a) Calculate the zero-point vibrational energy for this molecule for a harmonic potential.

Transcription

1 Problem set 4 Engel P7., P Problem 7. The force constant for a H 8 F molecule is 966 N m -. a) Calculate the zero-point vibrational energy for this molecule for a harmonic potential. The following relation gives the zero-point vibrational energy E/ hν where ν is the quantum vibrational frequency. We can calculate ν from the relation k µ / where m is the reduced mass and k is the force constant, 966 N m -. We can obtain the reduced mass from the relation mm µ m + m We must be careful here to stick with mks units. m mass of H/atom in kilograms g/mol 0.00 kg/g (atom/mol) kg/(atom H) Similarly, m The reduced mass is then µ kg, a little smaller than the mass of H (hence reduced mass). Substituting µ in the epression for the vibrational frequency, we obtain ν s - We now have what we need to calculate the zero-point energy E/ hν J-s s J. /4/005 :03 PM - - PChem05ProbSet4HO.doc

2 Problem 7.6 Evaluate the average vibrational amplitude, <>, of the quantum harmonic oscillator about its equilibrium value Common sense and symmetry tells us that the answer must be zero for any state of the quantum harmonic oscillator. The argument goes as follows: We want to evaluate integrals of the form Ψ Ψd ΨΨd We ve switched the order of multiplication in the integral to focus attention on the product ψψ. The HO wave functions are either symmetric or antisymmetric. Thus, the product ψψ must be symmetric for all wave functions. We multiply ψψ by an antisymmetric function,. We conclude that the product ψψ must be antisymmetric. Looking at the integral again, we see that it must have the value zero. Problem 7.7 Evaluate the average linear momentum <p >< quantum harmonic oscillator. d ih for various states of the d Again, common sense tells us the answer must be zero, although we have to do a little more work to show it. The common sense part here is that if the particle had a net momentum other than zero the whole assembly would be moving along through space. It s not. The oscillator is fied in place! To be rigorous, we will need to consider integrals of the general form Ψ d d Ψd We will not need to worry about the constant, as the idea is to show the integral is zero. a) Ground state. The ground state for the HO has the general form (/) /4 ep(- /). When we differentiate this function with respect to, we recover the same function (to within a constant) but multiplied by. Thus the integral (again to within a constant) becomes ΨΨd just as we had for the average amplitude. And by the same argument about the symmetry of the functions in the integrand, this integral is zero. Now let s generalize. Inspection of the other wave functions reveals that all the Hermite polynomials are either symmetric (even) or antisymmetric (odd) functions. When differentiated, each term of the polynomial loses a power of. If the eponent of the term was originally even, it becomes odd after differentiation. If the eponent of the term was odd, it becomes even. Thus differentiation of the Hermite polynomials converts even functions into odd functions, and odd functions into even functions. /4/005 :03 PM - - PChem05ProbSet4HO.doc

3 We also know from our treatment of ground state, above, that when the eponential term in the complete HO wave functions is differentiated, the leading eponential is converted from an even function to an odd one, i.e., from (/) /4 ep(- /) to - times the original even function. Putting all these arguments together for two cases Case. Hermite polynomial is even The derivative term must be odd because we have d/d (eponential (even) * Hermite (even))[d/d eponential] (odd) * Hermite (even) + eponential(even)*[d/d Hermite] (odd). Overall, the derivative is an odd function. To form the epectation value of the momentum we multiply this derivative (odd) by ψ, which by hypothesis is even. Thus the integrand is the product of an even function and an odd derivative, which is odd. The integral is zero. Case. Hermite is odd Here we find the derivative must be an even function. d/d (eponential (even) * Hermite (odd)) [d/d eponential] (odd) * Hermite (odd) +eponential(even)*[d/d Hermite] (even) Here, to form the epectation value of the momentum we multiply this derivative (even) by ψ, which by hypothesis is odd. Thus, once again, the integrand is the product of an even function and an odd derivative, which is odd. The integral is zero. Problem 7.8 Non-zero integral this time. We want < > <ψ ψ> for the following integrals ) (/) / ep(- ) d (/) / ep(- ) d Ground state n vib 0 ) (4 3 /) / ep(- ) d (4 3 /) / ep(- ) 4 d st ecited state n vib 3) (/[4]) / ( - ) ep(- ) d state n vib (/) / [4 ep(- ) 6-4 ep(- ) 4 d + ep(- ) d] nd ecited These integrals can be evaluated from the standard formula /4/005 :03 PM PChem05ProbSet4HO.doc

4 0 n ep(- )d n ep(- )d 3 5 ( n ) n + n using the values n, n, and n3. Note well that the n that appears in this equation is NOT the quantum number of the harmonic oscillator. For n (ground state - vibrational quantum number n vib 0), we have 3 Multiplying by the constant in () above, we obtain < > / hν/k For n (first ecited state - vibrational quantum number n vib ), we have Multiplying by the constant in () above, we obtain < > 3 3/ hν/k For n3 (second ecited state - vibrational quantum number n vib ), we have for the integral in The three terms in (3) are as follows Hmmm. There seems to be a pattern! + 5 < > 5/ hν/k Problem 7.9 Evaluate the average of the square of the linear momentum of the quantum harmonic oscillator < > for the ground state and the first two ecited states p /4/005 :03 PM PChem05ProbSet4HO.doc

5 We want d Ψ h d Ψ. The second derivative is a big nuisance. Although the work to be done is straightforward and you can solve the problem by brute force, here s some trickery that makes use of other things we already know. The average value > is related to the average value of the kinetic energy, T: >/m <T> Also, by conservation of energy <T> +<V> E where <V> is the average value of the potential energy. Re-writing, we have > m [E - <V>] Here s the trick: we know that V/ k so <V></ k >/ k< >, where k is the spring constant. This observation saves a lot of work because we already know < > from problem 7.8 and we know that for the harmonic oscillator E(n vib +/)hν with n vib 0,, where ν k m So our problem is solved without taking derivatives or doing any integrals!!!! Before we plug in, it will be convenient to epress ν in terms of. From the tet, we have km h so k ν h Now we can plug in. For n vib 0 k k > m hν k m For n vib mk / mhν > 3 3 3k 3k m hν k m 3 mk 3/ mhν /4/005 :03 PM PChem05ProbSet4HO.doc

6 And by etrapolation, for For n vib > 5/ mhν Problem 7.0 Using the results for problems 7.6 through 7.9, calculate the uncertainties in the position and momentum, σ p p p and σ for the ground state and the first two ecited states of the quantum harmonic oscillator. Compare your results with the predictions of the Heisenberg uncertainty pinricple. We will do the calculation for the ground state and leave the net two states to you. σ p p p h mk -0 h mk σ -0 h mk It turns out that the magnitude of σ is of the same order as the range of motion, calculated classically, for. multiplying, we get σ pσ h 4 And taking square roots, h σ pσ as epected. /4/005 :03 PM PChem05ProbSet4HO.doc