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2. Solving Schrödinger’s Equation Superposition Given a few solutions of Schrödinger’s equation, we can make more of them Let 1 and 2 be two solutions of Schrödinger’s equation Let c 1 and c 2 be any pair of complex numbers Then the following is also a solution of Schrödinger’s equation We can generalize this to an arbitrary number of solutions: We can even generalize to a continuum of solutions:

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2A. The Free Schrödinger’s Equation Consider the free Schrödinger equation: We know a lot of solutions of this equation: We know a lot of solutions By combining these, we can make a lot more solutions* *The factor of (2 ) 3/2 is arbitrary and inserted here for convenience. In 1D, it would be (2 ) 1/2 The function c(k) can be (almost) any function of k This actually includes all possible solutions

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We can (in principle) always solve this problem Goal: Given (r,t = 0) = (r), find (r,t) when no potential is present This just says that c(k) is the Fourier transform of (r) Substitute it in the formula above for (r,t) and we are done Actually doing the integrals may be difficult Set t = 0:

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Sample Problem A particle in one dimension with no potential has wave function at t = 0 given by What is the wave function at arbitrary time? Need 1D versions of these formulas: From Appendix: Find c(k):

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Sample Problem (2) A particle in one dimension with no potential has wave function at t = 0 given by What is the wave function at arbitrary time? Now find (x,t):

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2B. The Time Independent Schrödinger Eqn. Suppose that the potential is independent of time: Separation of Variables Conjecture solutions of the form: Substitute it in Divide by (r) (t) Left side is independent of r Right side is independent of t Both sides are independent of both! Must be a constant. Call it E

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Solving the time equation The first equation is easy to solve Integrate both sides By comparison with e -i t, we see that E = is the energy Substitute it back in:

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Given a potential V(r) independent of time, what is most general solution of Schrödinger’s time-dependent equation? First, solve Schrödinger’s time-independent equation You should find many solutions n (r) with different energies E n Now just multiply by the phase factor Then take linear combinations Later we’ll learn how to find c n The Time Independent Schrödinger Equation Multiply the other equation by (r) again: The Strategy for solving

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Why is time-independent better? Time-independent is one less variable – significantly easier It is a real equation (in this case), which is less hassle to solve If in one dimension, it reduces to an ordinary differential equation –These are much easier to solve, especially numerically

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2C. Probability current Recall the probability density is: This can change as the wave function changes Where does the probability “go” as it changes? –Does it “flow” just like electric charge does? Want to show that the probability moves around Ideally, show that it “flows” from place to place –A formula from E and M – can we make it like this? Probability Conservation To make things easier, let’s make all functions of space and time implicit, not write them

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Start with Schrödinger’s equation Multiply on the left by * : Take complex conjugate of this equation: Subtract: Rewrite first term as a total derivative Cancel a factor of i Left side is probability density The derivation (1)

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Consider the following expression: Use product rule on the divergence Substitute this in above Define the probability current j: Then we have: The derivation (2)

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Integrate it over any volume V with surface S Left side is P(r V) Use Gauss’s law on right side Change in probability is due to current flowing out Why is it called probability current? V j If the wave function falls off at infinity (as it must) and the volume V becomes all of space, we have

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This expression is best when doing proofs Note that you have a real number minus its complex conjugate A quicker formula for calculation is: Let’s find and j for a plane wave: Calculating probability current

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Sample Problem A particle in the 1D infinite square well has wave function For the region 0 < x < a. Find and j.

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Sample Problem (2) A particle in the 1D infinite square well has wave function For the region 0 < x < a. Find and j. In 1D:

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Sample Problem (3) A particle in the 1D infinite square well has wave function For the region 0 < x < a. Find and j. After some work …

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2D. Reflection from a Step Boundary The Case E > V 0 : Solutions in Each Region incident transmitted reflected I II A particle with energy E impacts a step-function barrier from the left: Solve the equation in each of the regions Assume E > V 0 Region I Region II Most general solution: –A is incident wave –B is reflected wave –C is transmitted wave –D is incoming wave from the right: set D = 0

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Step with E > V 0 : The solution Schrödinger’s equation: second derivative finite (x) and ’(x) must be continuous at x = 0 incident transmitted reflected I II We can’t normalize wave functions Use probability currents!

