[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
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
\[ \begin{align}
F & = -kx \rightarrow \text{ spring restoring force } \\
V(z) & = \int_{0}^{z} -F dz_0 \\
& = \int_{0}^{z} -K z_0 dz_0 \\
& = \frac{1}{2} K z^2 \\
& = \frac{1}{2} m \omega^2 z^2 \\
\end{align} \]
harmonic oscillator schrodinger equation
\[ \begin{align} V(z) = \frac{1}{2} m \omega^2 z^2 \\\ \end{align} \]
- schrodinger equation (time-independent) is
\[ \begin{align}
-\frac{\hbar^2}{2m} \frac{d^2 \psi}{dz^2} + \frac{1}{2} m\omega^2 z^2 \psi = E \psi
\end{align}
\]
- for convenience, we define a dimensionless distance unit \[ \begin{align} \xi = \sqrt{ \frac{m\omega}{h} } z \end{align} \]
- so the schrodinger’s equation becomes
\[ \begin{align}
\frac{1}{2} \frac{d^2 \psi }{d \xi^2} - \frac{\xi^2}{2} \psi = -\frac{E}{h\omega}\psi
\end{align}
\]
- one specific solution to this equation
\[ \begin{align}
\frac{1}{2} \frac{d^2 \psi}{d \xi^2} - \frac{\xi^2}{2} \psi & = -\frac{E}{ \hbar \omega} \psi \\
\psi & \propto \exp{\left( \frac{-\xi^2}{2} \right) } \\
\end{align} \] - with a corresponding energy \( E = \frac{\hbar \omega}{2} \)
- one specific solution to this equation
\[ \begin{align}
\frac{1}{2} \frac{d^2 \psi}{d \xi^2} - \frac{\xi^2}{2} \psi & = -\frac{E}{ \hbar \omega} \psi \\
- this suggests we look for solutions of the form
\[ \psi_{n}(\xi) = A_n \exp{ \left( \frac{-\xi^2}{2} \right) } H_n(\xi)
\]
- where: \( H_n(\xi) \): is a set of functions to be determined
\[ \begin{align}
\frac{d^2 H_n(\xi)}{d\xi^2} - 2 \xi \frac{dH_n(\xi)}{d\xi} + \left( \frac{2E}{h\omega} - 1 \right) H_n (\xi) & = 0 \\
\end{align}
\]
- this is the defining differential equation for the hermite polynomials
- solutions for hermite polynomials exist only if \[ \frac{2E}{\hbar\omega} - 1 = 2n \qquad \text{ n = 0,1,2,… } \]
- in terms of energy, solutions only exist for energy levels \[ E_n = \left( n + \frac{1}{2}\right) \hbar \omega \qquad \text{ n = 0,1,2,… } \]
- so, the allowed energy levels are equally spaced separated by an amount \( \hbar\omega \)
- \(\omega\): classical oscillation frequency
- like the potential well, there is a “zero-point energy”
- \( E_0 = \frac{\hbar\omega}{2} \)
hermite polynomial
- list of hermite polynomials
- \( H_0 = 1 \rightarrow \) even
- \( H_1(\xi) = 2\xi \rightarrow \) odd
- \( H_2(\xi) = 4\xi^2 - 2 \rightarrow \) even
- \( H_3(\xi) = 8\xi^3 - 12\xi \rightarrow \) odd
- \( H_4(\xi) = 16\xi^4 - 48\xi^2 + 12 \rightarrow \) 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. \( H_n(\xi) = 2\xi H_{n-1}(\xi) - 2(n-1)H_{n-2}(\xi) \)
- successive hermite polynomials can be calculated
- normalizing
\[ \psi_n(\xi)= A_n \exp{\left( \frac{-\xi^2}{2} \right) H_n(\xi) }
\]
- gives
\[ \begin{align}
A_n & = \sqrt{ \frac{1}{\sqrt{\pi} 2^n n!}} \\
\xi & = z \sqrt{ \frac{m\omega}{\hbar} } \\
E_n & = \left( n + \frac{1}{2}\right) \hbar \omega \qquad \text{ n = 0,1,2,… } \\
\\
\text{ expressing it } & \text{in terms of the original coordinates } \\
\psi_n(z) & = \sqrt{ \frac{1}{2^n n!} \sqrt{\frac{m\omega}{\pi \hbar}}} \exp{\left( -z^2 \frac{m\omega}{2\hbar} \right)} H_n\left( z \sqrt{\frac{m\omega}{\hbar}} \right) \end{align}
\]
- gives
\[ \begin{align}
A_n & = \sqrt{ \frac{1}{\sqrt{\pi} 2^n n!}} \\
- 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 this above plot, the inttersections of the parabola and the dashed lines
- 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