Problem 1:
Electrons of kinetic energy 10 eV travel a distance of 2 km. If the size of the initial wave packet is 10-9 m, estimate the size at the end of their travel.
Solution:
 | Concepts, principles, relations
that apply to the problem: The uncertainty principle |
| Why do they apply? We use the uncertainty principle to estimate the uncertainty in the momentum of the electrons. This translates into an uncertainty in the velocity. This Dv causes the wavepacket to broaden with time. |
| How do they apply? The electron energy is E = 10eV = 1.6´10-18J. The initial uncertainty in its position is Dx = 10-9m. Dp = h/Dx, Dv = h/(mDx) = 1.1´105m/s. After 10-3s the uncertainty in its position is on the order of Dv10-3s = 110m. |
| Details of the calculation: None |
Problem 2:
The wavefunction y of a particle is written as a linear combination of the three orthonormal eigenfunctions {fi} of the observable A with eigenvalues ai (i = 1,2,3).
Find <A>. What is the probability that the measurement of A yields a2? Find the wavefunction immediately after this measurement.
Solution:
| Concepts, principles, relations
that apply to the problem: The postulates of Quantum Mechanics, the mean value |
| Why do they apply? The expression for the mean value of an observable A in the normalized state |y> is <A> = <y|A|y>. When a physical quantity described by the operator A is measured on a system in a normalized state |y>, the probability of measuring the eigenvalue an is given by P(an) = Si=0gn|<uni|y>|2, where {|uni>} (i=1,2,...,gn) is an orthonormal basis in the eigensubspace En associated with the eigenvalue an. If a measurement on a system in the state |y> gives the result an, then the state of the system immediately after the measurement is the normalized projection of |y> onto the eigensubspace associated with an. |
| How do they apply? <y|y> = 1/6 + 1/3 + 1/2 = 1. |y> is normalized. <A> = <y|A|y> = a1/6 + a2/3 + a3/2. The probability that a measurement will yield a2 is |<f2|y>|2 = 1/3. Immediately after the measurement the wavefunction is f2. |
| Details of the calculation: None |
Problem 3:
In relativistic mechanics, energy and momentum of a free particle are related by the expression E2 = p2c2 + m2c4. Construct the relativistic analogue of the Schroedinger equation by introducing the appropriate operators.
Solution:
| Concepts, principles, relations
that apply to the problem: The momentum and energy operators, postulates of Quantum Mechanics |
| Why do they apply? How do we find the operator corresponding to a physical quantity that is classically defined? (a) Express the physical quantity in terms of the fundamental dynamical variables r and the conjugate momenta p. (b) Symmetrize the expression with respect to r and p, then replace the variables r and p with the operators R and P. Here r does not appear and no symmetrization is necessary. |
| How do they apply? (p2c2 + m2c4)y = E2y The momentum operator is (h/i)Ñ. The energy operator is ih¶/¶t. We therefore have (-c2h2Ñ2 + m2c4)y = -(h2¶2/¶t2)y, or (Ñ2 - m2c2/h2)y = (1/c2)(¶2/¶t2)y. |
| Details of the calculation: None |
Problem 4:
Consider a three-state quantum mechanical system with an orthonormal ‘color’ basis {|R>, |G>, |B>} (‘red,’ ‘blue,’ and ‘green’ respectively). Its evolution is governed by the Hamiltonian
.
(a) Construct the matrix representation of this
Hamiltonian using the {|R>, |G>, |B>} basis.
(b) Find the energy eigenvalues and normalized eigenstates
of the system. Express the latter as linear combinations of |R>, |G>, |B>.
(c) At time t = 0 the state vector is |y(0)>
= |G>. Find the state vector |y(t)> at
an arbitrary time t.
(d) After starting from the initial conditions of (c), the
‘color’ is measured at time t = t0 and found to be green. What are
the probabilities for the color to be measured as red, green, or blue at time t
= 2t0?
Solution:
| Concepts, principles, relations that apply to the
problem: Eigenvalues and eigenvectors, the evolution operator. |
| Why do they apply? We are asked to find the eigenvalues and eigenvectors of a given Hamiltonian H. We are then asked to find |y(t)> given |y(0)>. We expand |y(0)> in terms of eigenvectors of H and apply the evolution operator. |
| How do they apply? (a) The matrix of H in the {|R>, |G>, |B>} basis is
(b) The eigenvalues lE0 are found from  .
