Sunday, March 13, 2011

Mastering Physics: Applying the Harmonic Oscillator

The quantum harmonic oscillator is very important in analyzing the spectra of diatomic molecules. It can also be extended to the analysis of polyatomic molecules. In this problem, you will look at how the energy levels of the harmonic oscillator relate to the spectrum of carbon monoxide.

Part A
There is a strong line in the infrared spectrum of carbon monoxide with a wavelength of 4.61 \;\rm \mu m. What is the energy E of a photon in this line?
Express your answer in joules to three significant figures.

  E = 4.310×10−20
 \rm J

Part B
It can be shown that this line corresponds to a transition between adjacent energy levels in a harmonic oscillator. If this is true, what is the angular frequency omega of the oscillator? Use \hbar=1.055\times10^{-34}\; \rm J \cdot s.
Express your answer in inverse seconds to three significant figures.

  omega = 4.09×1014
 \rm s^{-1}

Part C
Find the value of k, the effective spring constant. Use 16.0 and 12.0 atomic mass units for the masses of oxygen and carbon, respectively. (1\;{\rm amu}=1.66\times 10^{-27}\;\rm kg.)
Express your answer in newtons per meter to two significant figures.

  k = 1900
 \rm N/m
This value is around an order of magnitude smaller than the spring constant for the springs in an average car suspension. Also, this is similar to the effective spring constant for a large trampoline.

Mastering Physics: Classical and Quantum Harmonic Oscillators

Consider a harmonic oscillator with mass m=0.100\; \rm kg and k=50\;\rm N/m. You may have worked similar problems before, as a mass on a spring using classical mechanics, but this time you will use the solution to the Schrödinger equation for the harmonic oscillator. Keep in mind that this system would be enormous by quantum standards, and in practice you would never expect to use quantum mechanics to describe a mass on a spring. Nonetheless, it is interesting to see what quantum mechanics predicts here.
Throughout this problem, use \hbar=1.055\times10^{-34}\; \rm J\cdot s.

Part A
Let this oscillator have the same energy as a mass on a spring, with the same k and m, released from rest at a displacement of 5.00 \;\rm cm from equilibrium. What is the quantum number n of the state of the harmonic oscillator?
Express the quantum number to three significant figures.

  n  = 2.650×1031

Part B
What is the separation DeltaE between energy levels in this harmonic oscillator?
Express your answer in joules to three significant figures.

  DeltaE  = 2.360×10−33
  \rm J
This energy is far smaller than you could possibly measure in an experiment with a mass on a spring. Just as for a classical harmonic oscillator, in experiments this huge quantum oscillator would appear as though its energy could take any value.

Part C
Nodes are the points where the wave function (and hence the probability of finding the particle) is zero. What is the separation between nodes of the wave function for the mass on a spring described in this problem? Assume that all of the nodes occur in the classically allowed region.
Express your answer in meters to three significant figures.

  \rm m
Since the diameter of an atomic nucleus is on the order of 10^{-15}\; \rm m, the separation that you've calculated is far too small to be measureable in any experiment. Just as for a classical harmonic oscillator, the position of this mass would seem to be able to take all values.
It is interesting to see that quantum mechanics reduces to classical mechanics on the scales of energy and size for which classical mechanics has been successful. However, to truly understand how the strange quantum world gives rise to the classical world of everyday experience requires the principle of decoherence, which describes how quantum states reduce to classical ones through the interactions of large systems with their environment.

Mastering Physics: Normalizing the Wave Functions for the Harmonic Oscillator

The wave function for the ground state of the harmonic oscillator is
where C is an arbitrary constant, hbar is Planck's constant divided by 2\pi, m is the mass of the particle, \omega=\sqrt{k/m}, and k is the "spring constant" for the harmonic oscillator.

Part A
Normalize this wave function. What is the (positive) value of C once this wave function is normalized? You will need the formula
\int _{-\infty}^\infty e^{-ax^2}=\sqrt{\frac{\pi}{a}}.
Express your answer in terms of omega, m, hbar, and pi.

