Lecture%2009

Lecture%2009 - EEE 352 Lecture 9 Examples of the Schrdinger...

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1 EEE 352: Lecture 9 Last time we had a barrier. What happes with 2 barriers? Quantum Confinement and Energy Quantization * Electron in a square potential well Energy quantization Wavefunctions Ψ = Ψ + Ψ E V m ) ( 2 2 2 r h x X = 0 V 0 →∞ X = L ? Examples of the Schrödinger Equation—Applications The Quantum Well Electron in a Box ¾ 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 ¾ 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. Electron in a Box ¾ 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 K , 3 , 2 , 1 , 2 = = n n L λ 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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2 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 ] 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: 0 ) ( 0 ) 0 ( = = L ψ We now can solve the Schrödinger equation in the central region. We must first set up the problem: regions of interest and boundary conditions: Electron in a Box x X = 0 V 0 →∞ X = L 2 2 2 2 2 2 2 2 2 , 0 ) ( ) ( ) ( ) ( 2 h h mE k x k dx x d x E dx x d m = = + = ) cos( ) sin( ) ( kx B kx A x + = ψ (0) = 0 forces B = 0 as the only possible value consistent with the boundary conditions. ψ ( L ) = 0 leads to kL = n π Electron in a Box x X = 0 V 0 →∞ X = L Earlier, we used waves So, if we write the wave as ikx e ± () ) sin( ) cos( ) sin( ) cos( ) ( 2 1 2 1 2 1 kx A kx B kx C C i kx C C e C e C x ikx ikx + = + + = + = We can also solve the problem directly: ) sin( 2 ) sin( 2 0 ) ( 0 0 ) 0 ( 1 1 2 1 2 1 kL A kL iC e e C L C C C C ikL ikL = = = = = + =
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3 Electron in a Box x X = 0 V 0 →∞ X = L = = = L x n A x L n k n kL n n π ψ sin ) ( L A L A dx L x n A dx x n n L n L n 2 2 sin 1 ) ( 2 0 2 2 0 2 = = = = We normalize the wave functions as: http://www.nanohub.org The density is the magnitude squared of the wave function (the probability density ): Since this is normalized, = L x n L x 2 2 sin 2 ) ( = = = = L L L dx L x n L dx L x
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Lecture%2009 - EEE 352 Lecture 9 Examples of the Schrdinger...

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