I.
THE FREE PARTICLE 040830
I think what Griffiths says about the non-normalizable wave-functions that occur for a free particle is a bit unfortunate and misleading. Below you find my view which I think is
1) simpler, 2) more general and 3) more standard.
A.
The Problem
The wave function for a free particle with momentum p =
~ k is
ψ ( x ) = Ae ikx
, (1) where A is a constant and k is any real number. (We consider the spatial part of the wave function here, the time dependence is obtained simply by multiplying with e
− iEt/ ~ , where, for a free particle, E =
~
2 k
2
/ 2 m .) These wave functions have definite momentum and energy, ie , σ p
= σ
E
= 0 and hence are the natural candidates to describe a free particle with definite energy and momentum. However, there is one problem—they are non-normalizable:
R
∞
−∞
ψ ( x ) | 2 dx = ∞ for any (non-zero) A . Hence, it is impossible to interpret | ψ ( x ) | 2 dx as the probability to find the particle in [ x, x + dx [.
B.
The Solution
We have already (almost) encountered the solution to this problem when we solved a particle in a box with length a and obtained the wave functions ψ n
( x ) = p
2 /a sin( nπx/a )
(and ψ = 0 outside of the box). These wave functions represent a particle that moves freely in the box—but not in all of the universe since it is constrained to stay in the box by forces that act at the walls. However, it is pretty clear that when the box becomes very large, ie a becomes very large, the particle is not much affected by the walls — at least not when it moves in a region of the box far away from the walls. Thus it moves as a free particle.
The solutions ψ n have definite energy—they are stationary states—but they do not (quite) have definite momentum since sin kx =
1
2 i
( e ikx − e
− ikx ). Hence, ψ n contains two momentum states ±
~ k — ψ n is a standing wave which is a superposition of two waves that move in opposite directions. Each of these waves are of the form (1), however they do not vanish at the walls and hence do not obey the necessary boundary conditions for a particle in a box.
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equation for a particle in a box since they do not have the correct boundary conditions— they do not vanish at the walls. However, let us modify the problem so that instead of looking at a particle moving in a box of length a we study a particle moving freely on a circle of length a . This means that we identify the points x = 0 and x = a . Since the wave function must be continuous this implies ψ ( a ) = ψ (0) and hence the wave function is a periodic function on the circle. Note that there is nothing outside of the interval 0 ≤ x ≤ a — we have identified the points 0 and a . The circle is our universe (or at least the universe of the particle) and the potential V vanishes everywhere.
tion for a free particle, are
ψ k
( x ) =
1
√ a e ikx
, where k = n 2 π a
, n = 0 , ± 1 , ± 2 , ± 3 .....
.
(2)
These states have definite energy and definite momentum and are normalized. Hence they describe a free particle with definite energy and momentum. Of course, it moves on a circle of finite length a and as a consequence the momenta p =
~ k and energies E =
~
2 quantized. Note that the wave function obeys ψ ∝ 1
√ k
2
/ 2 m are a and that the distance between nearby momenta is ∝ 1 /a and between energies ∝ 1 /a 2 . When a → ∞ , we obtain a continuum of momenta and energies. This describes a particle that moves freely in an infinitely large one-dimensional space. Note however that ψ → 0 in this limit—this is physically correct since the probability to find the particle is the same everywhere it has to go to zero when the space where the particle can be found goes to infinity. This is the problem we started with: the non-normalizability of the wave functions (1) over an infinitely large space. However, we have now defined this properly as a well-defined limit of a well-defined problem, namely the motion on a circle of radius a .
In the above, rather than looking at a particle moving on the real line we put the particle on a circle. The physical idea behind this is that if we are looking at a particle moving in a big space then it should not matter much what happens at a boundary that is far far away[1]. Thus when studying a free particle we may assume that it moves on a circle of length a .
Griffiths says that the solution is to form wave packets that are normalizable and describe a particle. Of course, this is correct and it is what one has to do to describe eg a particle
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that is localized in x -space. These wave packets will always contain different momentum states, by letting them become large in x -space one can of course make them arbitrarily narrow in p -space thus approaching states with definite momentum. However, what we have shown here is that there are well-defined states with definite momentum. They are non-normalizable, but if we put the system on a circle they become normalized.
C.
Fourier Analysis
Although ψ vanishes as a → ∞ , the functions e ikx play a very important role also in this limit—but now of course without the prefactor 1 /
√ a . They are basis functions in which any other function can be expanded f ( x ) = √
1
2 π
Z
∞
−∞
F ( k ) e ikx dk .
(3)
This is the Fourier decomposition of the function f ( x ) and the components F ( k ), which is called the Fourier transform of f , are obtained as
F ( k ) = √
1
2 π
Z
∞
−∞ f ( x ) e
− ikx dx .
(4)
Note that Griffiths of course uses the wave functions e ikx in this mathematical sense — it is just that he refuses to think of them as possible physical wave functions. But we have seen that they are almost OK. All we have to do is to remember that they are not normalizable— and hence we must be careful when interpreting them as probability densities. Fortunately, we can always make everything well-defined by letting space be a large circle.
D.
The Quantum Mechanical Formalism—A Comment on the Future
Also when discussing the formalism of quantum mechanics, we will see that it is useful to think of e ikx as physical states—it will bring out the unity of formalism. We will see that if we define
ψ k
( x ) =
1
√
2 π
~ e ikx
(5)
(in the limit a → ∞ where ψ → 0) then these wave functions obey a generalized orthonormality condition that make things look very similar to say particles in a box.
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E.
Comment of Griffiths
Griffiths says at the bottom of p. 60 ”A free particle cannot exist in a stationary state; or, to put it in another way, there is no such thing as a free particle with a definite energy.” similar statesments occur at other places in the book. From the discussion above, I hope it is clear that I find this misleading.
Anders Karlhede
[1] This is really an assumption and there are interesting exceptions to this.
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