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Roger's Equations

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The Hamiltonian - Quantum Mechanics

Posted April 09, 2007 10:19 AM by Bayes

Classical Lagrangian and Hamiltonian

The Lagrangian and the Hamiltonian exploit the principle of least action to determine the equations of motion for a system from its energy. The Lagrangian is expressed in terms of velocity, position, and time. The Hamiltonian is expressed in terms of momentum, position, and time. Both the Lagrangian and the Hamiltonian are functions that track the Kinetic Energy and Potential Energy of a closed system.

Presumably this means that if we could account for all the energy in the universe, we would get equations of motion for everything. Although completely impractical, this does have the important philosophical implication of a deterministic universe.

Quantum Mechanics

My purpose is not to present a lesson plan for Quantum Mechanics. I don't want to get into the history of the theory as it is outside of the scope of what I'm trying to do here. I also don't want to give a lesson on the abstract math involved. I don't care if you know commutation relations or Hilbert space. All I want to do is try an explain why there is quantum mechanics in the first place. What problem was so terrible to create such a solution.

The problem with quantum mechanics, and there really is only one problem when you get right down to it, is that you can't have exact position and exact momentum of a particle system exactly at the same time. Notice I didn't say "know the position and momentum of a particle at the same time", that's because it is more fundamental than that. Again, a particle does not have exact position and exact momentum. That's it, period. That's where all the trouble starts and ends. Why is that such a big deal? Lets go back to our classical Hamiltonian to find out.

The classical Hamiltonian uses the principle of least action to predict the equations of motion for a closed system from the energy of a system. The energy of a closed system consists of Kinetic Energy (Energy of Motion) and Potential Energy (Energy of Position). The Kinetic Energy depends on momentum and the Potential Energy depends on position. So if you don't have exact position or exact momentum, you can't express the energy of a system exactly, which means you can't determine the equations of motion exactly.

However all is not lost, for although there is no exact momentum or exact position, the exactness of position and the exactness of momentum are related. The more precisely you know position, the less precisely you know momentum, and vice versa. Taken to extremes, if you know the position of a particle exactly, the momentum could be anything, infinity, zero, who knows. Conversely, if you know momentum of a particle exactly, the particle could be anywhere, 5 cm away or 100,000 light years away, who knows. Please understand, the particle in question is in both scenarios above at the same time. Actually, it's both scenarios and all those that exist in-between (almost knowing position exactly/barely knowing momentum at all, somewhat knowing position/somewhat knowing momentum, barely knowing position at all/almost knowing momentum exactly). This is expressed by the Uncertainty Principle:

Δp ≥ (h/(4πΔx))

where h is Planck's constant, Δx is the uncertainty of position and Δp is the uncertainty of momentum. Notice the smaller Δp gets, the larger Δx must become to maintain the inequality.

or more generally expressed as:

Wave-functions

So now we have the terrible scenario where we don't know the position or momentum of a system exactly. The bright side is we do know the relation between the uncertainty of position and the uncertainty of momentum, that's something at least. Perhaps we can use the uncertainty relation above to express the system in a new way, one which doesn't need exact position and momentum. But how to find it? Lets go with what we know. The classical Hamiltonian of a single particle in a potential is:

KE = 1/2 mv2 = p2/2m (since p=mv)
PE = position dependent potential, lets call it V(x)

H = KE + PE = p2/2m + V(x)

Remember that the Hamiltonian is the total energy of a system (classically). So

H=E

so the classical expression for total energy is

p2/2m + V(x) = Etot

And now, exploiting the uncertainty relation, we find a relation between momentum and position is (for more details please see J. Phys. A 35 (2002) 3289-3303):

p=-(ih/2π)(d/dx)

p2= -(h2/4π2)(d2/dx2)

substituting we get

[-(h2/8mπ2)(d2/dx2) + V(x)] = Etot

[-(2/2m)(d2/dx2) + V(x)] = Etot

At this point we have a serious problem. In classical physics, equations of motion inherently assume exact knowledge of position and momentum. Think about it, the equations of motion will have and x and p in them, the idea being you put a number in for x and you'll get a solution for p at a given time t. This can't work anymore since we can't know position and momentum exactly at the same time. What we need is an expression of the state of the system that doesn't require exact position or momentum. We call this the wave-function.

Calling the wave-function ψ, we rewrite the Hamiltonian expression with the wave-function included.

