Physics 202H - Introductory Quantum Physics I Homework #12 - Solutions Fall 2004 Due 5:01 PM, Monday 2004/12/13

Transcription

1 Physics 0H - Introctory Quantu Physics I Hoework # - Solutions Fall 004 Due 5:0 PM, Monday 004//3 [70 points total] Journal questions. Briefly share your thoughts on the following questions: What aspects of this course do you think you are ost likely to use in the future, both in your physics existence and in your day-to-day life? Any coents about this week s activities? ourse content? Assignent? Lab?. Please coplete the anonyous end of course survey online on WebT. onstructive feedback will hopefully allow us to have the best possible courses in the future, and provide the instructor and departent with useful inforation about student reactions to any aspects of the progra. In addition to the bonus assignent arks, survey participation ay count towards overall class participation scores. [5.0-bonus] Solution: Do the survey - get the bonus arks.. Fro Eisberg & Resnick, Q 5-8 and 5-9, pg 69 Why is ψ necessarily an oscillatory function if V x < E? Why does ψ tend to go to infinity if V x > E? Liit your discussion to about 50 words or so. [0] Solution: The eigenfunction ψ is a solution to the tie-independent Schroedinger equation, and thus depending on the sign of V x R will have a solution of either positive and negative exponentials when V x > E or positive and negative iaginary exponentials. Exponentials go to infinity for large values of x, and of course negative exponentials go to infinity for large negative values of x, copared with iaginary exponentials, which are oscillatory. 3. Fro Eisberg & Resnick, P 5-4, pg 69 By evaluating the classical noralization integral in Exaple 5-6, show that the value of the constant B /π which satisfies the requireent that the total probability of finding the particle in the classical oscillator soewhere between its liits of otion ust equal one. [0] Solution: Fro Eisberg & Resnick, Exaple 5-6, pg 36, we can arrive at an expression for the probability density P based on the classical calculation of the fraction of the tie the oscillator is at each section of the range. This probability is inversely proportional to its velocity. v E x P B v B E x.

2 To find the value of B, we need to noralize this probability, insuring that the total probability of finding the particle in all allowed locations is unity. + + P + B B E x E x E, the total energy, is a constant, and can be stated in ters of a, the axiu value for x, since at this point, the velocity of the particle will be zero. E K + V v + x a This gives us the integral that we need to evaluate, naely B B B +a a +a a +a a x a a x a x. Either looking up the integral, or doing a trigonoetric substitution gives us the result that the integral is equal to arcsin x/a B B B B +a a a x [ x ] +a arcsin a a [ a arcsin arcsin a ] a a [arcsin arcsin ]. The angle who s sin is is π/, and the angle who s sin is is π/, so this expression becoes B B [arcsin arcsin ] [ π π ] B π.

3 Solving for B gives us B π 4. Fro Eisberg & Resnick, P 5-7, pg 70 B π π. a Use the particle in a box wave function verified in Exaple 5-9, with the value of A /a deterined in Exaple 5-0, to calculate the probability that the particle associated with the wave function would be found in a easureent within a distance of a/3 fro the right-hand end of the box of length a. The particle is in its lowest energy state. [0] Solution: The probability of finding the particle between points x and x is given by x x P x, t x x x x x Ψ x, tψx, t A cos x A cos A [ x πx a πx a e +iet/ A cos + sin πx/a 4π/a ] x x. πx e iet/ a For the situation described, x a/ a/3 a/6 and x a/, so we have x [ ] x P x, t A sin πx/a x + x 4π/a x [ ] x sin πx/a a/ + a 4π/a a/6 [ ] x sin πx/a a/ + a π a/6 [ sin π + ] sin π/3 π 6 π / π 3 3 4π The probability that the particle associated with the wave function would be found in a easureent within a distance of a/3 fro the right-hand end of the box of length a is about 0.955, or just under 0%. b opare with the probability that would be predicted classically fro a siple calculation related to the one in Exaple 5-6, for a particle with constant speed bouncing back and forth fro the ends of the box. [5] Solution: For a particle with constant speed, the probability of finding it at any point in the box is equal, and the probability density is just /a. Thus, x x P x, t a x x For a classical particle with constant speed, the probability that the particle would be found in a easureent within a distance of a/3 fro the right-hand end of the box of length a is /3 or a bit ore than 33%

4 c opare with the probability that would be predicted classically fro a calculation related to the one in Exaple 5-6, for a particle undergoing siple haronic otion with the value of B /π. [5] Solution: Building on the calculations carried out in the previous proble, with the sall change that the previous proble had a box size of a and this one has a size of a, the probability of finding the particle between points x and x for a particle undergoing siple haronic otion is given by x x P x, t x π x a/ x π a/ a/6 a/ x [ ] x a/ arcsin π a a/6 [ ] arcsin arcsin π 3 [ ] π π arcsin 3 arcsin / π For a classical particle undergoing siple haronic otion, the probability that the particle would be found in a easureent within a distance of a/3 fro the right-hand end of the box of length a is about or a bit less than 40%.

