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Next: Sum over paths picture Up: Electron diffraction Previous: Electron diffraction

Particle-wave picture

The first, proposed by Clinton Davisson and Lester Germer in 1927, was based on a hypothesis put forth earlier by Louis de Broglie in 1922. De Broglie suggested that if waves (photons) could behave as particles, as demonstrated by the photoelectric effect, then the converse, namely that particles could behave as waves, should be true. He associated a wavelength $\lambda$ to a particle with momentum $p$ using Planck's constant as the constant of proportionality:

\begin{displaymath}
\lambda = {h \over p}
\end{displaymath}

which is called the de Broglie wavelength. The Davisson-Germer experiment, which produced an electron diffraction pattern from electrons scattered off a nickel crystal, confirmed the de Broglie hypothesis. However, it was not until 1961 that an experiment in which electrons impinged on a slit apparatus was performed by Claus Jönsson. This was a five-slit set-up. The double-slit experiment was finally performed in the 1970s by Pier Giorgio Merli, Giulio Pozzi and GianFranco Missiroli. The point is, however, that through such experiments, the idea that electrons can behave as waves, creating interference patterns normally associated with light, is now well-established. The fact that particles can behave as waves but also as particles, depending on which experiment you perform on them, is known as the particle-wave duality.



In the following, we give a brief discussion of where the de Broglie hypothesis comes from. From the photoelectric effect, we have the first part of the particle-wave duality, namely, that electromagnetic waves can behave like particles. These particles are known as photons, and they move at the speed of light. Any particle that moves at or near the speed of light has kinetic energy given by Einstein's special theory of relatively. In general, a particle of mass $m$ and momentum $p$ has an energy

\begin{displaymath}
E = \sqrt{p^2 c^2 + m^2 c^4}
\end{displaymath}

Note that if $p=0$, this reduces to the famous rest-energy expression $E=mc^2$. However, photons are massless particles that always have a finite momentum $p$. In this case, Einstein's formula becomes $E=pc$. From Planck's hypothesis, one quantum of electromagnetic radiation has energy $E=h\nu$. Thus, equating these two expressions for the kinetic energy of a photon, we have

\begin{displaymath}
h\nu = {hc \over \lambda} = pc
\end{displaymath}

Solving for the wavelength $\lambda$ gives

\begin{displaymath}
\lambda = {h \over p}
\end{displaymath}

Now, this relation pertains to photons not massive particles. However, de Broglie argued that if particles can behave as waves, then a relationship like this, which pertains particularly to waves, should also apply to particles. Hence, we associate a wavelength $\lambda$ to a particle that has momentum $p$, which says that as the momentum becomes larger and large, the wavelength becomes shorter and shorter. In both cases, this means the energy becomes larger. That is, short wavelength and high momentum correspond to high energy.



If particles can behave as waves, then we need to develop a theory for this type particle-wave. We will do this in detail when we study the Schrdinger wave equation. For now, suffice it to say that the theory of particle-waves has some aspects that are similar to the classical theory of waves, but by no means can a classical wave theory, like that used to describe waves on a string on the surface of a liquid, be used to formulate the theory of particle waves. To begin with, what is the very nature of a particle wave? Here, we will give only a brief conceptual answer.



A particle-wave is still described by some kind of amplitude function ${\cal A}(x,t)$, but this amplitude must be consistent with the fact that we could, in principle, design an experiment capable of measuring the particle's spatial location or position. Thus, we seem to have arrived at a paradoxical situation: The electron diffraction experiment tells us that particle-waves can interfere with each other, yet it must also be possible to measure a particle-like property, the position, via some kind of actual experiment. The resolution of this dilemma is that the particle exhibits wave-like behavior until a measurement is performed on it that is capable of localizing it at a particular point in space. Making this leap, however, has a profound implication, namely, that the outcome of the position measurement will not be the same in each realization even if it is performed in the same way. The reason is that if it did yield the same result, then we could say that the particle was evolving in a particular way that would put it there when the measurement was made, and this negates the possibility of its ever having a wave-like character, since this is exactly how classical particles behave. Thus, if the result of a position measurement can yield different outcomes, then the only thing we can predict is the probability that a given measurement of the position yields a particular value. The quantum world is not deterministic but rather intrinsically probabilistic.



