# Can making a measurement change subsequent measurements?

1. Jan 21, 2012

### andrewkirk

Consider an experiment in which a closed box is set up with a system inside that is initialised at time t=0 with state psi(0). A measurement of quantity w of the box’s contents is made at time T, that is observed by scientist S1. There is another scientist S2 that can, by pressing a button, cause a measurement to be recorded of quantity q of the box’s contents at time T/2. q may or may not be the same as w, eg they may be the same variable, they may be non-commuting variables, or they may be commuting variables. S1 does not know whether S2 has pressed the button, and S2 does not look at the recorded measurement of q prior to when S1 looks at the measurement of w.

Say this experiment is conducted 200 times, with S2 pressing the button exactly half of those times, but S1 doesn’t know on which occasions the button was pressed. This gives us 100 trials in which w was measured at time T in a system at which q had been measured at time T/2 - call these ‘pre-measured trials’, and 100 trials that are ‘non-pre-measured’.

I have the following questions:
1. Does S2 pressing the button change the probability distribution of the measurement of w? ie does the difference between the distributions of w results from the 100 pre-measured trials and those from the 100 non-pre-measured trials have a nonzero expected value for any w?
2. Would the answer to 1 depend on whether S2 ever looks at the results of her q measurements (ie was the value recorded at T/2 a measurement that caused the wavefunction to collapse, if nobody looked at it?), or when she looks at them?
3. Does it depend on what the variables w and q are? eg what if they are
(i) the same variable
(ii) non-commuting variables such as position and momentum
(iii) commuting variables such as spin and position

What I’m trying to understand here is whether wavefunction collapse affects subsequent measurements. I would have thought that it would do so.
Also I’m wondering what makes something count as a measurement, in particular, whether recording a value with a recording device but never looking at it constitutes a measurement that will make the wavefunction collapse.

2. Jan 21, 2012

### San K

Not sure what you are exactly asking....however

- Wave function collapse would not affect the subsequent set of measurements..i.e. on the, say next, photon that follows after

however

- you can make a wave function collapse and then bring it back for the same photon, if I remember correctly

For example in a delayed choice quantum erasure ..there is more to be said on this...

- recording a value makes the wave function collapse whether we look at it or not, looking at an experiment (human consciousness) has nothing to do with wave function collapse

3. Jan 21, 2012

### jewbinson

andrewkirk, I read through the OP once. Before I read through it again, can i just say you have a lovely imagination. But as San k said, to "measure" a property of an electron, you must "interfere" with the electron. You can't "peek round the corner using a curved mirror" to see what the electron is doing. That just doesn't work. To see what state the electron is in you must "touch it directly" (which is what "measurement" means) with something. This "touching" is the only way to "measure" the property of the electron that you want to find. Moreover, the "touching" changes the state of the electron.

So really, if I want to know the spin of one of the electrons right in front of me RIGHT NOW then I can't. Spin is actually one of the best examples of this. So lets see what happens.

You choose an electron and you measure its spin.
You find that it spins up (spin +1/2). Wow! Surely that means that at the time just before the maesurement the spin was up.
Actually, no.
The spin of the electron before the measurement is not known and it is impossible to determine it.
It is certain that the electron was either in the spin up state or the spin down state just before the measurement was made. But which one? It is impossible to know.
This has been shown experimentally

4. Jan 21, 2012

### jewbinson

What do you mean by this?

A wavefunction has a set of possible results if you measure one of it's properties (e.g. spin or position).
If a measurement is made, then we get our results.

If you measure the position of an electron then you will find the position.

Okay, well actually you will not measure the exact position of the electron, as this is impossible. However you will know the eigenstate of the electron and this will tell you roughly where the electron is (in fact it will tell you where the electron can be and what the probabilities are of it being there at each small volume).

So once you take a measurement, the wavefucntion collapses into an eigenstate, and the eigenstate contains all the information you can possibly know about the electron. So for position, taking a measurement tells you where an electron is to a greater precision than before the measurement.

5. Jan 22, 2012

### maverick_starstrider

Your notion of measurement causing interference smells of the Heisenberg microscope which is a very incorrect interpretation of quantum mechanics. Particles do NOT have set positions and momentum if only we could find a way to get at it without imparting some energy to them. This is classical statistical mechanics, and it makes very different predictions than quantum. Quantum mechanics cannot be a local hidden variable theory, it's not that WE can't measure the value without "interfering" it's that the UNIVERSE hasn't assigned them a set value (simultaneously).

6. Jan 22, 2012

### andrewkirk

I'm just trying to get an understanding of the significance of wavefunction collapse. I understand what it means mathematically - that when a quantity w is measured for a system, the system's state instantaneously changes to an eigenket of the operator W corresponding to the quantity w, and then evolves from that point on according to the Schrodinger equation, until the next measurement. Prima facie, it sounds like the sudden change of system state would alter the probability distribution for a subsequent measurement, to be different from what it would have been if the first measurement had not been done. But I don't have a good enough feel for the implications of the Schrodinger equation to understand whether that is in fact the case, or whether the subsequent system evolution rapidly 'washes away' the impact of the measurement.

7. Jan 22, 2012

### The_Duck

Yes. Take the explicit example of a spin 1/2 particle like an electron. If the initial state is prepared to be spin-up in the z-direction, then measuring the z-component of spin will always give up. However, if an intermediate measurement of the x-component of spin is conducted, and then the z-component is measured, the z-component of spin will be found to be up or down with equal probability.

No. Human consciousness has nothing to do with the idea of "measurement" in QM.

If the two measured observables commute then the intermediate measurement does not affect the final measurement. Everything commutes with itself so that covers (i) too.

Yes; the fact that measurements affect subsequent measurements is the reason we need the idea of "collapse."

To conduct a measurement you cause some external thing to be correlated with the variable you are trying to measure. For example you might cause the position of a instrument pointer to become correlated with the z-component of the spin of an electron. Or you could cause a photon to be created if the spin is up and not if the spin is down, correlating the spin with the presence or absence of a photon. If you like you can cause your mental state to be correlated with the electron spin by for example looking at the instrument pointer. But your brain is not special, and quantum mechanics would work the same way if there were no humans around to look at things.

8. Jan 22, 2012

### jewbinson

Actually I think I was wrong before and have had a slight realization.

This is the main point.

If you act the observable A on the state |psi> then, say you get |phi>.

Using A again will not change the state, i.e. it will give you |phi>.

Now using B where [A,B] = 0 will not change the state, i.e. you will get |phi>.

If you now use C where [B,C] does not = 0, then we will (possible or definitely??) change the state.

So now write

DCBA|psi> and think about what can happen at each stage depending on whether or not each observable is compatible with another one.

9. Jan 27, 2012

### andrewkirk

Does 'external' here mean something that is not represented in the wavefunction of the system being studied?

Does that mean that, with Schrodinger's cat in a box, the cat's survival is a variable external to the quantum system of the particle that may or may not decay and release the poison gas, and hence the survival or nonsurvival of the cat constitutes a measurement, which implies that the wave function has collapsed, and the cat has either died or survived, before the box is opened by the scientists to look inside (rather than in a position of superposed life and death as some suggest)?