  C = \left(\frac{m{\omega}}{{\hbar}{\pi}}\right)^{\frac{1}{4}}

Mastering Physics: The Finite Square-Well Potential: Bound States

Learning Goal: To understand the qualities of the finite square-well potential and how to connect solutions to the Schrödinger equation from different regions.
The case of a particle in an infinite potential well, also known as the particle in a box, is one of the simplest in quantum mechanics. The closely related finite potential well is substantially more complicated to solve, but it also shows more of the qualities that are characteristic of quantum systems. The potential energy function for a finite square-well potential is
U(x)= \Biggl \{ \matrix{ 0,\hfill & 0 \le x \le L, \hfill \cr U_0, & x<0 \hbox{ or } x>L, \cr}
where U_0 is a positive number that measures the depth of the potential well and L is the width of the well. The figure is a graph of potential energy versus position, which shows why this is called the square-well potential. Inside the well (i.e., for 0 \le x \le L) the solutions take the form \psi(x) = A\cos kx + B\sin kx, where A and B are constants and k=\sqrt{2mE}/\hbar. Outside the well, the solutions take the form \psi(x)=Ce^{\kappa x}+De^{-\kappa x}, where C and D are constants and \kappa=[2m(U_0-E)]^{1/2}/\hbar. In this problem, you will consider a particle in a state with energy E<U_0. Such states are called bound states, because classically the particle would be trapped in the potential well.

Part A
For a one-dimensional wave function to be normalizable, it must go to zero as x goes to infinity or negative infinity. Consider the wave function \psi(x)=C'e^{\kappa x}+D'e^{-\kappa x} in the region x>L. As x goes to infinity, this must become zero. What does this imply about the constants C' and D'?

Notice that the wave function is nonzero in the entire domain x>L, even though this region would be forbidden by classical mechanics since E < U_0. This "tunneling" into the classically forbidden region is a key difference between classical and quantum mechanics. Also notice that, since \psi(x)>0 anywhere in the x>L region, there is some small, but nonzero, probability of finding the particle hundreds of kilometers away from the potential well.

Part B
Now, consider the wave function \psi(x)=Ce^{\kappa x}+De^{-\kappa x} in the region x<0. As x goes to negative infinity, this must become zero. What does this imply about the constants C and D? (Be careful of signs.)

Again, you see that the wave function is nonzero throughout the entire classically forbidden region.

Part C
Recall that a physical solution to the Schrödinger equation must be continuous everywhere. Consider the two branches
\psi(x)=\Biggl \{ \matrix{ Ce^{\kappa x},\hfill & x<0, \hfill \cr A\cos kx +B\sin kx, \hfill & 0 \le x \le L, \cr}
that meet at the point x=0. Use the continuity of the physical solution to find the value of C.
Express your answer in terms of A and B.

  C  = A

Part D
The derivative must also be continuous everywhere to have a physical solution (unless there are points where the potential energy suddenly becomes infinite, which never happens for the finite square well). Take the derivative of the two branches that you looked at in Part C to find the value of B.
Express your answer in terms of k, kappa, and C.

  B  = \frac{C{\kappa}}{k}

Part E
Since you found that C=A in Part C, you can now write the equations for the wave functions as
\psi(x)=\Biggl \{ \matrix{ A\cos kx + \frac{\kappa}{k} A \sin kx, \hfill & 0 \le x \le L, \hfill \cr D'e^{-\kappa x}, \hfill & x> L.\hfill \cr}
These two branches meet at x=L. Set them equal here to find an expression for D'e^{-\kappa L}.
Express your answer in terms of A, k, kappa, and L.

  D'e^{-\kappa L}  = A{\cos}\left(kL\right)+\frac{{\kappa}}{k}A{\sin}\left(kL\right)

Part F
The last boundary condition, continuity of the derivatives at x=L, yields a similar equation:
-kA\sin kL + \kappa A \cos kL = -\kappa De^{-\kappa L}.
Dividing this equation by your equation from Part E (and doing some algebra to simplify) gives
\tan(kL) = \frac{2\kappa k}{k^2 -\kappa^2}.
This is a transcendental equation, which must be solved numerically or graphically. However, since k and kappa both depend on E, the energy levels for the finite square-well bound states may be found from this equation. Instead of trying to do this, we will look at the behavior of this equation as U_0 \rightarrow \infty. Solving for kL, you find kL=\hbox{atan} [(2\kappa k)/(k^2-\kappa^2)]. In this limit, what value does the right-hand side of the equation approach? In other words, what is
\lim_{U_0 \to \infty} \hbox{atan}\left( \frac{2\kappa k}{k^2 -\kappa^2} \right)?

Part G
This result tells you that, as U_0 goes to infinity, the equation reduces to \tan(kL)=0. Substitute back in k=\sqrt{2mE}/\hbar and solve for the energy levels E_n in this limit. Use the fact that \tan(n\pi)=0 for any integer n.
Express your answer in terms of m, n, L, and hbar.

  E_n  = \frac{{\hbar}^{2}}{L^{2}}{\cdot}\frac{1}{2m}{\cdot}\left(n{\pi}\right)^{2}

This result shows that the finite square-well solution becomes the particle-in-a-box solution as U_0 goes to infinity. Notice that n=0 seems to be allowed here, though it wasn't for the particle in a box. However, substituting n=0 gives a solution that cannot be normalized and thus is not a physical solution.