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Continuing from my first question titled, Simple Quantum Mechanics question about the Free particle, (part1)

Griffiths goes on and says,

"A fixed point on the waveform corresponds to a fixed value of the argument and hence to x and t such that,"


x $\pm$ vt = constant, or x = $\mp$ vt + constant


What does this mean? I am so confused.

He goes on and says that psi(x,t) might as well be the following:

$$\psi(x,t) = A e^{i(kx-(hk^2/2m)t}$$

because the original equation,

$$\psi(x,t) = A e^{ik(x-(\hbar k/2m)t)} + B e^{-ik(x + (\hbar k/2m)t)}$$

only differs by the sign in front of k. He then says let k run negative to cover the case of waves traveling to the left, $k = \pm \sqrt{(2mE)}/\hbar$.

Then after trying to normalize psi(x,t), you find out you can't do it! He then says the following,

"A free particle cannot exist in a stationary state; or, to put it another way, there is no such thing as a free particle with a definite energy."

How did he come to this logical deduction. I don't follow. Can someone please explain Griffith's statement to me?

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Take for example a sine wave at 90degrees as the point of the function. It means that: the function moves/ changes value in time, when you sit on a specific x. The 90 degree passes over you, As I said in the comment to the previous: plot. –  anna v May 19 '11 at 4:00
    
I think he's just saying that a maximum of a wave is moving much like the particle itself, so its group velocity may be interpreted as the particle's velocity. Of course, this is not the proper canonical way to deal with the observable called "velocity" in QM. –  Luboš Motl May 19 '11 at 4:03
    
The "A free particle cannot exist in a stationary state; or, to put it another way, there is no such thing as a free particle with a definite energy." seems to me wrong, as far as experimental observations go. Otherwise the LHC is an exercise in futility. He probably means " a free particle cannot be represented by a psi function of the plane wave type shown". This is true, one needs wavepackets to represent a free particle mathematically. –  anna v May 19 '11 at 5:20
    
p.s. to my above: The plane wave represented by the psi above, once it starts it goes on to infinity inexorably by the functional form as t is not stopped. Thus it cannot represent a particle, stationary or not. Particles are all around us. –  anna v May 19 '11 at 5:37

5 Answers 5

up vote 5 down vote accepted

The first part about velocities says that we're looking at a function

$$\psi(x,t) = \psi(u)$$

for $u=x-vt$. For example, $\psi = \cos(x-vt)$.

Now pick some fixed value for $\psi$, say $0.4$. Find a place where $\psi = 0.4$ such as $x=1.16$. If you let some small amount of time $\textrm{d}t$ go by, then look at $x=1.16$ again, $\psi$ has changed a little, but if you look at the point $x = 1.16 + v\textrm{d}t$, you will find $\psi = 0.4$ there, so we could say that the place where $\psi$ is $0.4$ is moving at speed $v$.

This is not the same as saying that the particle is moving at that speed. It is saying that $v$ is the phase velocity of the wave.

The second part about normalizability says that a wavefunction must be an element in a vector space called a Hilbert space (by physicists; I think mathematicians call it $L_2$). The Hilbert space consists of wavefunctions that are square-normalizable; you can square them, integrate from negative infinity to infinity, and get a finite value. Things that die off exponentially on their tails do this, for example.

The sine function doesn't die off exponentially, or at all. If you square it and integrate from negative infinity to infinity, you get something infinite. Thus, the sine wave doesn't represent a reasonable probability density function for the location of a particle. The particle would be "infinitesimally likely to be observed everywhere in an infinite region", which physically does not make sense. Instead, for a real particle, we must have a normalizable wavefunction. Since a free particle with definite energy would have a pure sinusoidal wavefunction, a free particle with definite energy is physically not possible.

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I submit that the correct statement should be: "particles cannot be represented by plane waves." Particles exist, they are free have a definite energy given by the momentum measurement within the Heisenberg uncertainty principle but they are represented by wave packets because they cannot be represented by plane waves from the reasoning above. It is putting the cart in front of the horse to talk of plane waves as free particles. –  anna v May 19 '11 at 5:46
    
@anna I'm not sure I understand completely. If the free particle has a definite momentum, its wavefunction is a plane wave because that is the eigenfunction of the momentum operator. No plane waves and no exact momentum then go hand and hand, whichever one comes first. –  Mark Eichenlaub May 19 '11 at 6:15
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Yes, true. I object to the statement that the plane wave could represent a free particle ever. We choose the representations to fit reality, and should not a priory reality to fit the representations: because there exists a plane wave solution it must represent a particle. –  anna v May 19 '11 at 6:25
    
@anna I think it's pretty standard to find a solution to an equation and then ask whether that solution is physically meaningful. That's all I was trying to do. –  Mark Eichenlaub May 19 '11 at 6:28
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Fine, I understand that. The statement in the book quoted by OP (btw what are the words from which OP is derived?) assumes that the plane wave describes a particle and concludes that a particle cannot be stationary, (which our experimental data manifestly say is wrong), instead of concluding that "this wavefunction cannot describe a physical particle". –  anna v May 19 '11 at 8:59

I would normally just add this as a comment, but I still can not comment.

I would say the author is just saying that because the arguments are effectively the same, and the modulus of each number is arbitrary, one can use the distributive law to simplify the expression. If one tries to normalize using standard procedure, one gets: $$A^2$$ If one does not specify finite boundary conditions, one must find the integral of this constant value from -infinity to infinity which gives an infinite value, and thus is not normalizable.

