Question 1: suppose there is a quantum observable superposition of 4 eigenstates $\lambda_1e_1 + \lambda_2e_2 + \lambda_3e_3 + \lambda_4e_4$. Does a 'measurement' have to reduce this wavefunction to a single eigenstate with probability $\lambda_i^2$, or can it confine the variable to 2 or more eigenstates EDIT with unique eigenvalues?

Question 2: if the latter is possible then for example if the measurement confines the eigenstate to $i,j$, then is the probability of this measurement equal to $\lambda_i^2 + \lambda_j^2$?. If this is true, then is the measurement of $i,j$ just defined as the probability that all the 'other' eigenstates are zero? Now supposing this is true, let's suppose we had an incompatible operator, the measurement of the state would collapse this incompatible operator to a combination of eigenstates. Does the same 'formula' applied to this new operator predict the exact same probability?

Question 3: if the latter is true does a measurement have to be binary, i.e. turning on $i,j$ and turning off the rest?

Question 4: what information or definition do the eigenstates upon measurement have a 1 to 1 correspondence with? is it to do with what information is possibly extractable from the measurement? If so, does the information about eigenstates extractable from a measurement have to be binary? Can it not be a probability distribution in of itself, for example in $(i,j)$?

Think about this problem in the context of a screen 'measuring' the absence of a moving particle

  • $\begingroup$ If two or more eigenstates have the same eigenvalue, the resulting state after measurement is the projection of the state onto that eigenspace. $\endgroup$
    – Charlie
    Mar 5, 2021 at 14:42
  • 2
    $\begingroup$ This seems like a homework-like question to me. Could you please elaborate on the context and what specific concept you are having troubles with? Thanks! $\endgroup$
    – jng224
    Mar 10, 2021 at 17:39

4 Answers 4


When we say 'measurement' of an operator, we mean an operation that by definition collapses the operator into one of its eigenstates with some probability (the probability being given by $|\langle e_i|\psi\rangle|^2$). But of course, there can be other operators that has, as its eigenstate with same eigenvalue, two or more eigenstates of your first operator. Let's proceed with an example:

Consider a quantum die with the state $|\psi\rangle = a_1 |1\rangle + a_2 |2\rangle+ a_3 |3\rangle+ a_4 |4\rangle+a_5 |5\rangle+a_6 |6\rangle$. Where the states $|i\rangle$ are eigenstates of the operator $N$ say.

Now you may define another operator M (which measures if the state of the die is even or odd) with the following eigenvalues:

$$M |even\rangle = +1 |even\rangle \\ M |odd\rangle = -1 |odd\rangle$$

The eigenstate $|even\rangle$ is degenerate in terms of the eigenstates of N as the states $|2\rangle, |4\rangle, |6\rangle$ gives the same eigenvalue. Now if we perform an experiment to see if the state of the die is even or odd, we will be measuring the operator M. The probability that we get $|even\rangle$ for the above state IS $(a_2^2 + a_4^2 + a_6^2)$.

Now the question is: What is the 'state' of the system after the measurement (where we got the answer 'even') in terms of the eigenstates of $N$? The answer is again straightforward: It has collapsed to the 'even' subset of states with their relative weights intact:

$$ |\psi\rangle_{Collapsed} = \frac{a_2}{\sqrt{a_2^2 + a_4^2 + a_6^2}}|2\rangle + \frac{a_4}{\sqrt{a_2^2 + a_4^2 + a_6^2}}|4\rangle + \frac{a_6}{\sqrt{a_2^2 + a_4^2 + a_6^2}}|6\rangle$$

What about an incompatible operator then? The answer is guessable from the above expression. If we have a new operator $O$ with eigenstates $O |O_i\rangle = o_i |O_i\rangle$, then the probabilities for outcome of 'measurement' of this operator is simply calculated by $|\langle o_i|\psi\rangle_{Collapsed}|^2$.

Thus the answer to Question 3 also automatically follows. When the measurement of the operator $M$ takes place, it turns on only $|2\rangle, |4\rangle, |6\rangle$ and turns others off or vice versa. Question 4 is now asked in a different way. Now we talk about 2 different operators. $N$ which was the operator that specified the initial eigenstates and $M$ which specified the ${|even\rangle,|odd\rangle}$ states. The relation between them is apparent.

  • $\begingroup$ The question is, if we previously had a hermitian operator for which each of these eigenstates had a unique eigenvalue, then how may we construct another hermitian operator for which there is now degeneracy in these states. How does this work in the context of for example, the position operator? $\endgroup$
    – user86425
    Mar 11, 2021 at 11:10
  • $\begingroup$ It works more or less the same way as I explained above. For instance, for position operators in 1D, you can define an operator that gives -1 if the particle is at x<0 and gives +1 if it's at x>0. This operator can be shown to be Hermitian. $\endgroup$
    – Ari
    Mar 11, 2021 at 17:51

The measurement of the variable $\Lambda$ changes the state vector, which is in general superposition of the form $$|\psi\rangle =\sum_\lambda|\lambda\rangle\langle \lambda|\psi\rangle$$

into the eigenstate $|\lambda\rangle$ corresponding to the eigenvalue $\lambda$ obtained in the measure.

