously as a description of how the world works. $\endgroup$
    – alanf
    Oct 25, 2020 at 7:11

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