Lecture 8

Lecture 8 - EEE 352: Lecture 8 Examples of the Schrdinger

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EEE 352: Lecture 8 Last time we had a barrier. What happens with 2 barriers? Quantum Confinement and Energy Quantization * Electron in a square potential well Energy quantization Wavefunctions x X = 0 V 0 X = L ? Examples of the Schrödinger Equation—Applications The Quantum Well
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Electron in a Box Having looked at the motion of FREE electrons we now consider what happens when we CONFINE electrons in a VERY SMALL region of space A quantum mechanical BOX The box we consider is formed from an INFINITELY DEEP potential well Consequently NO barrier penetration of the wavefunction occurs (so the electron is TRAPPED COMPLETELY within the box) L We will address this problem in one dimension only.
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Electron in a Box Since the wavefunction VANISHES in the barrier regions (this is because V , hence the damping is infinitely fast), we conclude that: * The ALLOWED electron wavelengths within the box are QUANTIZED * This is similar to the case of a vibrating STRING that is TIED at both ends (the allowed wavelengths in the string depend on the DISTANCE BETWEEN the two points where the string is tied) L IN A VIBRATING STRING TIED AT TWO ENDS A DISTANCE L APART ONLY A LIMITED SET OF VIBRATIONAL MODES ARE POSSIBLE THESE MODES HAVE QUANTIZED WAVELENGTHS THAT ARE GIVEN AS It will turn out that our problem is precisely the same, and that these quantized modes will produce a limited set of “allowed” energies for the wave.
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A vibrating string has quite similar behavior: (1) it is fixed at both ends, (2) the differential equation leads to modes , and (3) any real vibration may be Fourier transformed in terms of these modes. [ web ] The key to our quantization is that we treat the electrons as waves —then we really don’t have anything new or different. But, it is the waves that are everything.
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Electron in a Box x X = 0 V 0 X = L In the regions x < 0, x > L , V 0 , so that ψ 0 in these areas. This gives the two boundary conditions: We now can solve the Schrödinger equation in the central region. We must first set up the problem: regions of interest and boundary conditions:
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x X = 0 V 0 X = L ψ (0) = 0 forces B = 0 as the only possible value consistent with the boundary conditions. ψ
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Lecture 8 - EEE 352: Lecture 8 Examples of the Schrdinger

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