Lecture 10: The Schrödinger Equation Lecture 10, p 1

Lecture 10: The Schrödinger Equation Lecture 10, p 1

Overview Probability distributions Schrödinger s Equation Particle in a Bo Matter waves in an infinite square well Quantized energy levels y() U= n=1 n=3 U= 0 L n= Nice descriptions in the tet Chapter 40 Good web site for animations http://www.falstad.com/qm1d/ Lecture 10, p

Matter Waves - Quantitative Having established that matter acts qualitatively like a wave, we want to be able to make precise quantitative predictions, under given conditions. Usually the conditions are specified by giving a potential energy U(,y,z) in which the particle is located. Eamples: Electron in the coulomb potential produced by the nucleus Electron in a molecule Electron in a solid crystal Electron in a nanostructure quantum dot Proton in the nuclear potential inside the nucleus U() For simplicity, consider a 1-dimensional potential energy function, U(). Classically, a particle in the lowest energy state would sit right at the bottom of the well. In QM this is not possible. (Why?) Lecture 10, p 3

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Act 1: Classical probability distributions Start a classical (large) object moving in a potential well (two are shown here). At some random time later, what is the probability of finding it near position? U() Ball in a bo: U() Ball in a valley: E KE Total energy E = KE + U() E P() P() a b c a b c HINT: Think about speed vs position. Lecture 10, p 5

The Schrödinger Equation (SEQ) In 196, Erwin Schrödinger proposed an equation that described the time- and space-dependence of the wave function for matter waves (i.e., electrons, protons,...) There are two important forms for the SEQ. First we will focus on a very important special case of the SEQ, the time-independent SEQ. Also simplify to 1-dimension: y(,y,z) y(). d y ( ) U( ) y( ) Ey( ) m d This special case applies when the particle has a definite total energy (E in the equation). We ll consider the more general case (E has a probability distribution), and also D and 3D motion, later. h QM entities don t always have a definite energy. Time does not appear in the equation. Therefore, y(,y,z) is a standing wave, because the probability density, y(), is not a function of time. We call y(,y,z) a stationary state. Notation: Distinguish Y(,y,z,t) from y(,y,z). Lecture 10, p 6

Time-Independent SEQ What does the time-independent SEQ represent? It s actually not so puzzling it s just an epression of a familiar result: Kinetic Energy (KE) + Potential Energy (PE) = Total Energy (E) d y ( ) U( ) y( ) Ey( ) m d KE term PE term Total E term Can we understand the KE term? Consider a particle with a definite momentum. Its wave function is: y() cos(k), where p = h/l = ħk. dy d y p ksin ( k) k cos( k) y ( ) d d So, the first term in the SEQ is (p /m)y. Note that the KE of the particle depends on the curvature (d y/d ) of the wave function. This is sometimes useful when analyzing a problem. Lecture 10, p 7

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Particle Wavefunctions: Eamples What do the solutions to the SEQ look like for general U()? Eamples of y() for a particle in a given potential U() (but with different E): y() We call these wavefunctions eigenstates of the particle. y() These are special states: energy eigenstates. y() The corresponding probability distributions y() of these states are: y y y Key point: Particle cannot be associated with a specific location. -- like the uncertainty that a particle went through slit 1 or slit. Question: Which corresponds to the lowest/highest kinetic energy? The particle kinetic energy is proportional to the curvature of the wave function. Lecture 10, p 9

Probability distribution Difference between classical and quantum cases U() Classical (particle with same energy as in qunatum case) U() Quantum (lowest energy state state) E E P() In classical mechanics, the particle is most likely to be found where its speed is slowest P() = y In classical mechanics, the particle moves back and forth coming to rest at each turning point In quantum mechanics, the particle can be most likely to be found at the center. In quantum mechanics, the particle can also be found where it is forbidden in classical mechanics.

Solutions to the time-independent SEQ d y ( ) U( ) y( ) Ey( ) m d Notice that if U() = constant, this equation has the simple form: d y Cy ( ) d where m C ( U E ) is a constant that might be positive or negative. For positive C (i.e., U > E), what is the form of the solution? a) sin k b) cos k c) e a d) e -a For negative C (U < E) what is the form of the solution? a) sin k b) cos k c) e a d) e -a Most of the wave functions in P14 will be sinusoidal or eponential. Lecture 10, p 11