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Well here is a +1 so you can reach the 50 needed for comments. –  anna v May 19 '11 at 10:57
    
thanks...so cool! –  Unassuminglymeek May 19 '11 at 23:54

I will answer this part of the question since it is important to distinguish what we are doing as physicists.

"A free particle cannot exist in a stationary state; or, to put it another way, there is no such thing as a free particle with a definite energy."

As Mark says in his comment above:

it's pretty standard to find a solution to an equation and then ask whether that solution is physically meaningful.

Which is fine. When the answer is "no", as in this case, one goes ahead to find physically meaningful solutions that will describe a free particle.

There is ample experimental evidence that particles exist, from atomic physics to nuclear physics to particle physics. Solving Schroedinger's equation for a potential well not only gives well defined state functions, the solutions describe the behaviour of the particle quite well to excellently. One then solves the "no potential" equation and finds the plane waves .

First thought: no potential means free particle.

second thought: are these plane wave functions well behaved so that they can be used to describe quantum mechanically a free particle?

Third thought should be: No, they cannot be normalized to a probability function.

Fourth thought: Could one use these solutions as a complete set to find a function that can be normalized and its probability describes a physical free particle ?

Bingo: wave packets.

The statement "there is no such thing as a free particle with a definite energy" is wrong for a physical particle. It should be : plane waves cannot describe free particles.

Free particles ( experiment) trump plane waves ( theory).

You were right to be confused.

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There is also experimental evidence that the existence of particles is not a simple matter. Bell, Kochen-Specker, the usual suspects. At high energies we see events (macroscopic or mesoscopic thermodynamic transitions in meticulously tuned experimental apparatus) that are obviously linked together --having a common cause, one might say-- but at the lower energies of non Particle Physics, the statistics of events that are separated by relatively large distances are more effectively modeled by probabilities over field configurations [which, sorry, is overcomplicated relative to the Question]. –  Peter Morgan May 19 '11 at 13:07

There seems to be a difference of opinion, unexpressed, as to what we mean when we say 'free particle'. It has to mean, at the very least, that it does not interact with any potential energy term. We can, then, consider it as part of a closed system in which there is no potential energy, so its Hamiltonian has no potential energy term. In fact, we usually mean to say that its Hamiltonian has the usual form from quantising a classical free Hamiltonian, and it is clear that that is what Mr. Eichenlaub, and some other contributors, meant. (The web interface here for some ungodly reason turns my backslash into yen...so I merely refer to the Hamiltonian written above.)

Now no such particles exist in Nature for the trivial reason that there are, indeed potential energy terms all over the place, no particle is really isolated from the rest of the universe. So the question is about a hypothetical situation, and is a reasonable question.

A free particle cannot be in an eigenstate of the Hamiltonian because the Hilbert space of states does not possess any such eigenstates: this is simply another way of saying what Mr. Eichenlaub wrote, and it cannot be really criticised. The next step is that if it had a definite energy it would be an eigenstate, so that is why it cannot have a definite energy.

It is rather unreal to first criticise this line of reasoning by saying one must look at reality only, since really there are no free particles at all. But it is then positively inconsistent to go on and talk about experimentally observed free particles.

Obviously there can be particles which are approximately free, but then, they could have a wave function that was approximately a plane wave: i.e., a narrow-band superposition in which their position would have a very large variance (but not infinite), and so the probabilities of finding it in any very small location would be practically zero (but not exactly zero). What would be the limits of this approximation? We would have to be justified in neglecting the potential energy, of which there are chiefly two kinds to worry about : forces exerted by other particles, and gravity. So if the particle were far away from all other particles, this might be justified. But the larger the variance of its position, the harder it is to arrange this...

I know that the practicalities of the existence of a free particle were not what the questioner was asking about, but anyone who wishes to make reality trump the formulation of a simple, sensible question about the free Hamiltonian is obligated to come to grips with reality and analyse the limits of this kind of approximation. If my tentative analysis in the previous paragraph is correct, this kind of approximation is justified when we are dealing with a more or less monochromatic wave....

The situation of a monochromatic wave is not what you might think. It is stationary, so nothing is moving. When regarded as a wave, it doesn't change its position, it is not moving that way. When regarded as a particle, it is not moving either. That is the whole point of being in a stationary state... and is the difference between phase velocity and particle velocity.

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It is the time-independent Schroedinger equation $$\frac {d^2 \psi} {d x^2} = - k^2 \psi$$ where $k^2 = \frac {2mE} {\hbar^2}$ that quantum mechanics uses to describe a free particle, not any other equation. Said equation has the following function as the solution: $$ \psi(x) = A e^{ikx} + Be^{-ikx}$$ which is a plane wave and not a wave packet. If the solution of the above equation were not a plane wave but a wave packet then it should be clearly shown and plane waves should not be discussed at all. A theory, however, should have a predictive quality. Quantum mechanics predicts the state of a free particle to be described by a plane wave (because, as said, that is what the solution of Schrodinger equation is) but that turns out to be non-physical (does not comply with experiment) for reasons already discussed. This is a major deficiency of quantum mechanics, along with a number of other deficiencies, which should be clearly spelled out and not sidetracked by offering constructs unrelated to Schrodinger's equation. Any comments?

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"Quantum mechanics predicts the state of a free particle to be described by a plane wave" is simply wrong. Plane waves form a (note the indefinite article here) basis for the solution space, but that is very much a different thing from saying that QM describes particles as plane waves. Wave packets (which are linear combination of plane waves, and so are also in the solution space) are the physically realized solutions. –  dmckee Jun 28 '13 at 17:50
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-1. Please stop talking about deficiencyies in QM.! It is spam already. –  Dimensio1n0 Jun 29 '13 at 13:03

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