What Are Some Common Misconceptions About Quantum Mechanics?

[unseensoul]

I've been struggling with the quantum mechanics' concepts for two weeks. It's been very hard to find any good resource (articles, books, etc) which explains the concepts of the quantum world in detail and in such a way so everyone can understand it.

However, here are some of my doubts...

Where does the uncertainty principle apply to? Elementary particles and photons?

Why is it impossible to measure both position and momentum precisely? Is it because the observer changes the nature of things (ie. the collapse of the wavefunction)? I do know that it has to do with Heisenberg's principle, but that doesn't answer my question.

According to de Broglie's postulate do particles whether behave literally like a wave or appear to act like one? Some people say that particles are moving up and down with a perpendicular direction of motion, is this correct? If so it doesn't make sense to me as it's unable to explain what happens in the double-slit experiment if we are really picturing the particles this way...

How do electrons behave in atoms? Why do their orbits' circumference has to be a multiple of their integral wavelength? In this situation when you're talking about their wavelength does it has to do with their wavefunction?

PS: Do not forward me to Wikipedia, please...

 

[Hurkyl]

Why is it impossible to measure both position and momentum precisely?

It's not. What's impossible is for both position and momentum to be precisely defined for a particle.

I.E. if you had a million identical quantum particles, and did a "measure position and momentum" experiment on all of them, and looked at the variance in your measurements, there is a theoretical lower limit as to how small the product of the variances can be, no matter how you designed your experiment.

This contrasts with the case of classical particles where, in principle, you could refine your measuring apparatuses to make both variances as small as you wanted. (And so the position and momentum of a particle could be defined as the result of an 'ideal' measurement)

 

[unseensoul]

If we are trying to measure the position of the particle... the more accurate we are about it's position the less we are about it's momentum and vice-versa, right? Why is this always true in the quantum world no matter how good our measuring devices are? This sounds a bit like counter-intuitive...

 

[malawi_glenn]

"Where does the uncertainty principle apply to? Elementary particles and photons?"

To every wavefunction which depends on position I would say. So not only elementary particles, you can define a wavefuntion for a composite system aswell. e.g an atom, the wavefunction for an atom in a solid or gas.

"Why is it impossible to measure both position and momentum precisely?"

As Hurkyl said, it is the simultaneous measurment of momentum and position that is impossible.
And that has to do with the commutator between the x and p-operator. And from that commutator you derive the Heisenberg uncertainty relation.

"According to de Broglie's postulate do particles whether behave literally like a wave or appear to act like one? Some people say that particles are moving up and down with a perpendicular direction of motion, is this correct? "

No, their "wave behavior" is a PROBABILITY wave, i.e the wavefunction. The de Brogile postulate is that the wavelenght of this wavefunction is related to the momentum of the particle. The particle does not oscillate within space as a waterwave, but the probability to find a particle at a certain place oscillate.

"How do electrons behave in atoms? Why do their orbits' circumference has to be a multiple of their integral wavelength? "

No this is just the bohr model, forget about it, you will eventually learn how to solve the Schrödinger Equation for the Hydrogen atom, thus obtain the wavefunctions of electrons in an atom.

I can push you to the result now:
http://hyperphysics.phy-astr.gsu.edu/hbase/quantum/hydsch.html
http://en.wikipedia.org/wiki/Hydrogen_atom

Here is a picture of the electron wavefunctions in hydrogen,
http://upload.wikimedia.org/wikipedia/en/e/e7/Hydrogen_Density_Plots.png

N.B that picture is a 2dimensional projection, here is another one:
http://www.wwnorton.com/college/chemistry/chemconnections/Stars/images/orbitals.jpg

 

[malawi_glenn]

If we are trying to measure the position of the particle... the more accurate we are about it's position the less we are about it's momentum and vice-versa, right? Why is this always true in the quantum world no matter how good our measuring devices are? This sounds a bit like counter-intuitive...

That is the whole issue of quantum mechanics, forget about your "common sense intution", do the formalism and compare to experiments.

As I wrote in my post just 5seconds ago, this "counter-intuitive" thing has to do with the commutator between x and p. Which is a quite intutive one, from mathematics, it is just the imaginary unit times the classical poisson-bracket of x and p...

 

[newbee]

Where does the uncertainty principle apply to? Elementary particles and photons?

It applies to all non-commuting observables. Now that may not mean much to you but if you learn the five (depending on how you count them) basic axioms of QM and a little linear algebra then it will.

Why is it impossible to measure both position and momentum precisely?

