= X = X ( ~) } ( ) ( ) On the other hand, when the Hamiltonian acts on ( ) one finds that

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1 6. A general normalized solution to Schrödinger s equation of motion for a particle moving in a time-independent potential is of the form ( ) = P } where the and () are, respectively, eigenvalues and normalized eigenfunctions of the Hamiltonian ˆ (a) Prove or demonstrate this statement. Solution: Thetimerateofchangeat of a function ( ) representedbysuchanexpansionis ( ) Multiplying through by ~ then shows that = X } () = X ( ~) } () = X } ~ () ( ) ~ = X } () (1) On the other hand, when the Hamiltonian acts on ( ) one finds that ˆ ( ) = ˆ X } () = X } ˆ () where we have moved the Hamiltonian past all quantities that are independent of since these act as constants with regard to the action of the linear operator ˆ But, by assumption, is an eigenfunction of ˆ with eigenvalue so ˆ = Thus, making this substitution and rearranging slightly shows that ˆ( ) = X } () (2) The right-hand sides of (1) and (2) are the same, so the left-hand sides must be as well, verifying that this expansion for the function ( ) satisfies the Schrödinger equation ( ) ~ = ˆ( ) It is, in fact, clearly that state that evolves out of the initial state ( ) = P () (b) Consider an initial state () = which contains exactly two terms corresponding to different energies 1 and 2 6= 1. For what times, ifany,isthestate () an eigenfunction of ˆ? Briefly explain your answer and comment on how one might respond to the question: "What is the energy of this state at time?" Solution: For this situation, or more generally when ( ) is an arbitrary superposition of eigenfunctions of different energy, the state, initially or at any time thereafter, is not an eigenstate of ˆ and the value of the energy is simply not well-defined. This is easily verified. If () = then () = 1 1} } 2 and for any value of ˆ = 1 1} ˆ } ˆ 2 = 1 1} } 2 i 2 6= h 1 1} } 2

2 (c) For an arbitrary initial state ( ) does there always exist a constant such that ( ) = } ( )? Part (b) shows that in general there does not exist such a constant, and the time dependent wave function cannot generally be so written. However, if ( ) = () is an eigenfunction of ˆ at time = then this single energy eigenfunction acquires a time-dependent phase factor ( ) = ~ () = ~ () that depends on its own (now well-defined) eigenenergy. The eigenenergy is, in this special case, the actual energy of the particle.

3 7. An otherwise free particle of mass moving on the axis is confined to a one-dimensional box of edge length with one end at the origin and the other at = The confining potential if () = otherwise vanishes inside the box, and is positively infinite outside the box. (a) Solve the energy eigenvalue equation = for this system and obtain a set of appropriately normalized energy eigenfunctions and eigenvalues. Solutions: The eigenfunctions clearly vanish outside the box. Although the wave function generally will not vanish inside the box, continuous solutions must vanish at each endpoint of the interval [] Inside this interval, where = the energy eigenvalue equation is ~2 2 = Multiplying through by 2~ 2 and introducing 2 =2~ 2 this is equivalent to a 1D version + 2 = of the Helmholtz equation. Being second order, this has two linearly independent solutions () = ± ± The general solution inside the box will be a linear combination of these two, and thus of the form () = + + For this to vanish at = we must have = + which is equivalent to writing () = sin () with = + 2 In order for it to vanish at = we must then have = for some non-zero integer (The solution with =vanishes everywhere and is not, therefore, a valid solution its like the wave function of a particle that you can t find anywhere at all). The solutions corresponding to negative integers are just the solutions corresponding to postive integers multiplied by 1 and so are not really different solutions. Thus, allowed values of the quantum number (sometimes referred to as the principle quantum number) are the positive non-zero integers, i.e., {1 2} For these solutions, the allowed wavevectors are = Normalization then requires that 1= () 2 = 2 sin 2 () = 2 2 Thus, we can take = p 2 giving the solutions () = p 2 sin () for [] and () =everywhere else. These are energy eigenfunctions with energies = ~2 2 2 = ~ = 2 1 {1 2} The spectrum, depicted schematically below, is entirely discrete. Accordingly, the wave functions are square-normalized to unity.

4 (b) Show that your normalized energy eigenfunctions form an orthonormal set of functions on the interval [] in the sense that R () () = () () =(2) sin ()sin() For = the integral is equal to unity as a result of our choice of normalization. When 6= we find () () = 2 sin ()sin() sin ()cos() cos ()sin() = 2 ( 2 2 ) The denominator does not vanish for this case, but since and are both integers, the factors sin () and sin () vanish for integer and Thus, for this set when 6= the integral vanishes. Combining these gives the desired result: () () = (c) Are the eigenfunctions () considered as functions on the whole real line, continuous? Show that the derivative () of any one of these eigenfunctions, considered as a function on the whole real line, is not a continuous function of and identify the points or regions where it is discontinuous. Solution: The solutions are everywhere continuous on the real line,since they go to zero smoothly at the end points of the interval and remain equal to zero for all values of outside the interval. Thus in all cases, for all positive infinitesimals For [] the derivatives are lim + [ ( + ) ( )] = () = p cos () For [] the wave functions are all equal to the constant value = the derivative of which vanishes. Thus, the wave functions have continuous derivatives everywhere except at the endpoints of the interval, where () = lim + () ( ) = p () = lim + ( + ) ( ) = p

