The harmonic oscillator is one of the most important models in quantum mechanics because any smooth potential $V(x)$ can be approximated as a harmonic oscillator near its stable equilibrium point (minimum). It describes atomic vibrations in molecules, phonons in solid state lattices, and even electromagnetic fields in quantum optics.
The classical potential for a harmonic oscillator is $V(x) = \frac{1}{2}m\omega^2 x^2$. In quantum mechanics, we solve the Schrödinger equation with this potential to find the quantized energy levels and wavefunctions.
The Time-Independent Schrödinger Equation (TISE) is:
$$ -\frac{\hbar^2}{2m}\frac{d^2\psi}{dx^2} + \frac{1}{2}m\omega^2 x^2 \psi = E\psi $$We introduce dimensionless variables to simplify. Let $\xi = \alpha x$ where $\alpha = \sqrt{\frac{m\omega}{\hbar}}$, and let $K = \frac{2E}{\hbar\omega}$. The equation becomes:
$$ \frac{d^2\psi}{d\xi^2} + (K - \xi^2)\psi = 0 $$For large $\xi$ ($\xi^2 \gg K$), the equation is approximately $\psi'' - \xi^2 \psi \approx 0$. The normalizable solution behaves as $e^{-\xi^2/2}$. Thus, we look for a solution of the form:
$$ \psi(\xi) = h(\xi)e^{-\xi^2/2} $$Substituting this back yields the Hermite differential equation for $h(\xi)$:
$$ \frac{d^2h}{d\xi^2} - 2\xi\frac{dh}{d\xi} + (K - 1)h = 0 $$We assume a power series solution $h(\xi) = \sum_{j=0}^{\infty} a_j \xi^j$. Substituting this into the Hermite equation yields a recursion relation for the coefficients:
$$ a_{j+2} = \frac{2j - (K - 1)}{(j+1)(j+2)} a_j $$For the wavefunction to be physically acceptable (normalizable), the series must terminate after a finite number of terms. If the series does not terminate, it behaves like $e^{\xi^2}$ at infinity, destroying normalizability.
To force the series to terminate at some highest power $n$, the numerator of the recursion relation must be zero for $j=n$. Therefore:
$$ 2n - (K - 1) = 0 \implies K = 2n + 1 $$Since $K = \frac{2E}{\hbar\omega}$, we find the quantized energy levels:
$$ \frac{2E_n}{\hbar\omega} = 2n + 1 \implies E_n = \left(n + \frac{1}{2}\right)\hbar\omega \quad \text{for } n = 0, 1, 2, \dots $$The resulting polynomial solutions $h(\xi)$ are the Hermite polynomials $H_n(\xi)$.
The expectation value of energy is $\langle E \rangle = P_0 E_0 + P_1 E_1$.
We know $E_0 = \frac{1}{2}\hbar\omega$ and $E_1 = \frac{3}{2}\hbar\omega$. Also, probability $P_0 = |c_1|^2$ and $P_1 = |c_2|^2$.
Normalization requires $|c_1|^2 + |c_2|^2 = 1 \implies |c_2|^2 = 1 - |c_1|^2$.
$$ \langle E \rangle = |c_1|^2 \left(\frac{1}{2}\hbar\omega\right) + (1 - |c_1|^2)\left(\frac{3}{2}\hbar\omega\right) = \frac{3}{4}\hbar\omega $$ $$ \frac{1}{2}|c_1|^2 + \frac{3}{2} - \frac{3}{2}|c_1|^2 = \frac{3}{4} $$ $$ -|c_1|^2 = \frac{3}{4} - \frac{3}{2} = -\frac{3}{4} \implies |c_1|^2 = \frac{3}{4} $$Thus, $|c_2|^2 = 1 - \frac{3}{4} = \frac{1}{4}$.
The ratio is $|c_1|^2 / |c_2|^2 = (3/4) / (1/4) = 3$.
First order correction is $E_0^{(1)} = \langle \psi_0 | H' | \psi_0 \rangle = \lambda \langle 0 | x^4 | 0 \rangle$.
Using ladder operators, $x = \sqrt{\frac{\hbar}{2m\omega}}(a + a^\dagger)$.
For $x^4$, we need terms in $(a+a^\dagger)^4$ that return to the $|0\rangle$ state, which are the terms with an equal number of $a$ and $a^\dagger$ operators. Specifically, we need terms that act on $|0\rangle$ without annihilating it.
Alternatively, using standard expectation values for the harmonic oscillator: $\langle 0 | x^4 | 0 \rangle = \frac{3}{4}\left(\frac{\hbar}{m\omega}\right)^2$.
Therefore, $E_0^{(1)} = \frac{3\lambda\hbar^2}{4m^2\omega^2}$.