# Quantum mechanics basics

1. Nov 2, 2008

### 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...

2. Nov 2, 2008

### Hurkyl

Staff Emeritus
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)

3. Nov 2, 2008

### 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...

4. Nov 2, 2008

### malawi_glenn

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

To every wavefunction wich 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,

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

Last edited by a moderator: May 3, 2017
5. Nov 2, 2008

### malawi_glenn

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...

6. Nov 2, 2008

### newbee

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.
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.
That is not part of the theory.

7. Nov 2, 2008

### unseensoul

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?

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.

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...

8. Nov 2, 2008

### unseensoul

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

9. Nov 2, 2008

### Hurkyl

Staff Emeritus
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 cant -- that is what the uncertainty principle forbids.

10. Nov 2, 2008

### borgwal

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

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?

Last edited: Nov 2, 2008
11. Nov 2, 2008

### 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?

12. Nov 2, 2008

### ZapperZ

Staff Emeritus
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.

13. Nov 2, 2008

### borgwal

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.

Last edited: Nov 2, 2008
14. Nov 3, 2008

### Fredrik

Staff Emeritus
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.

Last edited: Nov 3, 2008
15. Nov 3, 2008

### Hurkyl

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

16. Nov 3, 2008

### unseensoul

Would you guys stick in my doubts? :P

17. Nov 3, 2008

### malawi_glenn

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 $\psi (z)$ is made in QM as letting the operator $\hat{Z}$ act on that state:

$$\hat{Z} \psi (z) = z' \psi (z)$$

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

The commutator between operator A and B is defined as:

$$[A,B] = AB - BA$$

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

$$[x,p] = i \hbar$$

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 uncertanty relation.

The wave property of particles is related to its wavefunction wich 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.

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).

18. Nov 3, 2008

### Staff: Mentor

It's a consequence of using waves to represent particles, with $\lambda = h / p$, 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 ($\Delta x$) and the range of wavenumbers that it contains ($\Delta k$, where $k = 2 \pi / \lambda$):

$$\Delta x \Delta k \ge \frac{1}{2}$$

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

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

19. Nov 3, 2008

### borgwal

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

20. Nov 3, 2008

### Fredrik

Staff Emeritus
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.