d)p () = A = Z x x p b Particle in a Box 3. electron is in a -dimensional well with innitely high sides and width An Which of following statements mus

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1 4 Spring 99 Problem Set Optional Problems Physics April, 999 Handout a) Show that (x; t) =Ae i(kx,!t) satises wave equation for a string: (x; = Show that same wave function (x; t) =Ae i(kx,!t) satises Schrodinger equation with V (x; t) = : time-dependent In each case, we take derivatives of (x; t) and see left hand side equals right hand side of equation for some if of constants. choice space derivative is a) (x; =,k (x; (x; =,! (x; = k, left and right sides of equation are equal, and so For t) is a solution. (x; (x; =,k (x; (~k) =m = ~!, left and right sides of equation are equal, and For (x; t) is a solution. so Probability Density. time t =, wave function of an electron is (x; ) = Ae,x =b. At What is probability density P (x) for this electron, in terms of A a) b? and What must A be for this wave function to satisfy normalization (Note: you'll need to refer to a table of denite integrals to condition? What is probability of nding electron somewhere in interval d) < x < xwhere x << b? Use normalization condition to A. (Hint: Eir sketching P (x) or using Taylor's orem may eliminate with this one). help = = (this second equality is true because is real). (x) a)p wave function gives P (x) =A e,x =b. Squaring normalization condition, which says that probability of nding electron somewhere is, is + Z A p so p. = b A Z + probability density isaneven function of x (i.e., P (x) = P(,x)), c) electron is equally likely to be to left or right of origin. so probability of nding electron somewhere is, so probability nding it on right side is =. of. Time-dependent Schrodinger Equation solve this problem). (x; t) t) m What is probability of nding electron to right of origin c) at some position x>)? (i.e., time derivative is P (x)dx =, space derivative is Putting in P (x) from previous part, e,x =b dx = A b p =;, time derivative is =,i! (x; t)

2 d)p () = A = Z x x p b Particle in a Box 3. electron is in a -dimensional well with innitely high sides and width An Which of following statements must be true for a standing wave. solution? If decreases a factor of, ground state energy will increase (A). by If increases a factor of, ground state energy will increase (B). by probability of nding electron in box has a maximum (C) center of box. at probability of nding electron outside box is not (D) zero. identically (D) is not true because box has innitely high walls, (C) not true because electron wave function (if it is not in ground is but in rst excited state for example) can have a node in state of box, (B) is not true since as box width increases center state energy will decrease, and (A) is correct answer because ground fact, ground state energy is proportional to =. in Electron in a box 4. electron is conned in a -dimensional potential box by a potential An (x), where V (x) =;,a < x < aand V (x)!innity, x > aand V x<,a. Inside box in ground state, wavefunction is given by a) (x) = A cos(k x) where A ;k are constants. Using boundary R Setting R (a) =atx=agives relation k a = = k = =(a). or wavefunction for nth energy level can be written as R n (x) = n sin(k n x)+b n cos(k n x). What are B and k for rst excited state A rst excited state has a single interior node at x = gives B =. To satisfy boundary conditions R (a) =at which electron is initially in its ground state. n white light consisting c) a wide range of frequencies illuminates electron which can absorb of of photons that can be absorbed by electron in transitions energies ground state? Express your answers in terms of ground state from We know for square well potential, E n = h n =(8m(a) ). lowest energy photon that can be absorbed in ground state is that induces transition for electron from E to E.! = one, E =4E,E =3E where E = h =(3ma ). next lowest E photon electron in ground state can absorb is!3 = energy 3, E =9E,E =8E where E = h =(3ma ) E In real life potential can't be innite ourside box. Suppose d) V (x) =+V for x>aand x<,awhere V is large but nite. Will that energy of rst excited state for nite box be less than, greater or equal to value calculated for potential well with innite than, rst excited state energy for nite box will be less. reason is that wavefunction no longer has to go to zero at x = a, and P (x) isvery nearly constant for <x<x b p where x << b. Away of seeing this is to use Taylor's orem: (x) P() x x d x P + + () = P (), dp () P b dx dx that wavefunction is zero at x = a nd an expression for conditions. k P(): n P(x)dx Probability = n =? x=a,we n have A sin(k a) = which implies k a = or k = =a. photons and jump to excited energy states. What are two lowest energy E. (E) Two of above answers are correct. (F) None of above answers are correct. walls?

