Editing Quantum harmonic oscillator (section)

====Analytical questions====

The preceding analysis is algebraic, using only the commutation relations between the raising and lowering operators. Once the algebraic analysis is complete, one should turn to analytical questions. First, one should find the ground state, that is, the solution of the equation <math>a\psi_0 = 0</math>. In the position representation, this is the first-order differential equation
<math display="block">\left(x+\frac{\hbar}{m\omega}\frac{d}{dx}\right)\psi_0 = 0,</math>
whose solution is easily found to be the [[Gaussian_function|Gaussian]]<ref group="nb">The normalization constant is <math>C = \left(\frac{m\omega}{\pi \hbar}\right)^{{1}/{4}}</math>, and satisfies the normalization condition <math>\int_{-\infty}^{\infty}\psi_0(x)^{*}\psi_0(x)dx = 1</math>.</ref>
<math display="block">\psi_0(x)=Ce^{-\frac{m\omega x^2}{2\hbar}}.</math>
Conceptually, it is important that there is only one solution of this equation; if there were, say, two linearly independent ground states, we would get two independent chains of eigenvectors for the harmonic oscillator. Once the ground state is computed, one can show inductively that the excited states are Hermite polynomials times the Gaussian ground state, using the explicit form of the raising operator in the position representation. One can also prove that, as expected from the uniqueness of the ground state, the Hermite functions energy eigenstates <math>\psi_n</math> constructed by the ladder method form a ''complete'' orthonormal set of functions.<ref>{{citation|first=Brian C.|last=Hall | title=Quantum Theory for Mathematicians|series=Graduate Texts in Mathematics|volume=267|isbn=978-1461471158 |publisher=Springer|year=2013 |bibcode=2013qtm..book.....H | at = Theorem 11.4}}</ref>

Explicitly connecting with the previous section, the ground state  |0⟩  in the position representation is determined by <math> a| 0\rangle =0</math>,
<math display="block">  \left\langle x \mid a \mid 0 \right\rangle = 0 \qquad
  \Rightarrow \left(x + \frac{\hbar}{m\omega}\frac{d}{dx}\right)\left\langle x\mid 0\right\rangle = 0 \qquad
  \Rightarrow           </math>
<math display="block"> \left\langle x\mid 0\right\rangle = \left(\frac{m\omega}{\pi\hbar}\right)^\frac{1}{4} \exp\left( -\frac{m\omega}{2\hbar}x^2 \right) 
    = \psi_0  ~,</math>
hence
<math display="block"> \langle x \mid a^\dagger \mid 0 \rangle = \psi_1 (x) ~,</math>
so that <math>\psi_1(x,t)=\langle x \mid e^{-3i\omega t/2} a^\dagger \mid 0 \rangle </math>, and so on.