Hψ=Eψ

Note at this point ψ can be anything depending on what the system is. Expanding the H we get

[-(2/2m)(d2/dx2) + V(x)]ψ(x) = EtotΨ(x)

notice that we are now saying ψ is a function of position, that is because this is a quantity I'm interested in at the moment. If I wanted to know about the momentum of the system, I would express everything in terms of p. This is simply your choice of basis.

Simplifying

[-(2/2m)(d2/dx2)] Ψ(x)

= Etot - V(x) Ψ(x)
(d2/dx2)Ψ(x) = -(2m/2)(Etot - V(x)) Ψ(x)

we get a differential equation in terms of ψ(x). If we know the potential involved, we can solve for ψ(x).

What is a Wave-function

Now that we have the wave-function of a system, its important we figure out what it is we have. Simply put, the wave-function is an expression of all the possible states of a closed system defined by the Hamiltonian. A general solution for all possible energies. If you pick a particular energy, say the lowest (ground state), the corresponding wave-function ψ0(x) when taken as the modulus squared |ψ0(x)|2 expresses the probability of the position of the particle(s) in the closed system.

So you now can find the probability that the particle is in a certain location.

Or if you choose to express the wave-function in the momentum basis ψ(p), you can find the probability that the particle(s) have a certain momentum.

Notice the result is not an exact position or exact momentum, so the uncertainty principal is satisfied, yet since we are dealing with probabilities, so we can make predictions which can be tested in the real world.

Normalization

To have those predictions make more sense, we should probably normalize the wave-function for convenience. Since the wave-function is a complete description of all the possible states of a system with an associated probability densities, it is convenient that we make it so that if we add all of the probabilities up together for a system that the sum would equal one. That way if for a particular x you get a probability of .45, you know the chance of the particle being in that spot is 45% since .45/1 = 45%. Makes life a lot easier.

Think of it this way, the wave-function tells you the relative likelihood of each possible state of a system. If we set the total equal to one, that relativeness is easier to express and understand. That's what normalization does.

It's important to note that the wave-function described above is just a round about way to express a system at a moment in time. In order to see how a system changes in time, one must apply something else to the wave-function.

Conclusions

Notice above we haven't even brought in the principle of least action. That's because time must go by in order for the principle to take effect and everything we've done to this point is without regard for time. Everything we've done has been just an effort to find a new way to describe a system where you can't know position or momentum exactly. Once you have the wave-function and have normalized it, you must next apply different procedures in order to see how it changes in time, or changes when suddenly perturbed, etc.

Next time I'll talk about how a wave-function changes in time and we'll get into the principal of least action in quantum mechanics as presented by Richard Feynman.

Special thanks to the following websites:

http://en.wikipedia.org/wiki/Uncertainty_principle
http://hyperphysics.phy-astr.gsu.edu/hbase/uncer.html
http://hyperphysics.phy-astr.gsu.edu/hbase/quantum/schr.html

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Anonymous Poster
#1

Re: The Hamiltonian - Quantum Mechanics

04/10/2007 1:25 AM

Sorry, but i have to ask the stupid questions. if you know momentum and direction, you could calculate position. right? so it is a question of direction which is undeterminable? in quantum mechanics as in any other electromagnetic field, wouldn't it be logical to calculate direction according to the right hand rule due to the electromagnetic field? by using past position and present, the direction could be determined this way by calculating the lines of flux within the system?

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#2
In reply to #1

Re: The Hamiltonian - Quantum Mechanics

04/10/2007 5:33 AM

Roger, nice work, as usual.

Guest, how do you mean if you know momentum and direction, you know position?

All you know is the particle's momentum (or velocity) vector, with no idea where it is. That vector can be anywhere in the universe!

Jorrie

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The Engineer
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#3
In reply to #1

Re: The Hamiltonian - Quantum Mechanics

04/10/2007 8:17 AM

You Wrote: "if you know momentum and direction, you could calculate position. right?"

Actually, no you can't. momentum (p) is a first derivative of position with respect to time.

p=mv
p=m(dx/dt)

Just because you know dx/dt doesn't mean you know x.

To put it more clearly, say you had an object that wasn't moving. Since v=0, p=mv=0. So tell me, where is the object in space? Based on your statement, you should know since you know the momentum but as you can see, knowing momentum doesn't tell you the location.

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#4

Re: The Hamiltonian - Quantum Mechanics

04/10/2007 8:46 AM

Nice work, Roger!

(Btw, ever try to do business with a state agency? They also exploit the Principle of Least Action. )

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The Engineer
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#5
In reply to #4

Re: The Hamiltonian - Quantum Mechanics

04/10/2007 10:52 PM

If only they did, in my experience they're as close to an "immovable object" as I've ever seen.

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