In http://en.wikipedia.org/wiki/Measurement_in_quantum_mechanics#Degenerate_spectra, it is said that

If there are multiple eigenstates with the same eigenvalue (called degeneracies),..., The probability of measuring a particular eigenvalue is the squared component of the state vector in the corresponding eigenspace, and the new state after measurement is the projection of the original state vector into the appropriate eigenspace.

My question: Is the state vector after measurement when the eigenspace is degenerated a pure state or mixed state? And what is the mathematical formulation of the mentioned "projection" onto the eigenspace?


1) Let there be given a Hilbert space $H$ and a mixed state described by a density operator $\hat{\rho}:H\to H$, which is a positive operator $\hat{\rho}\geq 0$, and with trace ${\rm Tr}(\hat{\rho})=1$.

2) Let $V\subseteq H$ be an eigenspace of states for an Hermitian observable $\hat{A}:H\to H$ with eigenvalue $\lambda\in\mathbb{R}$. (We will ignore subtleties with unbounded operators, domains, selfadjoint extensions, etc., in this answer.)

3) Let $\hat{P}:H\to H$ be the projection operator onto $V$. It is the unique operator, such that

  1. $\hat{P}$ is Hermitian $\hat{P}^{\dagger}=\hat{P}$ (and hence $\hat{P}$ is diagonalizable in an orthonormal basis).
  2. An idempotent $\hat{P}^2=\hat{P}$ (and hence can only have eigenvalues $0$ and $1$).
  3. The eigenspace ${\rm ker}(\hat{P}-1)$ for $\hat{P}$ with eigenvalue $1$ is equal to the subspace $V$.

If $(\mid\psi_i\rangle)_{i\in I}$ is an orthonormal basis for $V$, then

$$\hat{P}=\sum_{i\in I}\mid\psi_i\rangle\langle\psi_i\mid. $$

4) Then the collapse $\hat{\rho}\longrightarrow \hat{\rho}^{\prime}$ of the density operator, due to the measurement, would be

$$ \hat{\rho}^{\prime}~=~ \frac{\hat{P}\hat{\rho}\hat{P}}{{\rm Tr}(\hat{P}\hat{\rho})}. $$

5) For a pure state $\hat{\rho}=\mid\psi\rangle\langle\psi\mid $, the collapse $\mid\psi\rangle\longrightarrow\mid\psi\rangle^{\prime}$becomes

$$ \mid\psi\rangle^{\prime}~=~\frac{\hat{P}\mid\psi\rangle}{\sqrt{\langle\psi\mid\hat{P}\mid\psi\rangle}}.$$

So if one starts from a pure state $\mid\psi\rangle$, the collapsed state $\mid\psi\rangle^{\prime}$ would also be pure.

  • $\begingroup$ But what is the exact form of that projection operator? $\endgroup$ – Siyuan Ren Oct 16 '11 at 15:55
  • $\begingroup$ @Karsus Ren: Explicit form of $\hat{P}$ added in v3. $\endgroup$ – Qmechanic Oct 16 '11 at 17:27

The exact projection which a measurment is doing depends on the details of the measurement process. An idealized position measurement is an impossible thing, because it would project to a state of infinitely indefinite momentum, and infinite energy.

Any interaction of a quantum particle with another quantum system which starts with a product state, where the particle is in the state $\psi$ and the system is in the state $\chi$ produces an entangled state, where different states of the particle are entangled with different states of the system. We say that the system is "measuring" the particle when some of the states of the system leave a classical imprint. If the measuring device stops interacting with the particle, the entangled state of the particle looks like a density matrix, and the particle will be left in the relative state to the appropriate state of the measuring device, as projected by the outcome.

The relative state concept is due to Hugh Everett, and it is central to the many-worlds interpretation. Leaving interpretation aside, it is the central tool for describing non-idealized notions of measurement in quantum mechanics.

So if you scatter a photon of wavelength $\lambda$ off a particle, you get different scattering angles at different positions. The scattering angle is entangled with the particle position. If you then let the photon get absorbed by a photosensitive device, the position of the photon reveals which particular state of the particle is present. The description of a particle when it is entangled is no longer by a wavefunction, but by a density matrix, and this continues to be true unless the outgoing photon comes back to the particle to undo the entanglement. When the measurement becomes macroscopic, this is just not going to happen, because the photon has entangled itself with everything else.

