# ARRGH Particle Wavefunction Questions

ARRGH!! Particle Wavefunction Questions...

Hello everyone at physicsforums.com!
Ook, I have a few questions, well, a lot actually. I guess I'll just ask two of them here. It is stated in quantum mechanics that a wave function (it's absolute square) tells the probability of finding a particle at a specific point, but what if one is dealing with an extended object, such as an atom or a basketball? Does the wavefunction tell us where we'll find the object's center of mass or what??
Now, if we have a point particle located at a specific point, then it's electric field is given as Cq/r^2, where r is the distance to that particle. But what if the particle is "smeared out in space"? I.e. the particle's wavefunction is an extended wavepacket rather than a delta function. Is the electric field the same in that case as the e. field of an extended charge distribution with charge density at each point proportional to the probability of finding the particle at that point? So, like, whould the e.field inside the wavepacket at some point be zero, increase as we move out of the wavepacket, and then start to decrease again? For example, when we solve the Shr. Equation for an atom, we assume that the nucleus is located at a specific point. But what if the nucleus is described as an extened wavepacket? When solving the Shr. Eq. for an electron bound to the nucleus, would our potential function (the one we substitute into the Shr. Eq) now be the same as the potential function of an extended charge distribution?? Don't wavepackets spread out in time? So even if the nucleus was located at a specific point at some time, wouldn't it become an extended wavepacket later? So, like, the whole atom spreads out and the electron cloud gets deformed? But back to the charged particle described as a wavepacket. If we actually measure the electric field at a point "inside" the wavepacket, would we get a small value ( the same value we'd get if we measured the field inside a charge distribution) or would we get a large value (the same value as we'd get if the charge was concentrated at a specific point) and therefore localize the particle with our measurement?? Isn't that how we localize particles and collapse their wavefunctions anyway, by measuring the electric field?? But if we are using the Shr. Eq. to describe the motion of charged particle A in the vicinity of charged particle B (the latter is described by a wavepacket) would the potential function in the Shr. Eq.(the equation which we're solving for particle A, the potential is created by particle B) be just Cq/r^2 (which of course would also change in time since the position of particle B changes), or would the potential function in the Shr. eq. describing particle A be the same as that of an extended charge distribution with charge density at each point equal to the probability of finding particle B at that point, or would it be something else entirely??

P.S.
Sorry for such a long question (two or three questions actually, although it looks like a lot more), I've tried to make myself clear, don't know if I have though. Thanks to anyone who answers.

ZapperZ
Staff Emeritus
Sorry, I'm sorta new here. Also, did anyone read the question? I was not asking about the solutions for the H atom!

dextercioby
Homework Helper
Inquisiter said:
Hello everyone at physicsforums.com!

As a part of the unofficial welcoming commitee:WELCOME! Inquisiter said:
It is stated in quantum mechanics that a wave function (it's absolute square) tells the probability of finding a particle at a specific point,

No.The square modulus gives u the probability density to find the point particle in a certain point from the physical space...

Inquisiter said:
but what if one is dealing with an extended object, such as an atom or a basketball? Does the wavefunction tell us where we'll find the object's center of mass or what??

An atom is not "spatially extended object"...Both an atom & a basketball fall under the "many-body system"...The wavefunction for such a system (if,by absurd,found) would give you the probability density for each particle from the system...

Inquisiter said:
Now, if we have a point particle located at a specific point, then it's electric field is given as Cq/r^2, where r is the distance to that particle. But what if the particle is "smeared out in space"? I.e. the particle's wavefunction is an extended wavepacket rather than a delta function.

No,no,you're confusing delta Dirac potential (or maybe field source) with the wavefunction.Incidentally,there's no such thing as a delta-Dirac wavefunction.

Inquisiter said:
P.S.
Sorry for such a long question (two or three questions actually, although it looks like a lot more), I've tried to make myself clear, don't know if I have though. Thanks to anyone who answers.

It's just that i got bored...Someone else's contribution is welcome... Daniel.

dextercioby said:
No.The square modulus gives u the probability density to find the point particle in a certain point from the physical space...
That's what I meant
dextercioby said:
An atom is not "spatially extended object"...Both an atom & a basketball fall under the "many-body system"...The wavefunction for such a system (if,by absurd,found) would give you the probability density for each particle from the system...
I thought a large object could be described by a wavefunction which does NOT give the probability density for each particle, only for the object as a whole.
dextercioby said:
No,no,you're confusing delta Dirac potential (or maybe field source) with the wavefunction.Incidentally,there's no such thing as a delta-Dirac wavefunction...
No, actually, I wasn't confusing them. Isn't the eigenfunctions of position with eigenvalue a: delta(x-a) ?
dextercioby said:
It's just that i got bored...Someone else's contribution is welcome... Daniel.
Sorry that I bored you, I didn't mean to. But your contribution wasn't very helpful...

dextercioby
Homework Helper

Daniel.

P.S.There's no such thing as "object as a whole" in QM...

dextercioby said:

Daniel.

