[QMSE] W03 - Harmonic Oscillator

• the WM harmonic oscillator
  • an exactly solvable problem
  • problem slightly more complicated than “particle-in-infinitely-deep-box”
• many kinds of quantum mechanical systems are explained based on this model
  • a good first approximation for many physical systems
• example one
  • tiny mass on spring to model atom vibration in a molecule or crystalline solids
  • these atom vibrations maybe described in QM as phonon modes
• example two
  • the idea of photons finds basis in the idea of harmonic oscillator

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quantum mechanical harmonic oscillator

• the classical simple harmonic motion is explored
  • with potential and potential differences instead of forces
  • this can be done within CM using hamiltonian mechanics
  • in QM working directly with forces is avoided
    • SE is also setup that way from the beginning

quantum mechanical solution

• we have a mass $m$ on a spring
  • $m$ is small
• we are going to use $z$ for our coordinate here
  • 
• the potential from the restoring force $F$ is $$\begin{array}{cc}F& =-kx\to \text{spring restoring force}\\ V\left(z\right)& ={\int }_0^z-Fd{z}_0\\ & ={\int }_0^z-K{z}_{0}d{z}_0\\ & =\frac{1}{2}K{z}^2\\ & =\frac{1}{2}m{\omega }^2{z}^2\end{array}$$
  • 

harmonic oscillator schrodinger equation

$$\begin{array}{c}V\left(z\right)=\frac{1}{2}m{\omega }^2{z}^2\\ \end{array}$$

• schrodinger equation (time-independent) is $$\begin{array}{c}-\frac{{\hslash }^2}{2m}\frac{d^2\psi }{d{z}^2}+\frac{1}{2}m{\omega }^2{z}^2\psi =E\psi \end{array}$$
  • for convenience, we define a dimensionless distance unit $$\begin{array}{c}\xi =\sqrt{\frac{m\omega }{h}}z\end{array}$$
• so the schrodinger’s equation becomes $$\begin{array}{c}\frac{1}{2}\frac{d^2\psi }{d{\xi }^2}-\frac{{\xi }^2}{2}\psi =-\frac{E}{h\omega }\psi \end{array}$$
  • one specific solution to this equation $$\begin{array}{cc}\frac{1}{2}\frac{d^2\psi }{d{\xi }^2}-\frac{{\xi }^2}{2}\psi & =-\frac{E}{\hslash \omega }\psi \\ \psi & \propto \exp \left(\frac{-{\xi }^2}{2}\right)\end{array}$$
  • with a corresponding energy $E=\frac{\hslash \omega }{2}$
• this suggests we look for solutions of the form $${\psi }_n\left(\xi \right)=A_n\exp \left(\frac{-{\xi }^2}{2}\right)H_n\left(\xi \right)$$
  • where: $H_n\left(\xi \right)$: is a set of functions to be determined

$$\begin{array}{cc}\frac{d^2{H}_n\left(\xi \right)}{d{\xi }^2}-2\xi \frac{d{H}_n\left(\xi \right)}{d\xi }+(\frac{2E}{h\omega }-1)H_n\left(\xi \right)& =0\end{array}$$

• this is the defining differential equation for the hermite polynomials
  • solutions for hermite polynomials exist only if $$\frac{2E}{\hslash \omega }-1=2n\text{n = 0,1,2,…}$$
  • in terms of energy, solutions only exist for energy levels $$E_n=(n+\frac{1}{2})\hslash \omega \text{n = 0,1,2,…}$$
• so, the allowed energy levels are equally spaced separated by an amount $\hslash \omega$
  • $\omega$: classical oscillation frequency
  • like the potential well, there is a “zero-point energy”
    • $E_0=\frac{\hslash \omega }{2}$

hermite polynomial

• list of hermite polynomials
  • $H_0=1\to$ even
  • $H_1\left(\xi \right)=2\xi \to$ odd
  • $H_2\left(\xi \right)=4{\xi }^2-2\to$ even
  • $H_3\left(\xi \right)=8{\xi }^3-12\xi \to$ odd
  • $H_4\left(\xi \right)=16{\xi }^4-48{\xi }^2+12\to$ even
• properties
  • they have “define parity” i.e. they are either even or odd
  • they satisfy a “recurrence relation”
    • successive hermite polynomials can be calculated
      • if the previous two are known
    • i.e. $H_n\left(\xi \right)=2\xi H_{n-1}\left(\xi \right)-2(n-1)H_{n-2}\left(\xi \right)$
• normalizing $${\psi }_n\left(\xi \right)=A_n\exp \left(\frac{-{\xi }^2}{2}\right)H_n\left(\xi \right)$$
  • gives $$\begin{array}{cc}{A}_n& =\sqrt{\frac{1}{\sqrt{\pi }{2}^{n}n!}}\\ \xi & =z\sqrt{\frac{m\omega }{\hslash }}\\ E_n& =(n+\frac{1}{2})\hslash \omega \text{n = 0,1,2,…}\\ & \\ \text{expressing it}& \text{in terms of the original coordinates}\\ {\psi }_n\left(z\right)& =\sqrt{\frac{1}{2^{n}n!}\sqrt{\frac{m\omega }{\pi \hslash }}}\exp (-z^2\frac{m\omega }{2\hslash })H_n\left(z\sqrt{\frac{m\omega }{\hslash }}\right)\end{array}$$
• harmonic oscillator eigensolutions are graphically seen as follows
  • in this above plot, the inttersections of the parabola and the dashed lines
    • give the “classical turning points”
    • where a classical mass of that energy turns round and goes back downhill
  • $\omega$ is the angular frequency of a classical oscillator
    • with the same mass and
    • the same parabolic potential energy
• in QM eigenstates found in this particle in a potential problem
  • there is no oscillation
  • they are static solutions, albeit, probability distributions
  • if the modulus squared of any of the wave functions
• to find and analyze the actual oscillations, schrodinger’s time equation has to be used