# I Wrong term for "measure"?

1. Nov 27, 2016

### FallenApple

Why do physicists use the word measure?

We determine the location of a particle by measurement. Before measurement it doesn't have a location. If there wasn't a measurement conducted, then it wouldn't have a location in the first place.

Therefore, measure should not be the term used. Wouldn't it be more accurate to use the term create instead?

Not saying that physicists created the particle, but certainly they create the location upon observing since the observer is the casual factor here.

2. Nov 27, 2016

### mikeyork

Detect is a better term in this situation. A detector has a location. It only detects particles which pass through the detector at that location. The measurement being made is, if you like, the time at which the particle was detected. Particles which do not pass through the detector are not detected. If you have an array of detectors and a single particle then you can measure the space-time location of such a particle only if it is detected.

3. Nov 28, 2016

### Staff: Mentor

No, although "interact with" might be - the words "observe", "measure", and "particle" are all used for historical reasons to describe quantum mechanical concepts that have very little to do with the ordinary English-language meanings of these words. Note that "particle" is in that list - you should not think of a quantum particle after a position measurement as if it were a little tiny object sitting at that position.

An interaction doesn't create a position for the particle. Before the interaction, we had a function that gave us the probability of detecting the particles at different points in space. After the interaction, that's still what we have - the only difference is that the function may have a different shape. Some interactions change the shape of the function in such a way that it becomes (for a while) denser in one region of space than another; we call these "detections" or "measurements" and say that we "detected" the particle in that region of space. But all we've really done is narrow down the region where it's likely to be detected next time we look.

4. Nov 30, 2016

### FallenApple

Oh ok I see. So the wavefunction collapses and we know its location at that instant, and when we don't look a moment later, we are confident that the wavefunction has changed.

So concious observations alters the wave function. And the interaction is merely due the act of knowing where a particle is at a given instant.

5. Nov 30, 2016

### DennisN

The wavefunction does not change simply because we are not looking... (see below)
No, definitely not in standard Quantum Theory. There is no consciousness involved at all in the basic theory. But I should add that there are many so-called interpretations of quantum mechanics, and one of them, a historical one, involves consciousness: the Von Neumann–Wigner interpretation.
No. You seem to equate/mix together interaction with knowledge. I will think about a quantum example to show you, and post about it later in this thread (maybe someone else beats me to it ).

Last edited: Nov 30, 2016
6. Nov 30, 2016

### DennisN

7. Nov 30, 2016

### Staff: Mentor

That's assuming that there is such a thing as "wavefunction collapse". The mathematical formalism of QM does not include any notion of collapse; instead collapse is something that shows up in some but not all of the many interpretations of quantum mechanics, and it is quite possible to do QM without ever invoking collapse.
The idea that conscious observation plays any part in QM was considered during the early days when it appeared to solve a serious problem with the then-current understanding of QM. However, it was largely abandoned after the discovery of decoherence (google for "quantum decoherence", but I have to caution you that the math is somewhat daunting - David Lindley's book "Where does the weirdness go" may be more helpful), and these days is somewhat fringe. Unfortunately, by then it had leaked into the popular imagination, where it's become as hard to kill off as any urban legend.

You can think of a "position measurement" as any interaction with anything that reduces the uncertainty in where the particle is. For example, if we can catch the particle in a deep potential well (loosely speaking, "put it in a box") that's a position measurement. The smaller the "box" the more precisely we have measured the position, but we will never get down to an exact position. Nor is any collapse required to explain the process: the particle has a wave function, that wave function is evolving according to Schrodinger's equation, and we just have to make sure that we properly include the potential well in that equation as we solve it. The solution will show the wave function naturally "bunching up" as the particle is in the potential well, and naturally spreading out again after we remove the well.
(This is a dangerously hand-wavey explanation. You've marked this as an I-level thread, so you should be able to work some of the exercises in a first-year QM textbook; work through some single-particle problems using the position basis and it will all make much more sense).

8. Nov 30, 2016

### mikeyork

No, its the other way round. Knowledge is due to interaction. Interaction involves the transfer of physical data.

9. Dec 2, 2016

### FallenApple

Oh ok. I think you are talking about the heisenberg uncertainty principle right? So measuring, are we creating a potential well, such that after the measurement, the well is removed?

btw, how far along a first year QM book will I need to get to? I'm thinking about a book like Griffins but I think my math background is good enough for me to attempt Shankar. Not sure my physical intuition will be caught up though for more linear algebra based books.

10. Dec 2, 2016

### Staff: Mentor

What's going on here is that we write the wavefunction as $\psi(x)$, a function of $x$; now the probability of finding the particle between points $x_0$ and $x_1$ is $\int_{x_0}^{x_1}\psi^*(x)\psi(x)dx$. We calculate the probability that the particle is at the exact location $X$ by setting $x_0=x_1=X$, but then the integral comes out zero; there's no state in which the particle can be at an exact position.

The uncertainty principle is something different. It comes in when we look at how the probability distribution for position is related to the probability distribution for momentum; the more sharply peaked one is, the more spread out the other will be.
The easiest position measurement is done by sending the particle through an opening in a screen. While the particle is passing through the screen, it is very tightly constrained in two dimensions (the edges of the screen are the walls) but once through the opening the wave will start spreading again.

Last edited: Dec 2, 2016
11. Dec 2, 2016

### FallenApple

Right. That makes sense. The probability at a single point is 0 when using a probability density function.

Ok I looked up wave function. So the wave function for position is written as the infinite sum of the fourier series of the position wave functions at every possible momenta.
So by using information on all the momentums, we are more certain of the truth of the position. But that logically means that we ascribe partial truths to each component of the sum, hence the extreme uncertainty of a particular position.