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Eample: Particle in a Bo As a specific important eample, consider a quantum particle confined to a region, 0 < < L, by infinite potential walls. We call this a one-dimensional (1D) bo. U = 0 for 0 < < L U = everywhere else U() We already know the form of y when U = 0: sin(k) or cos(k). However, we can constrain y more than this. 0 L The waves have eactly the same form as standing waves on a string, sound waves in a pipe, etc. The wavelength is determined by the condition that it fits in the bo. On a string the wave is a displacement y() and the square is the intensity, etc. The discrete set of allowed wavelengths results in a discrete set of tones that the string can produce. In a quantum bo, the wave is the probability amplitude y() ond the square y() is the probability of finding the electron near point. The discrete set of allowed wavelengths results in a discrete set of allowed energies that the particle can have. Lecture 10, p 13

Boundary conditions Standing waves A standing wave is the solution for a wave confined to a region Boundary condition: Constraints on a wave where the potential changes Displacement = 0 for wave on string E = 0 at surface of a conductor E = 0 If both ends are constrained (e.g., for a cavity of length L), then only certain wavelengths l are possible: n l f 1 L v/l L v/l 3 L/3 3v/L nl = L n = 1,, 3 mode inde L 4 L/ v/l n L/n nv/l

Boundary conditions We can solve the SEQ wherever we know U(). However, in many problems, including the 1D bo, U() has different functional forms in different regions. In our bo problem, there are three regions: 1: < 0 : 0 < < L 3: > L y() will have different functional forms in the different regions. We must make sure that y() satisfies the constraints (e.g., continuity) at the boundaries between these regions. The etra conditions that y must satisfy are called boundary conditions. They appear in many problems. Lecture 10, p 15

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Particle in a Bo (1) Regions 1 and 3 are identical, so we really only need to deal with two distinct regions, (I) outside, and (II) inside the well Region I: When U =, what is y()? U() d y m E U y ( ) ( ) ( ) 0 d I y I II I y I For U =, the SEQ can only be satisfied if: y I () = 0 0 L U = 0 for 0 < < L U = everywhere else Otherwise, the energy would have to be infinite, to cancel U. Note: The infinite well is an idealization. There are no infinitely high and sharp barriers. Lecture 10, p 17

Region II: When U = 0, what is y()? Particle in a Bo () d y m E U y ( ) ( ) ( ) 0 d d y ( ) me d y ( ) U() II y 0 L The general solution is a superposition of sin and cos: y ( ) B sink B cosk 1 where, k l Remember that k and E are related: E p k h m m ml because U = 0 B 1 and B are coefficients to be determined by the boundary conditions. Lecture 10, p 18

Particle in a Bo (3) U() Now, let s worry about the boundary conditions. Match y at the left boundary ( = 0). I II I Region I: y ( ) 0 I y I y II y I Region II: y II( ) B1sin k Bcos k 0 L Recall: The wave function y() must be continuous at all boundaries. Therefore, at = 0: y (0) y (0) I II ( ) B ( ) 0 B sin 0 cos 0 1 0 B because cos(0) = 1 and sin(0) = 0 This boundary condition requires that there be no cos(k) term! Lecture 10, p 19

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Particle in a Bo (4) Now, match y at the right boundary ( = L). At = L: This constraint requires k to have special values: k n y ( L) y ( L) n L I II 0 B sin 1 n 1,,... ( kl) Using k, we find : nl L l This is the same condition we found for confined waves, e.g., waves on a string, EM waves in a laser cavity, etc.: I y I U() II y II 0 L I y I n l (= v/f) 4 L/ 3 L/3 L 1 L For matter waves, the wavelength is related to the particle energy: E = h /ml Therefore Lecture 10, p 1

The Energy is Quantized Due to Confinement by the Potential E The discrete E n are known as energy eigenvalues : n n nl n L p h 1.505 ev nm m ml l n n h 8mL 1 where 1 E E n E electron n l (= v/f) E 4 L/ 16E 1 3 L/3 9E 1 L 4E 1 1 L E 1 Important features: Discrete energy levels. E 1 0 Standing wave (±p for a given E) n = 0 is not allowed. (why?) an eample of the uncertainty principle U = E n U = n=3 n= n=1 0 L Lecture 10, p

Quantum Wire Eample An electron is trapped in a quantum wire that is L = 4 nm long. Assume that the potential seen by the electron is approimately that of an infinite square well. 1: Calculate the ground (lowest) state energy of the electron. : What photon energy is required to ecite the trapped electron to the net available energy level (i.e., n = )? U= E n U= n=3 n= n=1 0 L The idea here is that the photon is absorbed by the electron, which gains all of the photon s energy (similar to the photoelectric effect). Lecture 10, p 3

Net Lectures Normalizing the wavefunction General properties of bound-state wavefunctions Finite-depth square well potential (more realistic) Lecture 10, p 4