(2 – l)3 - (2 –
l) = 0. The eigenstates are A1|R> +A2|B> +
A3|G>. (c) |B> = (|2> + |3>)/Ö2.
|G> = (|2> - |3>)/Ö2. |y(0)>
= |G> = (|2> - |3>)/Ö2. (d) |y(t0)>
= eif|G>
à sin(E0t0/h)
= 0, cos(E0t0/h)
= ±1, E0t0/h
= np. |
| Details of the calculation: None |
Problem 5:
Assume the wave function of a particle is
Here a and p0 are real constants and N is a
normalization constant.
(a) Find N so that y(x) is normalized.
(b) If the position of the particle is measured, what is the probability of
finding the particle between
and
?
(c) Calculate the mean value of the momentum of the particle.
Solution:
| Concepts, principles, relations
that apply to the problem: The postulates of Quantum Mechanics, the mean value |
| Why do they apply? When a physical quantity described by the operator A is measured on a system in a normalized state |y>, the probability of measuring the eigenvalue aa is given by dP(aa) = |<ua|y>|2da, where |ua> is the eigenvector corresponding to the eigenvalue aa; we assume aa is a non-degenerate continuous eigenvalue of A. The expression for the mean value of an observable A in the normalized state |y> is <A> = <y|A|y>. |
| How do they apply? (a)   .
(b) Let P denote the probability of finding the particle between
(c) Let <P> denote the mean value of the momentum of the particle.
|
| Details of the calculation: None |
Problem 6:
Consider the operators whose action is defined by the equations below:
Find the commutator [O1, O2].
Solution:
| Concepts, principles, relations
that apply to the problem: Commutator algebra |
| Why do they apply? We are asked to find the commutator of two given operators. |
| How do they apply? [O1, O2]y(x) = O1O2y(x) - O2O1y(x) = O1xdy(x)/dx - O2x3y(x) = x4dy(x)/dx - xd(x3y(x))/dx = x4dy(x)/dx - 3x3y(x) -x4dy(x)/dx = -3x3y(x) = -3O1y(x) for any y(x). Therefore [O1, O2] = -3O1. |
| Details of the calculation: None |
Problem 7:
(a) What is the wavelength of a 10 eV electron and what is the
energy of a photon with this same wavelength?
(b) Light with a wavelength of 300 nm strikes a metal whose work function
is 2.2 eV. What is the shortest de Broglie wavelength for the electrons that
are produced as photoelectrons?
(c) A surface is irradiated with monochromatic light whose
wavelength can be varied. Above a wavelength of 500 nm, no photoelectrons are
emitted from the surface. With an unknown wavelength, a stopping potential
of 3V is necessary to eliminate the photoelectric current. What is the unknown
wavelength?
Solution:
| Concepts, principles, relations
that apply to the problem: The de Broglie wavelength, the photoelectric effect |
| Why do they apply? We are asked to compare the energy of an electron and a photon with the same wavelength. Photoelectrons are only emitted if the photon energy is greater than the workfunction f of the surface. |
| How do they apply? (a) For the electron: p = h/l, l = h/(2mE)1/2 = 2p*1.054*10-34 Js/(2*9.1*10-31 kg*10 eV *1.6*10-19 J/eV)1/2 = 3.9*10-10 m For the photon: E = hc/l = (2p*6.58*10-16 eV-s*3*108 m/s)/(3.9*10-10 m) = 3180 eV |
| Details of the calculation: (b) Emax_electron = hc/l – 2.2 eV = 1.94 eV = 3.1*10-19 J. pmax_electron = (2mEmax_electron)½ = 7.52*10-25 kgm/s. le_min = h/pmax_electron = 8.8*10-10 m. (c) Ee(max) = hf – f = hc/l – f. Ee(max) = 0 if l > 500mn, therefore f = hc/(500 nm). With the unknown wavelength Ee(max) = 3 eV, so 3 eV = hc/l – hc/(500 nm). 3 eV = (1240 eV/nm)/l – (1240 eV/nm)/(500 nm). l = 226 nm. |
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0, odd function