# B Do we measure the particle or the wavefunction?

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1. Oct 30, 2016

### quantumfunction

It seems to me that we don't measure a particle because a particle doesn't have an objective existence independent of the wave function or does it? The wave function in this case would have to be real because you can't have probability without the underlying possibility of a specific outcome being real. So if I roll a dice, I may get a 1-6 but I can only get a 1-6 because these are real possibilities.

If a particles wave function is spread out in a box, of course the particle itself isn't everywhere in the box but the wave function is which carries the possible outcomes that can occur. We also know the wave function can exist without a particle. It's just evolving according to Schrodinger's equation until somone or something decides to disturb it.

So isn't it wrong to say we measure a particles x and shouldn't we say that you can measure the wave function which produces a particle which carries a possible quantum state of the wave function? Where am I going wrong?

2. Oct 30, 2016

### Staff: Mentor

Yes, we don't measure particles, we measure observables. See below.

No, because measurements can only give you real numbers, and the wave function is not a real number. It's a function with complex values.

Mathematically, to make a measurement on a quantum system with a particular wave function, we apply an operator to the wave function that corresponds to the measurement. This gives us a set of real numbers that correspond to the probabilities of all the different possible measurement outcomes. One of those outcomes will be the actual measurement result.

3. Oct 30, 2016

### zonde

You have to specify interpretation if you ask what's real in quantum mechanics.

4. Oct 30, 2016

### quantumfunction

Isn't part of the wave function real?

https://en.wikipedia.org/wiki/Complex_number

You said:

Doesn't these set of real numbers that correspond to probabilities of all different measurement outcomes, come from the wave function? If these probable states that can occur don't come from the wave function then where do they come from? When superposition occurs doesn't the wave function contain the information about each possible state the particle can be in?

It's like the dice. I can want to roll a 7 all day but it's not a possible outcome. So doesn't the wave function or something need to be real in order for these possible outcomes to be able to occur?

5. Oct 31, 2016

### quantumfunction

It's not really an interpretation. It's Unamendeed Quantum Mechanics, so no collapse and no particles, just the wave function and all it's different versions exist from the electron to Schrodinger's cat.

6. Oct 31, 2016

### Staff: Mentor

Sure, every complex number has a real part. But that doesn't make the wave function as a whole real.

Not just from the wave function. The numbers come from applying an operator to the wave function.

"Each possible state" depends on the operator; the wave function itself doesn't tell you which states are "possible states" when you make a particular measurement.

"Real" is not a scientific term. If you want to say that the wave function must be "real" for this reason, that's just your choice of words; it says nothing about the physics. The physics is what I said above.

7. Oct 31, 2016

### zonde

Then setup of experiment is real. What else is real depends on interpretation.

8. Oct 31, 2016

### quantumfunction

You said:

Wavefunction Properties

Image URL

http://hyperphysics.phy-astr.gsu.edu/hbase/imgmod2/wf2.gif

http://hyperphysics.phy-astr.gsu.edu/hbase/quantum/wvfun.html#c1

So the wave function contains measurable information. How can it not be real but contain measurable information. The wave function can transmit information without a particle as a medium.

The wave-function is real but nonphysical: A view from counterfactual quantum cryptography

https://arxiv.org/abs/1311.7127

You said:

I think real is a scientific term. I don't know any Scientist who wouldn't say the effects of Gravity isn't Real. We don't know if it's a fundamental force, an emergent property or if it comes from the entropy of entanglement but we do know it's a real effect.

Last edited: Oct 31, 2016
9. Oct 31, 2016

### Staff: Mentor

If you want to say that "real" means "contains measurable information", then that's your choice of words.

You can certainly choose to use "real" to mean something scientific. But saying something is "real" doesn't tell you what that scientific something is; different scientists use it to refer to different things. If I say "applying an operator to a wave function gives a set of real numbers which tell the probabilities of the possible measurement results", that is a definite scientific statement with testable content; saying "and that means the wave function is real" adds no new information, it's just a label you choose to put on it.

10. Oct 31, 2016

### quantumfunction

I didn't say the wave function was real in a vacuum. I said the wave function is real because of x,y and z. You can't debate the conclusion without the x, y and z that led to the conclusion.

The wave function contains all measurable information and establishes the probability distribution in three dimensions.

This is my point, the information has to be real in order for there to be possible outcomes. I can want to roll a 7 until I'm blue in the face but if the only possible outcomes that are real are 1,2,3,4,5 and 6 then I can't roll a 7. So the wave function contains real information about possible outcomes. I don't think you need anything but a wave function that contains this real information and no collapse or particles that mucks things up.