Well the question "why?" is being misused here. The "why?" question only makes sense in the context of a theory. That being said, it isn't impossible. What is impossible, according to QM theory, is to make such simultaneous measurements. In the language of the QM axioms the observables (position and momentum in this case but there are others) must share common eigenstates (ie their operators must commute) in order that measurements of the observables can be made simultaneously.

Is it because the observer changes the nature of things (ie. the collapse of the wavefunction)?

That is not part of the theory.

 

[unseensoul]

As Hurkyl said, it is the simultaneous measurment of momentum and position that is impossible.
And that has to do with the commutator between the x and p-operator. And from that commutator you derive the Heisenberg uncertainty relation.

What do you mean by "commutator between the x and p-operator"? Could you explain it in a simple (non-mathematical) way instead, if possible?

No, their "wave behavior" is a PROBABILITY wave, i.e the wavefunction. The de Brogile postulate is that the wavelenght of this wavefunction is related to the momentum of the particle. The particle does not oscillate within space as a waterwave, but the probability to find a particle at a certain place oscillate.

So where does that fit in wave-property of particles? I'm asking you this because if we think about the double-slit experiment an electron seems to interact with itself like a wave.

No this is just the bohr model, forget about it, you will eventually learn how to solve the Schrödinger Equation for the Hydrogen atom, thus obtain the wavefunctions of electrons in an atom.

...http://www.wwnorton.com/college/chemistry/chemconnections/Stars/images/orbitals.jpg

I think the last image you provided is easier to understand compared to the previous ones although I can't understand a thing of it lol

Nonetheless, I do fairly understand (I think :P) what the double-slit experiment produces (quantum mechanical behaviour so to speak) but I can't understand the quantum mechanical behaviour of electrons in an atom.
Regarding the double-slit experiment if we shoot an electron at a time we'll have an interference pattern by the end. There is 1/2 probability for the electron to pass through either slit, but according to the experiment it will pass through both slits which means there will be a superposition of the possibilities so the electron exists at multiple places at the same moment. However, if we try to look which slit the electron is passing through the wavefunction collapses. Did I wrongly mention something?
If my reasoning is right, good, but I still can't understand the behaviour of electrons within the atoms...

 

[unseensoul]

That is not part of the theory.

So what's the relationship between the uncertainty principle and the observer effect?

 

[Hurkyl]

"Why is it impossible to measure both position and momentum precisely?"

As Hurkyl said, it is the simultaneous measurment of momentum and position that is impossible.
And that has to do with the commutator between the x and p-operator. And from that commutator you derive the Heisenberg uncertainty relation.

That's not quite what I said... In principle, you can have a device that simtulaneosuly measures both position and momentum to arbitrary accuracy. The problem is that the quantum particle itself doesn't have a well-defined position and momentum. In other words, uncertainty isn't a limitation on measuring devices -- it's a description of the thing we are measuring.


To put it differently, suppose we are just measuring a particle's position, and ignoring momentum. No matter how precise our measuring device is, there will always be some minimum amount of variance due to the fact the particles themselves do not have a well-defined position. Of course, different particles have differing amounts of localization -- so if our ideal measuring device is measuring a well-localized particle, we will see little variance in the readings, but if it's measuring a poorly-localized particle, we will see large variance in the readings.

We might imagine that a particle could be in a state where it had a well-defined position and a well-defined momentum, but we can't -- that is what the uncertainty principle forbids.

 

[borgwal]

I agree with

The problem is that the quantum particle itself doesn't have a well-defined position and momentum.

Let me ask you what you really mean by the following statement, just to see if I agree or not:

In principle, you can have a device that simtulaneosuly measures both position and momentum to arbitrary accuracy.

Suppose I prepare a particle in one of four possible states:

1) well-defined position x=0 (meaning, with some small uncertainty), p not well defined (meaning, a relatively large uncertainty, such that the uncertainty principle is not violated)

2) well-defined position x=1 (in some units, irrelevant here), p similarly not well defined

3) well-defined momentum p=0, x not well defined

4) well-defined momentum p=1 (in some units, irrelevant here), x not well defined

Now I give you this one single particle without telling you which state I prepared. Do you think you can figure out with 100% certainty which state I prepared by measuring simultaneously both position and momentum to an arbitrary accuracy?

 

[borgwal]

"In principle, you can have a device that simtulaneosuly measures both position and momentum to arbitrary accuracy"

And to get to the real point: do you think that one can measure S_x and S_y simultaneously to an arbitrary accuracy on a single spin-1/2 system?