P.S.There's no such thing as "object as a whole" in QM...
I have. I've read Landau and Livschit's quantum mechanics (not the whole book, just parts of first few chapters, the basics) and some other books as well. Did you even read and understand my questions? I have a feeling you didn't. And yes, I was sort of suspecting that there's no such thing as "object as a whole." But I thought a billiard ball, for example, could be described by a wavepacket which gives us the probability of finding the ball here or there, but does not tell us anything about the individual atoms, only about the ball as a whole. That was one of my questions. So you're saying that a billiard ball can NOT be described by a "simple" wavepacket, i.e. a superpositiong of plane waves exp(ipx/h), where p is the momentum of the ball. My second question was... well, I don't think you read it.

vanesch
Staff Emeritus
Gold Member
Inquisiter said:
Does the wavefunction tell us where we'll find the object's center of mass or what??

The quantities in quantum theory are "expectation values of operators, given a quantum state".
For the time being, we only have (confirmed) quantum theories of the following situations:
*) a finite, and fixed number of POINT particles (that's non-relativistic quantum mechanics)
*) a variable number of point particles (creation, annihilation) which turns out to be equivalent to the quantum theory of fields

There are attempts to write down quantum theories of extended objects, the most famous one the quantum theory of 1-dimensional extended objects, a.k.a. string theory.

Let us limit ourselves to the first case (non-relativistic quantum theory). As we can only describe POINTS, for your baseball there are two options: or you consider it to be a point (say, when you only look at it in a very coarse-grained way ; like you do with planets in the solar system when calculating naively the Kepler orbits), or you will have to consider each constituent particle as a point. In the first case, you have (just as in classical mechanics) made an approximation which can, or cannot, be appropriate ; in the second case you will have a wave function which depends on all the coordinates (and spins) of the constituent POINT particles.

You can now define an operator (well, 3 actually: X, Y and Z) which measures the position of the center of gravity. As an example, imagine you consider 3 point particles, A, B and C ; we have now a wavefunction depending on 9 coordinates psi(xA, yA, zA, xB, ..., zC).
You have the position operators XA, YA, ZA, XB, YB...
It is now possible to define the "x-coordinate of the center of gravity" as:

XM = 1/(mA + mB + mC) ( mA xA + mB xB + mC xC)

In a similar way you can define YM and ZM (as operators).
For a given wavefunction psi, you can then calculate the expectation value of XM and call that its associated center of gravity.

But note that there is not necessarily a WAVEFUNCTION corresponding to the center of gravity. It MIGHT be. Indeed, it might be that you can rewrite the wavefunction as a PRODUCT:
psiM(xM, yM, zM) x psi(rAx,...)
where rAx... are RELATIVE coordinates with respect to the center of gravity.
Note that this is in general not possible: your wavefunction will have to take on a special form to do so.
Also, in general, the DYNAMICS will not allow you to keep this factorized form during time evolution EXCEPT if your hamiltonian can be written as a sum of something that only acts on the center of gravity part, and another term that only acts on the relative positions.
It is only under these conditions that you can talk strictly about the "wave function of the center of gravity".

Now, if we have a point particle located at a specific point, then it's electric field is given as Cq/r^2, where r is the distance to that particle. But what if the particle is "smeared out in space"? I.e. the particle's wavefunction is an extended wavepacket rather than a delta function. Is the electric field the same in that case as the e. field of an extended charge distribution with charge density at each point proportional to the probability of finding the particle at that point?

This reasoning shows that it is indeed not possible to use classical fields and quantum particles. In fact, your electric field becomes in a superposition of states in the same way as your particle is in a superposition of position states and you have a non-classical electric field (a quantum field).

However, for many applications, what you propose above IS done, it is called the semi-classical approach (classical EM field and quantum particles). It is an acceptable approximation in many cases often used in quantum chemistry. There, it is called the "effective potential" method.

For example, when we solve the Shr. Equation for an atom, we assume that the nucleus is located at a specific point. But what if the nucleus is described as an extened wavepacket? When solving the Shr. Eq. for an electron bound to the nucleus, would our potential function (the one we substitute into the Shr. Eq) now be the same as the potential function of an extended charge distribution?? Don't wavepackets spread out in time? So even if the nucleus was located at a specific point at some time, wouldn't it become an extended wavepacket later? So, like, the whole atom spreads out and the electron cloud gets deformed? But back to the charged particle described as a wavepacket. If we actually measure the electric field at a point "inside" the wavepacket, would we get a small value ( the same value we'd get if we measured the field inside a charge distribution) or would we get a large value (the same value as we'd get if the charge was concentrated at a specific point) and therefore localize the particle with our measurement?? Isn't that how we localize particles and collapse their wavefunctions anyway, by measuring the electric field?? But if we are using the Shr. Eq. to describe the motion of charged particle A in the vicinity of charged particle B (the latter is described by a wavepacket) would the potential function in the Shr. Eq.(the equation which we're solving for particle A, the potential is created by particle B) be just Cq/r^2 (which of course would also change in time since the position of particle B changes), or would the potential function in the Shr. eq. describing particle A be the same as that of an extended charge distribution with charge density at each point equal to the probability of finding particle B at that point, or would it be something else entirely??

You touch upon a lot of questions that bothered physicists in the 30-ies and 40-ies. The short answer is that a QUANTUM FIELD solves these issues (more or less :-)