Last edited: Oct 31, 2016
11. Oct 31, 2016

### rubi

The wave function is just an improved version of a probability distribution. A probability distribution just stores the information about what is going to happen, but it doesn't itself "physically exist". It can nevertheless be objective. Here's an example: The probability distribution of a die is $p(n)=\frac{1}{6}$. This distribution just tells you how often you would expect $n$ if you throw a die a large number of times. The number $\frac{1}{6}$ doesn't have a "physical existence", but it is objective, since it best describes the experiment of throwing a die a large number of times. The situation is exactly analogous for the wave function. It's a just a specification of probability distributions. The difference between classical and quantum mechanics is that classical mechanics is deterministic and all probability distributions can in principle be derived from an underlying theory about things that physically exist. In quantum mechanics however, the world is not deterministic and there is a genuine element of randomness, so the best description you can have is a probability distribution. These probability distributions still just describe the relative frequencies of outcomes and they are still objective, since only one probability distribution will match the observed relative frequencies. But also, they still don't "exist physically", just like the probability distribution for a die doesn't "exist pysically".

(Disclaimer, so I don't have to bother with interpretations: Some people prefer alternative interpretations.)

12. Oct 31, 2016

### Staff: Mentor

Exactly. And for different situations, "real" will refer to a different x, y, and z. So why even use the term "real" at all? Why not just talk about the specific x, y, and z, since those are what contain the actual physics?

The wave function plus an operator contains information about possible outcomes. But without an operator, the wave function itself does not.

This fact is obscured in many sources because they always write down the wave function in a particular basis (for example, the position basis or the momentum basis). But choosing a basis is equivalent to picking an operator and applying it to the wave function. So when you talk about a wave function written in a particular basis, you are not just talking about the wave function; you are talking about the wave function plus the operator that defines that basis. And all the information about possible outcomes which you read off from the wave function written in that basis is really information from the wave function plus the operator.

To really understand what it means to have just the wave function without choosing a basis, the Dirac bra-ket notation is much better. In this notation, the term "state vector" is usually used, not "wave function", to make it clear that we are talking about a vector in an abstract space (called Hilbert space) that can be defined and manipulated mathematically without choosing any particular basis.

13. Oct 31, 2016

### A. Neumaier

It's rather the other way around. A wave function has no objective existence, whereas particles exist at least under certain conditions. For example, a single proton is a very real particle. We measure properties of single particles or collective properties of multi-particle systems.

14. Oct 31, 2016

### quantumfunction

Where did this probability distribution come from if the wave function isn't the underlying reality that contains all measurable information about the system. You can't have a probability distribution without the possible outcomes being real. So in this case, the wave function is like the dice that contains all measurable information that can occur. The dice contain the numbers 1-6 and the wave function contains all measurable information.

15. Oct 31, 2016

### quantumfunction

There's no need for particles. Waves are enough to explain particle like properties. Here's 2 good papers on this.

There are no particles, there are only fields

https://arxiv.org/ftp/arxiv/papers/1204/1204.4616.pdf

No Evidence for Particles

https://arxiv.org/ftp/arxiv/papers/0807/0807.3930.pdf

Like I said, all you need is the wave function to describe particle like properties.

16. Oct 31, 2016

### Staff: Mentor

It's not at all clear what you mean by a "real possibility", but if you mean "has a non-zero probability" then this statement is correct.

However, you wouldn't claim that this means that the axioms of probability theory are real solid tangible things that we can measure, nor that the abstract probability space that contains abstract elements corresponding to the six possible outcomes is real. It just means that these mathematical abstractions can be used to make useful calculations, including that the outcome of rolling a die modelled by these abstractions will always be 1, 2, 3, 4, 5, or 6.

The situation is similar with the mathematical formalism of quantum mechanics. We have an abstract mathematical object (called a "wave function" for historical reasons) in an equally abstract infinite-dimensional Hilbert space, and if we perform certain abstract mathematical manipulations on this object we will get results that match the results of measurements quite well. That doesn't mean that we're measuring the wave function, it means that we're using the wave function to calculate the results of measurements performed on the system it describes (and yes, that system may be a single particle).

So..... If we are looking for a B-level answer to the question in the thread title, it will be, consistent with @PeterDonis's and
@A. Neumaier's posts above: We measure observables of the quantum system, which may but need not be a single particle. We don't measure the wave function; it's just a piece of math that we use to calculate the likelihood of various measurement results.

There are many subtleties here, but unfortunately none of them can be discussed effectively in a B-level thread, so this thread is closed. As always, anyone can a PM any mentor to ask that it be reopened of there is more to say at this level.