 

[ZapperZ]

There's nothing to stop you from measuring, to arbitrary accuracy, the SINGLE measurement of both position AND THEN momentum! The accuracy of such measurement depends simply on the accuracy of the instrument and measurement technique! This isn't the HUP. I mean, just look at how the HUP for, say, position uncertainty is defined. It involves a statistical spread in a NUMBER of repeated measurements, not just ONE measurement.

Now, this is different than saying "I've measured the position x, now can you tell me what the value of the momentum would be?" This is where the HUP kicks in, because even in an identically-prepared system, the value that p can acquire depends very much on how accurate the value of x was measured. If x is measured very accurately, then p can attain a very large range of values, so your ability to predict what p is will be less certain.

But no where here does it prevent you from making a very accurate measurement of p. It is just that if you keep repeating the experiment and measure many, many of these p's, the SPREAD in values will be large when the spread in x is very small.

Zz.

 

[borgwal]

There's nothing to stop you from measuring, to arbitrary accuracy, the SINGLE measurement of both position AND THEN momentum! The accuracy of such measurement depends simply on the accuracy of the instrument and measurement technique! This isn't the HUP. I mean, just look at how the HUP for, say, position uncertainty is defined. It involves a statistical spread in a NUMBER of repeated measurements, not just ONE measurement.

Now, this is different than saying "I've measured the position x, now can you tell me what the value of the momentum would be?" This is where the HUP kicks in, because even in an identically-prepared system, the value that p can acquire depends very much on how accurate the value of x was measured. If x is measured very accurately, then p can attain a very large range of values, so your ability to predict what p is will be less certain.

But no where here does it prevent you from making a very accurate measurement of p. It is just that if you keep repeating the experiment and measure many, many of these p's, the SPREAD in values will be large when the spread in x is very small.

Zz.

Are you reacting to my post? I know all of this, but I'm asking Hurkyl what s/he means by *simultaneous* measurements: I don't have a problem with arbitrarily accurate sequential measurements, but I do with simultaneous arbitrarily accurate measurements of noncommuting observables. In a Socratean way I was going to arrive at a contradiction.

 

[Fredrik]

Simultaneous measurements of two non-commuting observables don't make sense in QM. Hurkyl knows that. He may have taken it a little too far by claiming that we can imagine a device that performs both measurements at once, but he was definitely right to point out that the uncertainty principle isn't about limitations of measuring devices. It's about a property of state vectors / wave functions.

A few thoughts about the possibility of a device that performs "a simultaneous measurement of two non-commuting observables" on a system... Let's take a spin-1/2 system as an example, and first consider an apparatus that "measures" the z component of the spin. Such a device only needs to produce an inhomogeneous magnetic field and detect the position of the spin-1/2 particle after it's deflected in one direction or the other by the inhomogeneous magnetic field. (Stern-Gerlach experiment).

A device that performs "a simultaneous measurement of the x component and the y component of the spin" would just be a device that produces two inhomogeneous magnetic fields in a location that's surrounded by detectors. It's easy to see what would happen in this case: A magnetic field is a vector quantity, so the "two fields" produced by the device would be equivalent to one that's stronger and pointing in the direction defined by the vector sum of the two field vectors.

I expect something similar to hold in all cases, not just in the case of the spin components of spin-1/2 particles. A "simultaneous measurement of two non-commuting observables" would always turn out to be just one measurement of one observable that isn't one of the two intended.

 

[Hurkyl]

Simultaneous measurements of two non-commuting observables don't make sense in QM. Hurkyl knows that.

Yeah. I made an arithmetic mistake -- one I've made before even, so I should know better!

 

[unseensoul]

Would you guys stick in my doubts? :P

 

[malawi_glenn]

What do you mean by "commutator between the x and p-operator"? Could you explain it in a simple (non-mathematical) way instead, if possible?

If you don't know any QM formalism, maybe you could read a bit on it since the mathematical language is essential for QM.

A measurment of a variable z on a state [itex]\psi (z) [/itex] is made in QM as letting the operator [itex] \hat{Z} [/itex] act on that state:

[tex] \hat{Z} \psi (z) = z' \psi (z) [/tex]

Where z' is one of the possible values of z.

The commutator between operator A and B is defined as:

[tex] [A,B] = AB - BA [/tex]

Now x is the operator for position and p is the operator for momentum, one can show that the commutatur between those is:

[tex] [x,p] = i \hbar [/tex]

i.e measuring the momentum first THEN the position of a state gives another result that what you would get if you FIRST measured the position then the momentum. From that commutator above, you derive the Heisenberg uncertainty relation.