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Summary: Step with E > V 0 incident transmitted reflected I II

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incident reflected I II evanescent What if V 0 > E? Region I same as before Region II: we have Step with E < V 0 Most general solution: –A is incident wave –B is reflected wave –C is damped “evanescent” wave –D is growing wave, can’t be normalized (x) and ’(x) must be continuous at x = 0: No transmission since evanescent wave is damped

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Step Potential: All cases summarized For V 0 > E, all is reflected Reflection probability: R T Note that it penetrates, a little bit into the classically forbidden region, x > 0 This suggests if barrier had finite thickness, some of it would bet through

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2E. Quantum Tunneling Setting Up the Problem Barrier of finite height and width: Solve the equation in each of the regions Particle impacts from left with E < V 0 General solution in all three regions: x V(x)V(x) V0V0 - d/2 + d/2 I II III Why didn’t I include e -ikx in III ? Why did I skip letter E? Match and ’ at x = -d/2 and x = d/2 Solve for F in terms of A

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Skip this Slide – Solving for F in terms of A Multiply 1 by ik and add to 2 Multiply 3 by and add to 4 Multiply 3 by and subtract 4 Multiply 5 by 2 and substitute from 6 and 7

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Barrier Penetration Results We want to know transmission probability For thick barriers, Exponential suppression of barrier penetration

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Unbound and Bound State For each of the following, we found solutions for any E –No potential –Step potential –Barrier This is because we are dealing with unbound states, E > V( ) Our wave functions were, in each case, not normalizable Fixable by making superpositions: We will now consider bounds states These are when E < V( ) There will always only be discrete energy values And they can be normalized Usually easier to deal with real wave functions

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2F. The Infinite Square Well Finding the Modes Infinite potential implies wave function must vanish there In the allowed region, Schrödinger’s equation is just x V(x)V(x) a The solution to this is simple: Because potential is infinite, the derivative is not necessarily continuous But wave functions must still be continuous: 0

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Normalizing Modes and Quantized Energies We can normalize this wave function: Note that we only get discrete energies in this case Note that we can normalize these Most general solution is then

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The 3D Infinite Square Well a b c In allowed region: Guess solution: Normalize it: –This is product of 1D functions Energy is –This is sum of 1D energies

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2G. The Double Delta-Function Potential Finding Bound States Bound states have E < V( ) = 0 Within each region we have: x V(x)V(x) a/2-a/2 I II III General solution (deleting the parts that blow up at infinity): First, write out Schrödinger’s Equation:

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Dealing with Delta Functions To deal with the delta functions, integrate Schrödinger’s equation over a small region near the delta function: –For example, near x = +a/2 Do first term on right by fundamental theorem of calculus Do second term on right by using the delta functions V(x)V(x) a/2-a/2 I II III Take the limit 0 –Left side small in this limit

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Simplifying at x = ½a Since there is a finite discontinuity in ’, must be continuous at this boundary V(x)V(x) a/2-a/2 I II III On the right side of the equation above, is that I, II, or III ? Write these equation out explicitly: Substitute first into second:

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Repeating at x = – ½a Repeat the steps we did, this time at x = –½a Note these equations are nearly identical: The only numbers equal to their reciprocal are 1

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Graphical Solution Right side is two curves, left side is a straight line Right side, plus Right side, minus Left side Black line always crosses red curve, sometimes crosses green curve, depending on parameters –Sometimes two solutions, sometimes one Normalize to finish the problem Note one solution symmetric, one anti-symmetric

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