# If measurement causes entanglement with the observer, is any further measurement possible?

When an observer measures a system, the systems wavefunction collapses and they become entangled. Does this mean that any further measurement by the observer on the system is impossible, as the system no longer has a wavefunction that is independent of the observer to collapse?

"When an observer measures a system, the systems wavefunction collapses and they become entangled."

That's not quite correct. When systems interact, they become entangled. They remain entangled even if separated. When an observer observes one of the entangled systems, the joint wavefunction of both collapses, 'instantaneously', faster-than-light, in some vaguely-defined and unobservable sense. They are no longer in an entangled state - they are each in a pure state - the pure states they each collapse to are correlated, but now in an ordinary, classical, non-spooky manner. For interpretations of quantum mechanics that posit wavefunction collapse, that is.

[For other interpretations, like the Everett interpretation, systems that interact become entangled. They remain entangled even when separated. When an observer observes one part, the observer becomes entangled with both but no wavefunction collapse happens. Nothing travels faster than light. Nothing happens to the other unobserved part of the system. When it too is observed and the observers compare notes, the observers end up entangled in such a way that each only perceives a partner making the consistent, correlated observation.]

Either way, they're entangled first, before the measurement, which is why the outcomes are correlated.

"Does this mean that any further measurement by the observer on the system is impossible, as the system no longer has a wavefunction that is independent of the observer to collapse?"

In either type of interpretation, further measurements can certainly be made. If the observer makes exactly the same sort of measurement (same observable), and does so fast enough that the state hasn't evolved, she gets the same result. (Collapse the second time around has no effect - the first collapse left it in a pure state for which the measurement gives the same outcome with 100% probability.) If the observer makes a different measurement but for a commuting observable, this gives a different result but the pure eigenstate is undisturbed. Going back to the first observable gives the same result you got originally. If the observer instead makes a complementary, non-commuting sort of measurement (e.g. momentum instead of position) this tends to scramble the entanglement, and then another measurement of the first sort gives a new random outcome.

• Just for completeness, it may be worth noting that this is only true for continuous measurements or measurements where the outcomes are good quantum numbers. Making a measurment with a commuting observable (if this is what you mean by same sort of measurement) could lead to a different outcome, while retaining the same state. Time evolution between measurements can also make a difference if the outcomes are not good quantum numbers. Commented Oct 1, 2022 at 23:50
• Fair point on completeness. I had thought about adding more caveats when I was writing it, but wanted to keep it simple - trying to match the level of understanding of the questioner, and not get diverted onto complications not related to the question. But it would probably be helpful. I have edited. Commented Oct 2, 2022 at 0:08
• Instead of saying nothing is traveling faster than light, wouldn't it be better to say that nothing is traveling at all (at any speed)? (no information at least) Commented Oct 2, 2022 at 0:58
• While it is true that nothing is travelling at all, that there is nothing travelling faster than light makes a point about the pros/cons of the two interpretations, which is what I was really getting at. Implying FTL and backwards-in-time effects would normally be enough to kill a hypothesis, if there is a viable non-FTL explanation available. (That wavefunction collapse remains popular despite this disadvantage I find interesting.) Essentially, I was trying to say 'but there is no collapse' without saying it directly, because it would be off-topic for the Q and metaphysically controversial. Commented Oct 2, 2022 at 1:25

When an observer measures a system, the systems wavefunction collapses... Does this mean that any further measurement by the observer on the system is impossible...

No. You can keep measuring at much as you want. (Please see my update below. I am discussing measurement of a pure state. This is textbook quantum mechanics.)

The "collapse" means that the wavefunction is projected onto the eigenstates of the measured operator that correspond to the measured eigenvalue. The "collapsed" state is still a wavefunction, just a different wavefunction.

For example, if you measure that the position is a $$\vec x$$, the wavefunction "collapses" to state that is typically denoted by $$|\vec x\rangle$$. This ket is still a wavefunction and its further evolution is still governed by whatever Hamiltonian/Schrodinger equation previously governed the evolution.

Update

To clarify, I am referring to measurement of a pure state, which is a well-known concept in Quantum Mechanics. See Volume 1, Chapter V, Section 13 of Messiah's Quantum Mechanics textbook, an excerpt of which is summarized below.

Supposed the wavefunction is given by: $$\Psi = \sum_p\Psi_p\;,$$ where $$\Psi_p = \sum_r c_{pr}\phi_{pr}\;$$

Here, $$p$$ means the part of the wave function that corresponds to the eigenvalue $$a_p$$ of the observable $$A$$.

The probability to measure $$a_i$$ is: $$\sum_r |c_{ir}|^2\;.$$

After an ideal measurement, if the value $$a_i$$ is measured the wavefunction "collapses" to: $$\Psi_i = \sum_r c_{ir}\phi_{ir}\;.$$

This is really not debatable. So, I'm not sure why this is downvoted...

If the measurements is not "ideal" the wavefunction is still projected onto the same subspace, it is just that the coefficients $$c_{ir}$$ can be modified. The wavefunction still lives in the projected subspace after measurement: $$\Psi \to \Psi' = \sum_{r}b_r \phi_{ir}$$

• We can only measure a quantum exactly once. That's one of the main differences between quantum mechanics and classical mechanics. "The system" is gone after we have performed the measurement on it. Quantum mechanics is an ensemble theory. A wave function doesn't predict the behavior of the individual system. It only predicts the average behavior of the ensemble. Commented Oct 2, 2022 at 9:05
• @FlatterMann I'm talking about ideal measurement. It is well know that the ideal measurement projects onto the portion of the wave function corresponding to the measured eigenvalue.
– hft
Commented Oct 2, 2022 at 19:18
• @FlatterMann Measuring a pure state is well know. In the ideal case it projects directly onto the subspace. In the non-ideal case it still projects onto the subspace, but the coefficients of the vectors in the subspace can change... This is textbook quantum mechanics, so I'm not sure why this is downvoted...
– hft
Commented Oct 2, 2022 at 19:28
• An ideal measurement is an irreversible energy transfer and it destroys the quantum system. A non-ideal measurement simply gives the wrong number for the total transferred energy, momentum, angular momentum and charge in addition. What you are citing is von Neumann's mathematics (which is not wrong) to describe the quantum mechanical ensemble, but it does not explain the physics that is actually going on to get there. I don't know if von Neumann actually understood the physics behind quantum mechanics. Sometimes I have my doubts. Commented Oct 2, 2022 at 20:19
• @FlatterMann Ok, I guess you know better than Johnny von Neumann.
– hft
Commented Oct 2, 2022 at 20:44