So where does that fit in wave-property of particles? I'm asking you this because if we think about the double-slit experiment an electron seems to interact with itself like a wave.

The wave property of particles is related to its wavefunction which gives you the probability amplitude to find a partice at a certain place at a given time.

In the Double slit experiment you basically do the same thing as for ordniary waves, same math. But the difference lies in what the wave resembles. In QM the wave resembles the probability to find a particle at a certain position. So it is the wavefunction of the electron that splits into two wavefunctions, one for eash slit, then on the other side of the slit, you will get destructive and constructive interference - just as with light!

The only difference lies in what the wave resembles. Don't think of the particle as a small ball flying around, think of it as a probability wave spread in space and time... forget about intuition and drawing pictures, workout the math ;-) Sorry, but that is the way nature is.

Nonetheless, I do fairly understand (I think :P) what the double-slit experiment produces (quantum mechanical behaviour so to speak) but I can't understand the quantum mechanical behaviour of electrons in an atom.
Regarding the double-slit experiment if we shoot an electron at a time we'll have an interference pattern by the end. There is 1/2 probability for the electron to pass through either slit, but according to the experiment it will pass through both slits which means there will be a superposition of the possibilities so the electron exists at multiple places at the same moment. However, if we try to look which slit the electron is passing through the wavefunction collapses. Did I wrongly mention something?
If my reasoning is right, good, but I still can't understand the behaviour of electrons within the atoms...

In the atom, you get a wavefunction for the electron by solving the schrödinger equation, which is the equation of motion for a quantum particle. Solving the Schrödinger equation, with the atomic potential, you'll get the wavefunctions for the electron.

You'll get the same "effect" by solving the schrödinger equation for a "particle in a box" (google if you want). And the double slit experiment is just solving the schrödinger equation for a free particle (i.e no potential).

 

[jtbell]

If we are trying to measure the position of the particle... the more accurate we are about it's position the less we are about it's momentum and vice-versa, right? Why is this always true in the quantum world no matter how good our measuring devices are? This sounds a bit like counter-intuitive...

It's a consequence of using waves to represent particles, with [itex]\lambda = h / p[/itex], and Fourier analysis. When you form wave packets by superposing waves with different wavelengths, you have a general constraint on the product of the spatial width of the packet ([itex]\Delta x[/itex]) and the range of wavenumbers that it contains ([itex]\Delta k[/itex], where [itex]k = 2 \pi / \lambda[/itex]):

[tex]\Delta x \Delta k \ge \frac{1}{2}[/tex]

This relationship applies to all wave packets, including classical ones like water waves, sound waves, electrical signals, etc.

Substituting [itex]k = 2 \pi p / h[/itex] gives you the QM uncertainty principle for x and p.

 

[borgwal]

Yeah. I made an arithmetic mistake -- one I've made before even, so I should know better!

Sorry to have brought it up: I hoped you meant something *interestingly* wrong :-)

 

[Fredrik]

I think it is interesting, and it isn't really wrong. My conclusion was that it is possible to build a device that subjects the system to the same external influences that a measurement of observable A and a measurement of observable B would do. I think that's an interesting fact, and it's even more interesting that what such a device really does is to measure some other observable (which I assume is a linear combination of A and B). I hadn't really thought these things through before, so I learned something new here.

 

[DrChinese]

Would you guys stick in my doubts? :P

What do you have doubts about?

 

[Gerenuk]

I've been struggling with the quantum mechanics' concepts for two weeks. It's been very hard to find any good resource (articles, books, etc) which explains the concepts of the quantum world in detail and in such a way so everyone can understand it.

I'm afraid there is only one way to understand QM without further questions. You need to work through a good book with equations, like "Quantum Mechanics" by Claude Cohen-Tannoudji. No word-based explanation can explain QM without seeming inconsistent and posing new questions.

Where does the uncertainty principle apply to? Elementary particles and photons?

It applies to everything that obey quantum mechanics and has wave functions. Therefore to all particles.

Why is it impossible to measure both position and momentum precisely? Is it because the observer changes the nature of things (ie. the collapse of the wavefunction)?

For that you need to understand the wavefunction and how position and momentum representation follow as Fourier transforms from each other.
An analogy would be: a coin is either heads up or tails up.

According to de Broglie's postulate do particles whether behave literally like a wave or appear to act like one?

The probability wavefunction of the particle behaves like a wave, but in the end the particle will be found at a certain position if measured.