  • $\begingroup$ Nice explanation. $\endgroup$ – Mike Dunlavey Oct 16 '11 at 14:38
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    $\begingroup$ I don't see how this actually answers the question, though... $\endgroup$ – David Z Oct 16 '11 at 17:16
  • $\begingroup$ @David: the question is what is the projection operator you apply during a measurement. This projection operator is the same as the one that extracts the pure state of the particle entangled with the outgoing photon state, in the case of a photon measurement. If you make an ideal measurement on the photon, you get a non-ideal projection for the particle. You have to go far enough up on the chain of interactions so that the ideal measurement is a perfect approximation in order for the formalism of Copenhagen quantum mechanics (ideal measurements) to apply. $\endgroup$ – Ron Maimon Oct 16 '11 at 19:22

Suppose you are in the state

$$|\Psi \rangle = a|\alpha_1\rangle + b|\alpha_2\rangle + c|\beta \rangle$$

$|\alpha_1\rangle$ and $|\alpha_2\rangle$ are eigenvectors of the observable $A$, both with eigenvalue $\alpha$. $|\beta\rangle$ is also an eigenvector of this observable but with the different eigenvalue $\beta$.

If you make a measurement of $A$ and the result is $\alpha$, the state of the system becomes the state

$$N\left(a|\alpha_1\rangle + b|\alpha_2\rangle\right)$$

with $N$ a factor chosen to keep the state normalized.

If you do not know the state of $|\Psi\rangle$ before making the measurement, then you make the measurement and get the value $\alpha$, you don't know the wavefunction after the measurement. Assuming the three states $|\alpha_1\rangle, |\alpha_2\rangle, |\beta\rangle$ form a basis, all you can say is that after the measurement

$$|\Psi\rangle = p|\alpha_1\rangle + q|\alpha_2\rangle$$

with $p^*p + q^*q = 1$, but you don't know what $p$ and $q$ are. If you want to determine them, you should find some other operator $B$ which commutes with $A$, so that $|\alpha_1\rangle$ and $|\alpha_2\rangle$ are eigenvectors of $B$, but have different eigenvalues. Then measuring $B$ will allow you to determine the state completely.

  • 2
    $\begingroup$ a|α1⟩+b|α2⟩ is also a pure state. Mixed state is what only density matrix can represent. $\endgroup$ – Siyuan Ren Oct 16 '11 at 10:09
  • $\begingroup$ @Karsus Okay. I guess I was unsure the definition. $\endgroup$ – Mark Eichenlaub Oct 16 '11 at 13:39

The projection operator on to a subspace $E$ spanned by a set of orthogonal states $\{|e_k\rangle\}$ is

$$P_E = \sum_k \frac{|e_k\rangle\langle e_k|}{\langle e_k|e_k\rangle}$$

That was the easy part. But figuring out whether you have a pure state or a mixed state after a measurement is, in some sense, an issue of definitions.

Consider a single quantum system which starts in a pure state $|\psi\rangle$ or $\rho$ (if you prefer density operators). Suppose you measure an observable corresponding to the operator $A$. As a Hermitian operator, $A$ has eigenstates $|a^i_k\rangle$, associated with eigenvalues $a^i$, which form an orthonormal basis. Let's say the result you get is $a^0$; then this particular measurement projects the state of the system on to the subspace $A_0$ spanned by the eigenstates $|a^0_k\rangle$:

$$\begin{align}|\psi\rangle &\to P_{A_0}|\psi\rangle & \rho &\to P_{A_0}\rho P_{A_0}\end{align}$$

Evidently if you're dealing with a single measurement on a single system, assuming you started with a pure state, you can express the post-measurement state as a pure state.

But what if you instead have a large ensemble of systems, each prepared identically and then measured? After the measurement, for each possible result $a^i$, a fraction $p(a^i)$ of the systems will have produced the result $a^i$ and thus will be in the state $P_{A_i}|\psi\rangle$ or $P_{A_i}\rho P_{A_i}$. This is a mixture of systems in different pure states, and thus it corresponds to a mixed state

$$\rho \to \sum_i p(a^i)P_{A_i}\rho P_{A_i}$$

The key difference is that with a single system, you were able to know the result of the measurement, which was the knowledge you needed to "extract" a single pure state from the mixture. But in this case, you can't do that because you have many different result of the measurement.

  • $\begingroup$ "This quantity is equal to one if and only if there is only a single term in the sum, with P(ah∣∣ψ)=1 and all the other probabilities zero." But the trace of ANY density matrix should be one. $\endgroup$ – Siyuan Ren Oct 16 '11 at 15:48
  • $\begingroup$ Ah, for some reason I remembered the pure-state criterion wrong. That's what I get for writing this up at 4 in the morning... I'll fix it later. $\endgroup$ – David Z Oct 16 '11 at 17:09

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