lecture 7

# lecture 7 - EEE 352: Lecture 7 Electron Motion and...

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EEE 352: Lecture 7 Electron Motion and Probability Velocity * Propagation of a wave packet expansion of amplitude and phase recognition of current Classical analogy Ernst Madelung David Bohm

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T = k 2 k 1 ψ 2 inc 2 = k 2 k 1 C A 2 = 4 k 1 k 2 k 1 + k 2 ( ) 2 Why did we put this here? This is because both the reflection coefficient and the transmission coefficient are defined in terms of the probability current. Last time, we introduced the idea of probability current. We need to pursue this a little further in order to garner the ideas of current in quantum mechanics.
Now, we normally think of velocity as the description of the motion of a particle. Here, we are talking about waves. Previously, we connected the particle velocity to the group velocity of the wave, and used the momentum operator from which we can infer Hence, the momentum is an operator (it defines a differential operation on the wave function). x e i ( kx ω t ) ( ) = ike i ( kx t )

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In a similar manner, we could extract the position operator as and, once again, we find that this is a differential operator. We will see later that, since x and k are Fourier transform pairs, we can work in a position representation or a momentum representation, such that k e i ( kx ω t ) ( ) = ixe i ( kx t ) x i k
OK. That’s all well and good, but what do we mean by a “representation.” We have written the Schrödinger equation as Everything here is a function of position! Hence, we call this the “position representation.” We work in the position representation most of the time because it is hard to describe the potential energy (and the barriers it represents) in momentum space. Rather than Fourier transform the above equation, we begin with our energy equation as where 2 k 2 2 m ϕ ( k ) + d k V ( k k ) ( k ) = E ( k )

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ϕ ( k ) = 1 2 π Ψ ( x ) e ikx dx V ( k ) = 1 2 V ( x ) e ikx dx are the Fourier transforms of the wave function and the potential. If we now replace the k 2 bits with the derivatives, we have Now, we have the equation in the “momentum representation.” This is inconvenient because of the convolution integral on the potential term. 2 2 m d 2 dk 2 ( k ) + d k V ( k k ) ( k ) = E ( k )
By Fourier transform pairs, we mean that the two wave functions (one in the position representation and one in the momentum representation ) are related by a Fourier transform:

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an operator but, if we are in the momentum representation, then it is the position which is an operator
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## This note was uploaded on 09/28/2009 for the course EEE 352/333 taught by Professor Allee during the Fall '09 term at ASU.

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lecture 7 - EEE 352: Lecture 7 Electron Motion and...

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