Some people say that particles are moving up and down with a perpendicular direction of motion, is this correct?

No. Note that neither electrons nor electromagnetic wave oscillate in spatial directions. Analogy: imagine a string with a modulated colour.

How do electrons behave in atoms? Why do their orbits' circumference has to be a multiple of their integral wavelength?

The electron wavefunction is a single entity around the nucleus. Like a string around the nucleus, when going round one full turn it has to attach to the point there it started. That only works if its periodicity is a fraction of the total circumference.

In this situation when you're talking about their wavelength does it has to do with their wavefunction?

The wavefunctions is a mathematical functions which repeats itself every one wavelength.

PS: Do not forward me to Wikipedia, please...

Wise choice. Your questions will never end unless you really carefully read through the mathematical concepts of QM (university undergrad level).

 

[unseensoul]

It's a consequence of using waves to represent particles, with [itex]\lambda = h / p[/itex], and Fourier analysis. When you form wave packets by superposing waves with different wavelengths, you have a general constraint on the product of the spatial width of the packet ([itex]\Delta x[/itex]) and the range of wavenumbers that it contains ([itex]\Delta k[/itex], where [itex]k = 2 \pi / \lambda[/itex]):

[tex]\Delta x \Delta k \ge \frac{1}{2}[/tex]

This relationship applies to all wave packets, including classical ones like water waves, sound waves, electrical signals, etc.

Substituting [itex]k = 2 \pi p / h[/itex] gives you the QM uncertainty principle for x and p.

That's confusing...

Isn't there a easier way to explain it without the use of mathematics?

 

[unseensoul]

i.e measuring the momentum first THEN the position of a state gives another result that what you would get if you FIRST measured the position then the momentum. From that commutator above, you derive the Heisenberg uncertainty relation.

Imagining we were the same size as the smallest particles... Would we be able to measure them (position and momentum) precisely?
Does this uncertainty has anything to do with the particles' wavefunction?

The wave property of particles is related to its wavefunction which gives you the probability amplitude to find a partice at a certain place at a given time.

Do the particles always behave like a wave when they aren't under any measurements? Is that why the quantum world is probabilistic?

 

[newbee]

Do the particles always behave like a wave when they aren't under any measurements?

Yes. According to quantum mechanics. See the Stearn Gerlach experiment. The wikipedia page lokks good but I have not read it yet.

 

[Fredrik]

Imagining we were the same size as the smallest particles... Would we be able to measure them (position and momentum) precisely?
Does this uncertainty has anything to do with the particles' wavefunction?

No, we wouldn't. The uncertainty is a property of the wavefunction. A particle doesn't have a really well-defined position or a momentum at any time. Its position is always spread out over some range of values, and the same goes for the momentum. A measurement "squeezes" the range of values of one of those variables to make it as small as the measuring device can, and this process makes the range of values of the complementary variable spread out even more. (It's hard to explain it better than that without mathematics).

Do the particles always behave like a wave when they aren't under any measurements? Is that why the quantum world is probabilistic?

Yes and no. The time evolution of an isolated system (like a particle that we're not doing measurements on) is described by the Schrödinger equation. I guess you could call that "behaving like a wave". The probabilistic stuff is what you get when the system isn't isolated from the environment. It evolves in time in a different way, a way which can be described as "probabilistic".

 

[Gerenuk]

That's confusing...

Isn't there a easier way to explain it without the use of mathematics?

Indeed it is confusing if you don't know the details of the explanation above

But there is really no other way than reading through a mathematical book.
Analogy: You could explain lightning with gods, but this explanation is only putting it off.
Or you could read the more detailed mathematical explanation and really understand what's going on.

 

[JakeStan]

I've had an idea in my head about HUP and could use some clarification. I don't remember which book it came from (possibly Greene's "The Elegant Universe") but the basic idea was that a wave taking a measurement has a precision based on its wavelength. That the shorter the wavelength, the more accurate a measurement could be and as a corollary, no measurement could be made upon an object smaller than the wavelength of the measuring wave (i think the example was a photon colliding with an electron).

Now if in that example you wanted to increase the precision of position measurement of the electron you would need to continually decrease the wavelength of the measuring, thereby increasing the energy of the photon. Unfortunately, the more energy you put into the measuring photon, the more that photon would effect the resultant momentum of the electron which would put an increase in the spread of values for your electron's momentum.

Now is this in excruciatingly simplified terms how the math of HUP works? Or do I need to do more reading? :)

 

[malawi_glenn]

well no, since then it would imply that it is the probe that is altering the system - not that it is inherent by the system itself as QM formalism will show you!