Consider an example of a degenerate case $$|\psi\rangle =2^{-1}|\lambda,1\rangle +2^{-1}|\lambda,2\rangle +2^{-1/2}|\lambda_3\rangle$$

where we have assumed $\lambda_1=\lambda_2=\lambda$. Suppose the measurement gives a value $\lambda$, the normalized state after the measurement is known to use to be $$|\psi\rangle =2^{-1/2}(|\lambda,1\rangle +|\lambda,2\rangle)$$ If the initial state is unknown, and the measurement gives $\lambda$ then $$|\psi\rangle =\frac{\alpha|\lambda,1\rangle +\beta|\lambda,2\rangle}{\sqrt{\alpha^2+\beta^2}}$$


Consider it geometrically: the measurement projects the state to its (renormalized) projection in a subspace, with probability equal to the length of that projection (pre-renormalizaton).


Question 1: Eigenstates of which operator? The measurement operator has to give the same eigenvalues to two (orthogonal) states for it to be able to project the initial state to a two-dimensional subspace. After such a measurement result, the resulting vector in that subspace could then be resolved by another measurement operator into one of two states/measurement results.

Question 2: One would have a state like $a_1 |1, \lambda_1 \rangle+a_2 |2, \lambda_1 \rangle + a_3 |\lambda_2\rangle$. Then by Pythogoras' theorem, the length in the 2-dimensional subspace is indeed $|a_1|^2+|a_2|^2$, which is the probability of measuring $\lambda_1$. If this was the measurement result, then post-measurement there would be no component of the state in the third, orthogonal subspace, spanned by $|\lambda_2\rangle$, and the process of measurement can be considered as "erasing" that component from the vector (and then renormalizing).

If we now consider an incompatible operator, then the original state would have a different spectral decomposition under it, say $\tilde{a}_1 |1\rangle+\tilde{a}_2 |2\rangle+\tilde{a}_3 |3\rangle$. The same general formula (Born rule) for the probabilities applies for it, but of course the resulting probabilities would in general be different because the decomposition's amplitudes $\tilde{a}_i$ will be different.

Question 3: Yes, essentially. An ideal measurement always completely projects into a subspace, so that after it there is no component for the state in the other directions (corresponding to different measurement results).

Question 4: The measurement provides information that one eigenvalue was selected, rather than all the others. It does not constitute a probability distribution. One can of course construct an estimate of a probability distribution by repeated measurements, which encapsulates information about the amplitudes $a_i$ in the initial state.

I hope I helped, I found your phrasing sometimes difficult to follow.

  • $\begingroup$ could you reconsider the edited question? $\endgroup$
    – user86425
    Mar 6, 2021 at 1:52
  • $\begingroup$ No, an ideal measurement cannot confine the state to two vectors with different eigenvalues (in the measurement operator's eigenbasis). It can only project to a subspace, and the subspace will have the same eigenvalue for all vectors in it. $\endgroup$ Mar 6, 2021 at 6:08
  • $\begingroup$ i didn't downvote. but surely real measurements are not ideal? for example in continuous variables, the measurement of a particle cannot collapse the wavefunction to an exact dirac delta like function $\endgroup$
    – user86425
    Mar 6, 2021 at 11:01
  • $\begingroup$ Well, for continuous variables you can consider ideal measurements into finite regions, such as the resolution of your measurement device. But yes, real measurements probably aren't ideal ones. Nevertheless, measurements are generally treated as if they were ideal, or POVM (which is a slight generalization, that doesn't change the above). $\endgroup$ Mar 6, 2021 at 14:22
  • $\begingroup$ what about a screen measuring the absence of a particle? This collapse is nothing like an ideal measurement. What defines the 'actual resolution', surely not the experimenter's opinion? $\endgroup$
    – user86425
    Mar 6, 2021 at 14:24

Let me suggest an unverified answer to question 3 and 4 first, let's begin by defining measurement as a change in the wavefunction as it interacts with a classical environment. This change causes the eigenstates distribution to suddenly match the probability distribution that is defined in some sense by the amount of classical information one can deduce about the observable. (The answer to how or why this occurs is unclear). For most cases, the effect of the distribution in $i,j$ is so negligible that most of these 'collapses' can be modelled via projection operators. If these are used then the effect of measurement can indeed be seen as 'binary' operation, however there is a scaling up of eigenvalues in the projection space.

Now let's move on to question 1 and 2. The probability of 'getting eigenvalue m' is $p(m) = \langle \psi | P_m | \psi \rangle $, where $P_m$ is the projection operator for a single eigenvalue. We can extend this result to a projection onto a larger space, just as suggested in the question. I suspect that whatever operation this corresponds to in the incompatible operator space will produce the same result, however I cannot prove this.


Your Answer

By clicking “Post Your Answer”, you agree to our terms of service and acknowledge you have read our privacy policy.