Mixture of States Proof

A particle of mass m sits in a one-dimensional square well with infinitely high walls and width L . The particle is in a 50–50 mixture of states, half in the ground state and half in the first excited state. I want to show how to derive a formula for the complete, time-dependent wave-function of the particle.

I begin by the using the normalizing condition, that \int|\Psi|^{2}dx=1 . This is because the probability of finding the particle somewhere in space must equal 1 at all times. Because the particle is in a mixture of states, my wave function will take the form \Psi = A(\Psi_{1}+\Psi_{2}) . Combining this with the normalizing condition, I get the equation A^{2}\int(\Psi_{1}+\Psi_{2})(\Psi_{1}^{*}+\Psi_{2}^{*})dx = 1 . The individual wave-functions will take the forms \Psi_{1} = \sqrt{\frac{2}{L}}\sin(\frac{\pi x}{L})e^{-i\omega_{1}t} and \Psi_{2} = \sqrt{\frac{2}{L}}\sin(\frac{\pi x}{L})e^{-i\omega_{2}t} . I can then plug these two equations in the the above normalizing condition for a particle in mixed states, convert all sine functions to cosine functions and cancel out like terms until I get down to the simple expression that 2A^{2}=1 which implies that A = \frac{1}{\sqrt{2}} . Now I can use this constant to write down my mixed wave-function which will look as follows \Psi = \frac{1}{\sqrt{L}}(\sin(\frac{\pi x}{L})e^{-\omega_{1}t} + \sin(\frac{2\pi s}{L})e^{-i\omega_{2}t}) . Now, I want to show that the probability of finding the particle between positions \frac{1}{4}L and \frac{1}{4}L+dx , as measured from the left-hand side of the well, as a function of time, is [\frac{3}{2}+\sqrt{2}\cos(\frac{3E_{1}t}{\hbar})]\frac{dx}{L} . This is done by finding the probability amplitude which is the square modulus of the wave function. This will look as follows |\Psi_{1}+\Psi_{2}|^{2} = (\Psi_{1}+\Psi_{2})(\Psi_{1}^{*}+\Psi_{2}^{*}) . After filling in each wave function and multiplying out each term I obtain \frac{1}{L}(\sin(\frac{\pi x}{L}))^{2} + \frac{1}{L}(\sin(\frac{2\pi x}{L}))^{2} + \frac{1}{L}\sin(\frac{2\pi x}{L}) \sin(\frac{\pi x}{L}) ( e^{-i\omega_{1} t}e^{i\omega_{2} t}+e^{-i\omega_{2} t}e^{i\omega_{1} t}) . After converting the sine terms to cosines and converting the exponentials to trigonometric functions, I can plug in x for L , which will give me\frac{1}{4}L and \frac{1}{4}L+dx which is what I wanted to show.

Phase Velocity Proof

Using the relativistic energy equation E^{2} = p^{2}c^{2}+m^{2}c^{4} , I want to show that the resulting phase velocity for the de Broglie wave of an electron is greater than the speed of light.

I start by deriving an expression for Energy in terms of the phase velocity, which is the rate at which the phase of the wave propagates in space, and the momentum of the electron. Using the equation v_{p}=f\lambda , I can then make two substitutions namely that f =\frac{E}{h} which gives the energy required or released when electrons change their energy levels and \lambda=\frac{h}{p} which is the wavelength associated with a particle as postulated by Louis DeBroglie. After canceling like terms this gives me E=v_{p}p . Now I can plug this into my relativistic formula to obtain v_{p}^{2}p^{2} = p^{2}c^{2} + m^{2}c^{4} . Next I can make a substitution using the equation p=mv_{p} , and cancel out the masses. This leaves me with the bi-quadratic equation y^{2}-c^{2}y-c^{4} = 0 where y=v_{p}^{2} . This equation has only one real solution which is \pm \sqrt\frac{c^{2}+\sqrt{5c^{4}}}{2} . Taking the positive root, I obtain the velocity 3.8*10^{8} m/s which is greater than the speed of light.

Reduced Mass Proof

In a two-bodied system of hydrogen, an electron and a proton orbit each other about a shared center of mass. If I wanted to analyze the atomic motion of just the electron, the situation would become a one-body problem, and I would have to replace the mass of the electron with its corresponding reduced mass which is expressed in terms of the mass of the nucleus and the mass of the electron. In this proof I want to suppose that this reduced mass changes by a small amount \Delta \mu when the electron jumps from the energy level with quantum number n_{i}=3 to the one with quantum number n_{f} = 2 . From this I want to show that the wavelength changes by a corresponding amount \Delta \lambda that approximately satisfies \frac{\Delta \lambda}{\lambda} = -\frac{\Delta \mu}{\mu} . In this situation, I am going to assume that \Delta \mu is small in comparison with \mu .

I begin by stating that \Delta \lambda \approx \frac{d\lambda}{d\mu} \Delta \mu , which represents a change in wavelength. This is because in calculus \Delta \mu and d\mu are practically the same for small \Delta . I can now use the equation \frac{1}{\lambda} = R(\frac{1}{n_{f}^{2}} - \frac{1}{n_{i}^{2}}) where \lambda is the wavelength of an emitted photon during electronic transition from E_{i} to E_{f} . R is known as the Rydberg constant which is the physical constant relating to atomic spectra and each n refers to a quantum number in a particular energy level. I can call everything on the right of this equation b for the sake of simplification and write the formula as \lambda = R^{-1}b^{-1} , where b represents everything that was on the left side fo the equation other than the Rydberg constant. Now, I can differentiate each side of the equation with respect to \mu , and because R is a function of \mu I must use the chain rule which looks as follows: \frac{d\lambda}{d\mu} = -R^{-2}\frac{dR}{d\mu}b^{-1} . I can now approximate \frac{dR}{d\mu} as \frac{R}{\mu} . This will cause certain terms to cancel and I will be left with the formula \frac{d\lambda}{d\mu} = -\frac{1}{R\mu b} . Next, I can use the formula derived above, that \lambda = \frac{1}{Rb} , and plug this in for R which will give me \frac{d\lambda}{d\mu} = -\frac{\lambda}{\mu} . I can then write the left side of the equation as \frac{\Delta\lambda}{\Delta\mu} and rearrange terms to give me \frac{\Delta\lambda}{\lambda} = -\frac{\Delta\mu}{\mu} which is what I wanted to show.