3 can leak out of box. electron is eectively conned to a larger and and so its momentum spread and its energy will be less. Anor way box think about it is that wavelength of wavefuntion is larger since to particle can leak into previously forbidden regions; since energy is proportional to square of wavelength, energy will be inversely less. Particle in a Box: Probabilities 5. a particle in a -dimensional box of width and innitely high sides, For estimate probability that in lowest energy state, particle a) in interval =4 to3=4. is estimate probability that n=3 excited state and n=9 state are in interval =4 to3=4. excited Justify answers to parts (a) and ( (hint: draw a picture and c) qualitatively why your answers came out way that y did). explain Z 3=4 n =3: For 3=4 Z Z b = n(x)dx = = (x)dx = 3(x)dx = + =, 3 r nx sin( ): Z b x, 3 8, 8, ( nx sin )dx nx sin 4n b 8 sin + sin for n = 9: and 3=4 Z = 9(x)dx 3 8, 8, =, 9 (6)(9) ()(9) + sin sin 4 (4)(9) 4 (4)(9) lowest energy state wavefunction squared is shown below c)for clearly re is a higher probability (area under P (x) curve) to and squared (shown below) is oscillating very rapidly and wavefunction value of probability density inany section of box that is average compared to wavelength is just average of a sinusoid over large cycles (/) times normalization constant =, which equals many =4 nd particle in center of box since (x) is largest re. For a particle in a box, n(x) = In general, probability to nd particle in interval a to b is a a a a)for n =, above formula becomes: 3 8, 8, 4 sin sin However, for a particle in a highly excited state (i.e. n = 9) =4 =. =4 3

4 time independent Schrodinger equation in dimension is given by, m ~ (E, U) = region of interest in our problem E = and hence rearranging In yields d dx Ae x x, A, = = 4x,6 4 x 4 =, 6 A parabola with U() =, 3~ m (, x Axe 4 x, 6 q located at x = 3. zeros and Hence x e, d x e,, 4A x x e,,a = x dx +4A x3 x 4e, (x) Thus ~ U(x) = m 6. Serway, Chapter 4, pg 48, Problem 4 d dx ~ d = U dx m 7. Uncertainty Principle in Waves a) Waves on a String 4

5 we want to create a localized wave on a string, as a superposition Suppose of standing waves. et's try to make all standing waves add to a sharp peak in middle of string. Consider a string that up from x =,= to=. (That is, x = is at center of stretches fth harmonic) have peaks at x = : se have form harmonic, our variables) (in, X M+m m=m cos((m, )x=) =, X M+m m=m a plot, with = 6, using eir a spreadsheet program or Mat- Make from x =,= tox=+= inintervals of., of y(x), for M =, ab, 4 string where we want peak, and not at edge.) standing corresponding to all odd harmonics ( fundamental, third waves y m,(x) = cos((m, )x=) = cos(k m,x) m =;;3;::: and k n = n= is called wave number of where wave. So, to get a big peak in middle, we want to add up standing - M waves -4 y(x) = cos(k m,x) Plot of y(x) vsxfor M =5 5, and. To see wave envelope clearly, start sum at m = Plot of y(x) vsxfor M = Plot of y(x) vsxfor M = 5

6 width of peaks x, dened as distance from x = Find rst zero in envelope of transverse displacement y, and ll to???? 5???? that as we use larger and larger ranges of wave numbers k n Notice get a narrower and narrower peak! wave numbers range from we ' m = to k ' (M + m )=, so spread k ' M=. Show k we have a position{wave number uncertainty relation that is constant C? constant C depends on our denition of What x and x. ( uncertainty relations discussed in Serway are widths notice from our table above that M x. Hence We instead of waves on a string, se waves were quantum Suppose, probability waves associated with a particle of momentum mechanical p = h = ~k. By using de Broglie formula to get range of mo- from range in wave numbers, show that it satises position{ menta uncertainty relation momentum de Broglie formula p = ~k means that spread in momentum & ) x p xk & ~ Making a laser pulse think about a laser beam: nearly monochromatic light. asers et's xk & C: for root-mean-square widths. We use full widths in this problem.) xk M M = Plot of y(x) vsxfor M = in following table: xp & h: M x is p = ~k. From previous section ) xp & ~ = h M x are not quite monochromatic... et's model a laser in a way nearly 6

7 as our calculation above: numerically add up some cosine waves same in frequency between f, f= and f +f=: spread that this is just generalization of beats, with M waves of dierent Note rar than just two. Make a plot, with f = 5 and f =:5, frequencies, eir a spreadsheet program or Matab, from t =,4 tot=4in using of :4, of E(t), with M =. How wide is pulse in time t, intervals from t = to rst zero of envelope? (Note t should be i.e. if measured from that rst zero to second zero in envelope). same energy formula for a photon, show that energy spread Using this pulse E satises energy{time uncertainty relation in f =E=h and hence Since ft = E h + m (f + f)t M X M= Et & h: cos E(t) = m=,m= t & ) Et & h Show that it satises frequency{time uncertainty relation ft & : 5 5 E(t) t We note from gure t =. Hence ft =:5&: 7