# ch04 - CHAPTER 4 1 The solution to the left side of the...

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CHAPTER 4. 1. The solution to the left side of the potential region is ψ ( x ) = A e ikx + B e - ikx . As shown in Problem 3-15, this corresponds to a flux j ( x ) = h k m | A | 2 - | B | 2 ( 29 The solution on the right side of the potential is ψ ( x ) = C e ikx + D e - ikxx , and as above, the flux is j ( x ) = h k m | C | 2 - | D | 2 ( 29 Both fluxes are independent of x . Flux conservation implies that the two are equal, and this leads to the relationship | A | 2 + | D | 2 = | B | 2 + | C | 2 If we now insert C = S 11 A + S 12 D B = S 21 A + S 22 D into the above relationship we get | A | 2 + | D | 2 = ( S 21 A + S 22 D )( S 21 * A * + S 22 * D * ) + ( S 11 A + S 12 D )( S 11 * A * + S 12 * D * ) Identifying the coefficients of | A | 2 and | D | 2 , and setting the coefficient of AD* equal to zero yields | S 21 | 2 + | S 11 | 2 = 1 | S 22 | 2 + | S 12 | 2 = 1 S 12 S 22 * + S 11 S 12 * = 0 Consider now the matrix S tr = S 11 S 21 S 12 S 22 The unitarity of this matrix implies that

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S 11 S 21 S 12 S 22 S 11 * S 12 * S 21 * S 22 * = 10 01 that is, | S 11 | 2 + | S 21 | 2 = | S 12 | 2 + | S 22 | 2 = 1 S 11 S 12 * + S 21 S 22 * = 0 These are just the conditions obtained above. They imply that the matrix S tr is unitary, and therefore the matrix S is unitary. 2. We have solve the problem of finding R and T for this potential well in the text.We take V 0 < 0. We dealt with wave function of the form e ikx + Re - ikx x <- a Te ikx x a In the notation of Problem 4-1, we have found that if A = 1 and D = 0, then C = S 11 = T and B = S 21 = R .. To find the other elements of the S matrix we need to consider the same problem with A = 0 and D = 1. This can be solved explicitly by matching wave functions at the boundaries of the potential hole, but it is possible to take the solution that we have and reflect the “experiment” by the interchange x - x . We then find that S 12 = R and S 22 = T . We can easily check that | S 11 | 2 + | S 21 | 2 = | S 12 | 2 + | S 22 | 2 = | R | 2 + | T | 2 = 1 Also S 11 S 12 * + S 21 S 22 * = TR * + RT * = 2R e ( TR * ) If we now look at the solutions for T and R in the text we see that the product of T and R* is of the form (-i) x (real number), so that its real part is zero. This confirms that the S matrix here is unitary. 3. Consider the wave functions on the left and on the right to have the forms ψ L ( x ) = Ae ikx + Be - ikx ψ R ( x ) = Ce ikx + De - ikx Now, let us make the change k - k and complex conjugate everything. Now the two wave functions read
ψ L ( x ) ' = A * e ikx + B * e - ikx ψ R ( x ) ' = C * e ikx + D * e - ikx Now complex conjugation and the transformation k - k changes the original relations to C * = S 11 * ( - k ) A * + S 12 * ( - k ) D * B * = S 21 * ( - k ) A * + S 22 * ( - k ) D * On the other hand, we are now relating outgoing amplitudes C *, B * to ingoing amplitude A *, D *, so that the relations of problem 1 read C * = S 11 ( k ) A * + S 12 ( k ) D * B * = S 21 ( k ) A * + S 22 ( k ) D * This shows that S 11 ( k ) = S 11 * ( - k ); S 22 ( k ) = S 22 * ( - k ) ; S 12 ( k ) = S 21 * ( - k ) . These result may be written in the matrix form S ( k ) = S + ( - k ) . 4. (a) With the given flux, the wave coming in from x =-∞ , has the form e ikx , with unit amplitude. We now write the solutions in the various regions x < b e ikx + Re - ikx k 2 = 2 mE / h 2 - b < x <- a Ae κ x + Be - κ x κ 2 = 2 m ( V 0 - E )/ h 2 - a < x < c Ce ikx + De - ikx c < x < d Me iqx + Ne - iqx q 2 = 2 m ( E + V 1 )/ h 2 d < x Te ikx (b) We now have x < 0 u ( x ) = 0 0 < x < a A sin kx k 2 = 2 mE / h 2 a < x < b Be κ x + Ce -

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• Spring '05
• mokhtari
• wave function, bound state, wave functions, odd solution

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