If we are only able to predict the probability that a measurement of position will yield a particular value, then how do we obtain this prediction. Recall that the particle-wave is described by an amplitude function ${\cal A}(x,t)$. Let us suppose that the particle-wave reaches the screen at time $T$ when the amplitude is ${\cal A}(x,T)$, which we denote as $A(x)$ (when $T$ is fixed, the amplitude is a function of $x$ alone). Here, $x$ denotes the position along the screen. The screen, itself, acts as an apparatus for measuring the particle's position. The amplitude $A(x)$ can be either positive or negative, and we could even choose it such that it is a complex number. Thus, $A(x)$ is not a probability because it is not positive-definite, and it is not real. However, we can turn $A(x)$ into a probability by taking the square magnitude of $A(x)$. We define a new function

\begin{displaymath}
P(x) = \vert A(x)\vert^2 = A^*(x)A(x)
\end{displaymath}

where $A^*(x)$ denotes the complex conjugate of $A(x)$. The function $P(x)$ is positive-definite, and if $A(x)$ is appropriately chosen, $P(x)$ will satisfy the normalization condition

\begin{displaymath}
\int_a^b P(x)dx = 1
\end{displaymath}

where $a$ and $b$ denote the endpoints of the screen. $P(x)$ is an example of a probability density or probability distribution function. Because $x$ is continuous, we cannot define a discrete probability since the probability to be at any particular value of $x$ is zero (there are an infinite number of values for $x$). What $P(x)$ tells us is the probability to measure the particle's position on the screen within a small interval $dx$. In particular, the probability that a measurement of position yields a value in an interval $dx$ centered on the point $x_0$ is $P(x_0)dx$. The amplitude $A(x)$ is called a probability amplitude.



So why does an interference pattern arise? Because we do not observe the particle's position until it reaches the screen, we have to consider two possibilities for the particle-wave: passage through the uppoer slit and passage through the lower slit, and we assign to one possibility a probability amplitude $A_1(x)$ and other an amplitude $A_2(x)$. Just as with ordinary waves, we must add the amplitudes to obtain the total wave amplitude, to the total amplitude is $A_{\rm tot}(x) = A_1(x) + A_2(x)$. The probability that we observe something at a given point $x_0$ on the screen in an interval $dx$ is

$\displaystyle P(x_0)dx$ $\textstyle =$ $\displaystyle \vert A_{\rm tot}(x_0)\vert^2 dx$  
       
  $\textstyle =$ $\displaystyle \vert A_1(x_0) + A_2(x_0)\vert^2 dx$  
       
  $\textstyle =$ $\displaystyle \left(A_1^*(x_0) + A_2^*(x_0)\right)\left(A_1(x_0) + A_2(x_0)\right)dx$  
       
  $\textstyle =$ $\displaystyle \vert A_1(x_0)\vert^2 + \vert A_2(x_0)\vert^2 +
\left(A_1^*(x_0)A_2(x_0) + A_1(x_0)A_2^*(x_0)\right)dx$  

If the two possibilities were completely independent, then their corresponding probabilities would simply add, and we we just have the first two terms in the last expression. However, there is a cross term that is generally nonzero, and it is this cross term that gives rise to the inteference between the two possibilites, which leads to the observed interference pattern. Let us look at this term in greater detail. Recall that a complex number $A = a+ib$ can be expressed as

\begin{displaymath}
A = \vert A\vert e^{i\theta}
\end{displaymath}

where $\vert A\vert$ is the magnitude of the number

\begin{displaymath}
\vert A\vert = \sqrt{a^2 + b^2}
\end{displaymath}

and $\theta$ is the phase of the number

\begin{displaymath}
\theta = {\rm tan}^{-1}{b \over a}
\end{displaymath}

Letting

\begin{displaymath}
A_1 = \vert A_1\vert e^{i\theta_1},\;\;\;\;\;\;\;\;\;\;
A_2 = \vert A_2\vert e^{i\theta_2}
\end{displaymath}

so that

\begin{displaymath}
A_1^* = \vert A_1\vert e^{-i\theta_1},\;\;\;\;\;\;\;\;\;\;
A_2^* = \vert A_2\vert e^{-i\theta_2}
\end{displaymath}

and substituting into the probability distribution expression, we obtain
$\displaystyle P(x_0)$ $\textstyle =$ $\displaystyle \vert A_1\vert^2 + \vert A_2\vert^2 +
\left(\vert A_1\vert e^{-i\...
...^{i\theta_2} +
\vert A_1\vert e^{i\theta_1}\vert A_2\vert e^{-i\theta_2}\right)$  
       
  $\textstyle =$ $\displaystyle \vert A_1\vert^2 + \vert A_2\vert^2 +
\vert A_1\vert\vert A_2\vert\left(e^{i(\theta_2-\theta_1)} + e^{-i(\theta_2-\theta_1)}\right)$  

Recognizing that

\begin{displaymath}
e^{ix} + e^{-ix} = 2\cos(x)
\end{displaymath}

the probability finally becomes

\begin{displaymath}
P(x_0) = \vert A_1(x_0)\vert^2 + \vert A_2(x_0)\vert^2
+ 2\v...
...vert A_2(x_0)\vert\cos\left(\theta_2(x_0)-\theta_1(x_0)\right)
\end{displaymath}

The oscillatory nature of the cosine is suggests of the oscillating pattern of bright and dark fringes in the interference pattern.


next up previous
Next: Sum over paths picture Up: Electron diffraction Previous: Electron diffraction
Mark E. Tuckerman 2011-12-12