5 (d) Are the energy eigenfunctions for this system also eigenstates of the 1d momentum operator = ~? Solutions: Theeigenvalueequationforthemomentumoperatoris () = () which requires of any eigenfunction that ~ = i.e., that the function must be proportional to its derivative. These energy eigenfunctions of the particle in the box are not eigenstates of the momentum operator since, for [] () = ~ hp i 2 sin () = ~ p cos () Clearly, ~ () 6= () for any constant

6 8. A certain observable ˆ of a quantum mechanical particle has a discrete spectrum { =1 2} of eigenvalues each with a single-square normalized eigenfunction () Suppose that the discrete set of functions { ()} forms an orthonormal set of functions on R 3 i.e., R () () 3 = Suppose the particle is in a square normalized state () that can be expanded in the form () = P () (a) Use the orthonormality of the set { } and the fact that the state is square normalized to unity to show that P 2 =1 which justifies identifying ( )= 2 as a probability (the sum of which havetoadduptoone). The normalization condition on means that 1= () () 3 = [ X ()][ X ()] 3 wherewehaveinsertedourexpansionfor () and its complex conjugate (). Note that it is important to use separate dummy summation variables so as to avoid index clash. Pulling the summation signs to the left (which you can always do provided there is no conflict with the limits on the sum and the limits on the integral), and pulling any multiplicative constants outside of the integral, we deduce that 1 = X () () 3 = X = X = X 2 where we have followed the standard prescription for evaluating a sum containing a Kroncker delta function. Thus, under these conditions P 2 =1 (b) Multiply the expansion for () by the function () integrate over all space, and use the orthonormality relation to obtain a prescription for calculating the expansion coefficients. " # X () () = () () Note that we cannot bring the function inside the sum, without changing the sumation variable to avoid index clash. Changing the summation variable to and distributing over the remaining terms we have X () () = () () =1 Integrating both sides over all of space, and interchanging the order of integration and summation then shows that X () () 3 = () () 3 = =1 =1 X =1 = Thus, we find that to obtain the expansion coefficient we just need to perform the integral = () () 3

7 (c) If a very large number 1 of measurements of are made on an ensemble of systems initially in this same state (), the values obtained can be sorted into subsets having the same value. If is the number of times the eigenvalue is obtained, then the average value (or mean value) that will be obtained is, by definition, given by the relation hi = 1 X = 1 X = X =1 =1 ( ) = X 2 where ( )= = 2 is the probability (or relative frequency with which) is obtained in the limit that. Solution: From the assumed expansion for () is is clear that " # X () = ˆ () = ˆ () = X ˆ () = X () wherewehaveusedthefactthat ˆ = since the are eigenfunctions of ˆ Thus () ˆ () = X () () and () ˆ () 3 = X () () 3 = X where =[ ] = () () 3 = () () 3 Thus, () ˆ () 3 = X 2 = hi

8 9. A particle of mass is confined to the interval [ 2] by an infinite 1D potential well. [This is like an earlier problem, the results of which you should use here, except with =2]. At a certain moment the particle is in a state associated with the wave function () = (2 ) 2 (a) Sketch this wave function. What is the value of that makes this wave function square normalized to unity? Solution: We require that 1 = 2 () 2 2 = (2 ) so choosing real and positive gives r 3 = 2 3 The non-zero parts of the wave function form the two sides of equal length of an isoceles triangle, with a base of length 2 running along the -axis, and the two sides of equal length rising and falling linearly to the vertex at = whereittakesthevalue = = p x/a (b) If a measurement of the particle s position is made, what is the probability that it will be found in the interval [2] Solution: The probability density to find the particle at vanishes outside the interval [ 2]. Inside this interval it 2 2 () = () 2 2 (2 ) 2 2 The probability of finding the particle in [2] is the integral of () over this interval, i.e., or a little greater than 6% = = = 3 48 ' 625

9 (c) If, instead, an energy measurement is made when the particle is in this state, what values can be obtained? With what probability will those values occur? Calculate numerically to four significant figures the sum of those probabilities for the 3 lowest energies that could be obtained during an energy measurement performed on a particle in this state. Solution: The standard orthonormal set of energy eigenfunctions for this system vanish outside the well and are given for [ 2] by the expression r 1 ³ () = sin {1 2} 2 These are energy eigenfunctions associated with the discrete set of energy eigenvalues = 2 2 ~ These are the only values that can be obtained in a measurement of the energy on any state. For any such state, the probability of measuring the energy is the squared magnitude of the amplitude in the expansion X () = () =1 of the state in energy eigenfunctions. For the state of interest, we have (breaking the integral into two parts): = = = 2 r 3 () () 2 2 sin ³ [4 sin (2) + 2 sin ()] 2 r sin ³ (2 ) 2 After a little work, and after dropping terms that vanish when is an integer, one finds that = sin (2) 2 This vanishes for even Thus, there is zero probability of measuring an energy with even. (This occurs because the even wave functions are antisymmetric about = whilethewavefunction is even about that point.) For odd =2 1, with =12 the sine function can be evaluated explicitly giving = ³ (2 1) 2 sin 4 6 = 2 2 (2 1) 2 2 ( 1) The probability of measuring an energy with odd =2 1 therefore is = (2 1) 4 4 For the lowest three energies with nonzero probability, we find that 1 = 96 4 =986 the sum of which add up to = 5 = 96 (3) 4 = (5) 4 =