5 5. Fro Eisberg & Resnick, P 6- and 6-, pg 9 a Verify by substitution that the standing wave general solution, Eisberg & Resnick, Equation 6-6, pg, satisfies the tie-independent Schroedinger equation, Eisberg & Resnick, Equation 6-, pg 78, for the finite square well potential in the region inside the well. [5] Solution: The tie-independant Schroedinger equation is d ψx + V x, tψx Eψx. 5. For the region inside the well, the potential V x 0, so the tie-independent Schroedinger equation becoes d ψx Eψx. 5. The standing wave general solution is ψ a x A sin k I x + B cos k I x where k I The derivatives of ψ a x are: E, a < x < +a. 5.3 If we ultiply 5.4 by / we get dψ a x k I A cos k I x k I B sin k I x d ψ a x ki A sin k I x ki B cos k I x ki A sin k I x + B cos k I x ki ψ a x E ψ ax. 5.4 d ψ a x E ψ a x Eψ a x, which is 5., the tie-independent Schroedinger equation in the region inside the well. Thus we have shown that 5.3 is a solution to the tie-independent Schroedinger equation 5. in the specified region.

6 b Verify by substitution that the exponential general solution, Eisberg & Resnick, Equation 6-63 and 6-64, pg, satisfy the tie-independent Schroedinger equation, Eisberg & Resnick, Equation 6-3, pg 86, for the finite square well potential in the regions outside the well. [5] Solution: For the region outside the well, the potential V x V 0, so the tieindependent Schroedinger equation becoes The exponential general solutions are d ψx + V 0 ψx Eψx. 5.5 ψ b x e k IIx + De k IIx where k II V0 E, x < a, 5.6 and ψ c x F e k IIx + Ge k IIx where k II V0 E, x > + a. 5.7 The derivatives of ψ b x are: dψ b x k II e kiix k II De k IIx d ψ b x kiie kiix + kiide k IIx kii e kiix + De k IIx k IIψ b x V 0 E ψ c x. 5.8 If we ultiply 5.8 by / and add V 0 ψ b x we get d ψ b x + V 0 ψ b x V 0 E ψ b x + V 0 ψ b x Eψ b x, which is 5.5, the tie-independent Schroedinger equation in the region outside the well. Thus we have shown that 5.6 is a solution to the tie-independent Schroedinger equation 5. in the specified region. Siilarly, for ψ c x we have: d ψ c x k IIψ c x V 0 E ψ c x, 5.9 and if we ultiply 5.9 by / and add V 0 ψ c x we get d ψ c x + V 0 ψ c x V 0 E ψ c x + V 0 ψ c x Eψ c x, which is 5.5, the tie-independent Schroedinger equation in the region outside the well. Thus we have shown that 5.7 is a solution to the tie-independent Schroedinger equation 5. in the specified region.

7 6. Fro Eisberg & Resnick, Q 6-5, pg 7 A particle is incident on a potential barrier of width a, with total energy less than the barrier height, and it is reflected. Does the reflection involve only the potential discontinuity facing its direction of incidence? If the other discontinuity were oved by increasing a, is the reflection coefficient changed? What if the other discontinuity were reoved, so that the barrier was changed into a step? Liit your discussion to about 50 words or so. [0] Solution: The reflection does not only involve the potential discontinuity facing the direction of incidence. The width of the barrier has an effect on the reflection and transission coefficients, as the barrier width is increased the aount of reflection in general decreases thought there are soe particular widths and energies for which the aount of reflection is particularly high or low e to constructive or destructive interference between the waves reflected off of each discontinuity. If the second discontinuity were reoved, changing the barrier into a step, then there would be 00% reflection, copared to saller reflections for barriers of finite thickness. 7. Fro Eisberg & Resnick, Q 6-34, pg 3 Verify the eigenfunction and eigenvalue for the n state of a siple haronic oscillator by direct substitution into the tie-independent Schroedinger equation, as in Eisberg & Resnick, Exaple 6-7, pg 4. [0] Solution: The tie-independant Schroedinger equation is d ψx + V x, tψx Eψx. 7. For the siple haronic oscillator, the potential V x x /, so the tie-independent Schroedinger equation becoes d ψx + x ψx Eψx. For u x /, and E 5/ /, this gives us d ψ The n eigenfunction is The first derivative of ψ x is: dψ x x ψ E ψ u ψ 5 ψ u A u e u / dψ u d A A u e u / u ψ 5 ψ ψ u 5 ψ. 7. where d d u e u / + u d u /4 / x. 7.3 /4 / x /4 / /4 e u / A 4ue u / + u u e u / u 4A e u / A u e u / /4 / / u 4A e u / /4 ψ 4A ue u / /4 uψ. / /

8 The second derivative of ψ is: d ψ x d dψx d ψ u d dψ u d 4A ue u / uψ /4 / d 4A ue u / + d uψ d 4A d e u u / e u / + 4A u + d u ψ + u d ψ 4A e u / 4A u e u / + ψ + u 4A ue u / uψ 4A e u / 4A u e u / + ψ 4A u e u / u ψ 4A e u / + 8A u e u / ψ + u ψ 4A u e u / ψ + u ψ 4ψ ψ + u ψ u 5 ψ, which is 7., the tie-independent Schroedinger equation for the haronic potential. Thus we have shown that 7.3 is a solution to the tie-independent Schroedinger equation 7. for the specified potential. Headstart for next week, Week 3, starting Monday 004//3: Review notes, review texts, review assignents, learn aterial, do well on exa