The HUP is derived from the axioms of QM and there is nowhere that the probe which you use to determine the position of the system (particle) shows up in the derivation. HUP is obtained by simply using the anticommutator of the position and momentum operator.

 

[JakeStan]

Cool, thanks for the answer, I'll keep digging.

 

[malawi_glenn]

Cool, thanks for the answer, I'll keep digging.

You can go through the derivation by yourself, it is found in (almost) every intro textbook in QM.

 

[nfelddav]

That's confusing...

Isn't there a easier way to explain it without the use of mathematics?

Sorry... but no. as you've probably noticed by know, QM is basically a purely mathematical theory. The physical representation (if there is one) not part of the theory at all. The only way to explain it is in the mathematics, because that's really all there is.

well no, since then it would imply that it is the probe that is altering the system - not that it is inherent by the system itself as QM formalism will show you!

The HUP is derived from the axioms of QM and there is nowhere that the probe which you use to determine the position of the system (particle) shows up in the derivation. HUP is obtained by simply using the anticommutator of the position and momentum operator.

But for this I have to go the other direction: You seem to be dismissing a physical representation which sounds very reasonable. As far as I know, many/most people think there must be some physical explanation for the mathematics of QM.
HUP is obtained by simply using the anticommutator of the position and momentum operator, but it is not unreasonable to assume that something physical makes it happen. I've seen the argument that the probe alters the system put forth a lot.

 

[malawi_glenn]

Sorry... but no. as you've probably noticed by know, QM is basically a purely mathematical theory. The physical representation (if there is one) not part of the theory at all. The only way to explain it is in the mathematics, because that's really all there is.




But for this I have to go the other direction: You seem to be dismissing a physical representation which sounds very reasonable. As far as I know, many/most people think there must be some physical explanation for the mathematics of QM.
HUP is obtained by simply using the anticommutator of the position and momentum operator, but it is not unreasonable to assume that something physical makes it happen. I've seen the argument that the probe alters the system put forth a lot.

So in the first section in your answer you stresses that QM theory is just math math math.. then you switch position?

You mean that the commutator itself has a 'physical' relation? And with "physical" here, you mean some kind of "collision"? Why is not the description of the commutator itself physical? Is it physical to talk about that an operatoration on the wavefunction with an operator makes the wavefunction collapse into one of its eigenstates to the operator? Further definitions of what "physical" means is required.

 

[Tac-Tics]

QM seems strange because in every experiment, there are particles flying towards each other, a black box where "something magic happens", but you're not sure what, and then particles flying back out, whose properties we can predict with the theory.

It seems this is no different than any other physical theory. In mechanics, if you roll two billiard balls into each other, you know their incoming speeds and you know their outgoing speeds, but you don't know jack about the mechanism which cause their speeds to change. It isn't encompassed by the theory. (Something to do with electrons interacting or some nonsense). But no one seems confused by this!

Billiard balls collide and magically trade momenta. Electrons collide and trade spin. Same deal. The mechanism isn't well understood given the theory used to predict the correct and useful results.

The annoying part of QM, from what I can tell, seems to be that the black box in the experiment seems significant.

In the billiard ball example, we pretend that the world acts the same inside the box and outside, and that the ball it just made of quadrillions of smaller balls, all acting under the same physical rules.

With QM, inside the black box, we know that things obey a different set of rules. At least, according to our theory. Everything outside of the box is measured "classically" and everything inside is in a superposition of classical states. We of course realize that the box isn't really there at all, and the rules of physics should apply just as well to the outside. Why don't we see the billiard ball splitting into an infinity of balls when we strike it with a cue stick? Why do measurements seem to cause all those "ghost" balls to vanish, leaving only a single classical state?

The answer is, shut up and calculate. Physicists don't deal with questions that are impossible to measure, so they just stick to electrons and photons and slits and detectors and pretend it's all just magic, because no one quite understands the truth.

 

[nfelddav]

So in the first section in your answer you stresses that QM theory is just math math math.. then you switch position?

Yeah that's just about it...
I was stressing that if you want to understand anything about QM, you have to do the math, because as Tac-Tics says, we don't know what happens in that "black-box."
Then the second part was the observation that something does.

A collision would be an example of something physical, yes. Sorry, I hate imprecise language as well, but I'm having trouble on precise formulation. Physical as in not just the mathematical formulation: that which causes the system to behave in the way described by the operator.
Something which "has material existence" (dictionary definition).