Editing Quantum harmonic oscillator (section)

===Ladder operator method===
[[Image:QHarmonicOscillator.png|right|thumb|Probability densities <nowiki>|</nowiki>''ψ<sub>n</sub>''(''x'')<nowiki>|</nowiki><sup>2</sup> <!--or in pseudoTeX: <math>\left |\psi_n(x)\right |^2</math> --> for the bound eigenstates, beginning with the ground state (''n'' = 0) at the bottom and increasing in energy toward the top. The horizontal axis shows the position {{mvar|x}}, and brighter colors represent higher probability densities.]]

The "[[ladder operator]]" method, developed by [[Paul Dirac]], allows extraction of the energy eigenvalues without directly solving the differential equation.{{sfnp|Zwiebach|2022|pp=246-249}} It is generalizable to more complicated problems, notably in [[quantum field theory]].  Following this approach, we define the operators {{mvar|a}} and its [[Hermitian adjoint|adjoint]] {{math|''a''<sup>†</sup>}},
<math display="block">\begin{align}
          a &=\sqrt{m\omega \over 2\hbar} \left(\hat x + {i \over m \omega} \hat p \right) \\
  a^\dagger &=\sqrt{m\omega \over 2\hbar} \left(\hat x - {i \over m \omega} \hat p \right)
\end{align}</math>Note these operators classically are exactly the [[Generator (mathematics)|generators]] of normalized rotation in the phase space of <math>x</math> and <math>m\frac{dx}{dt}</math>, ''i.e'' they describe the forwards and backwards evolution in time of a classical harmonic oscillator.{{clarify|reason=Probably correct, e.g. phase space is incompressible and therefore this is a rotation, Obscure reference, expand argument about time evolution operator, representation theory and unitarity on [[Ladder operators]] page and link here|date=June 2024}}

These operators lead to the following representation of <math>\hat{x}</math> and  <math>\hat{p}</math>,
<math display="block">\begin{align}
  \hat x &=  \sqrt{\frac{\hbar}{2 m\omega}}(a^\dagger + a) \\
  \hat p &= i\sqrt{\frac{\hbar m \omega}{2}}(a^\dagger - a) ~.
\end{align}</math>

The operator {{mvar|a}} is not [[Hermitian operator|Hermitian]], since itself and its adjoint {{math|''a''<sup>†</sup>}} are not equal. The energy eigenstates {{math|{{ket|''n''}}}}, when operated on by these ladder operators, give
<math display="block">\begin{align}
  a^\dagger|n\rangle &= \sqrt{n + 1} | n + 1\rangle \\
          a|n\rangle &= \sqrt{n} | n - 1\rangle.
\end{align}</math>

From the relations above, we can also define a number operator {{mvar|N}}, which has the following property:
<math display="block">\begin{align}
                        N &= a^\dagger a \\
  N\left| n \right\rangle &= n\left| n \right\rangle.
\end{align}</math>

The following [[commutator]]s can be easily obtained by substituting the [[canonical commutation relation]],
<math display="block">[a, a^\dagger] = 1,\qquad[N, a^\dagger] = a^{\dagger},\qquad[N, a] = -a, </math>

and the Hamilton operator can be expressed as
<math display="block">\hat H = \hbar\omega\left(N + \frac{1}{2}\right),</math>

so the eigenstates of {{mvar|N}} are also the eigenstates of energy.
To see that, we can apply <math>\hat{H}</math> to a number state <math>|n\rangle</math>:

<math display="block">
\hat{H} |n\rangle = \hbar \omega \left(\hat{N} + \frac{1}{2}\right) |n\rangle.
</math>

Using the property of the number operator <math>\hat{N}</math>:

<math display="block">
\hat{N} |n\rangle = n |n\rangle,
</math>

we get:

<math display="block">
\hat{H} |n\rangle = \hbar \omega \left(n + \frac{1}{2}\right) |n\rangle.
</math>

Thus, since <math>|n\rangle</math> solves the TISE for the Hamiltonian operator <math>\hat{H}</math>, is also one of its eigenstates with the corresponding eigenvalue:

<math display="block">
E_n = \hbar \omega \left(n + \frac{1}{2}\right) .
</math>

QED.

The commutation property yields
<math display="block">\begin{align}
  Na^{\dagger}|n\rangle &= \left(a^\dagger N + [N, a^\dagger]\right)|n\rangle \\
                        &= \left(a^\dagger N + a^\dagger\right)|n\rangle \\
                        &= (n + 1)a^\dagger|n\rangle,
\end{align} </math>

and similarly,
<math display="block">Na|n\rangle = (n - 1)a | n \rangle.</math>

This means that {{mvar|a}} acts on  {{math|{{!}}''n''⟩}}  to produce, up to a multiplicative constant,   {{math|{{!}}''n''–1⟩}}, and {{math|''a''<sup>†</sup>}} acts on   {{math|{{!}}''n''⟩}} to produce {{math|{{!}}''n''+1⟩}}. For this reason, {{mvar|a}} is called an '''annihilation operator''' ("lowering operator"), and {{math|''a''<sup>†</sup>}} a '''creation operator''' ("raising operator"). The two operators together are called [[ladder operator]]s. 

Given any energy eigenstate, we can act on it with the lowering operator, {{mvar|a}}, to produce another eigenstate with {{math|''ħω''}} less energy. By repeated application of the lowering operator, it seems that we can produce energy eigenstates down to {{math|1=''E'' = −∞}}. However, since
<math display="block">n = \langle n | N | n \rangle = \langle n | a^\dagger a | n \rangle = \Bigl(a | n \rangle \Bigr)^\dagger a | n \rangle \geqslant 0,</math>

the smallest eigenvalue of the number operator is 0, and
<math display="block">a \left| 0 \right\rangle = 0. </math>

In this case, subsequent applications of the lowering operator will just produce zero, instead of additional energy eigenstates. Furthermore, we have shown above that
<math display="block">\hat H \left|0\right\rangle = \frac{\hbar\omega}{2} \left|0\right\rangle</math>

Finally, by acting on  |0⟩ with the raising operator and multiplying by suitable [[Wave function#Normalization condition|normalization factors]], we can produce an infinite set of energy eigenstates 
<math display="block">\left\{\left| 0 \right\rangle, \left| 1 \right\rangle, \left| 2 \right\rangle, \ldots , \left| n \right\rangle, \ldots\right\},</math>

such that
<math display="block">\hat H \left| n \right\rangle = \hbar\omega \left( n + \frac{1}{2} \right) \left| n \right\rangle, </math>
which matches the energy spectrum given in the preceding section.

Arbitrary eigenstates can be expressed in terms of |0⟩,{{sfnp|Zwiebach|2022|p=248}}
<math display="block">|n\rangle = \frac{(a^\dagger)^n}{\sqrt{n!}} |0\rangle. </math>
{{math proof|<math display="block">\begin{align}
   \langle n | aa^\dagger | n \rangle &= \langle n|\left([a, a^\dagger] + a^\dagger a\right) \left| n \right\rangle = \langle n| \left(N + 1\right) |n\rangle = n + 1 \\[1ex]
     \Rightarrow a^\dagger | n\rangle &= \sqrt{n + 1} | n + 1\rangle \\[1ex]
                    \Rightarrow|n\rangle &= \frac{1}{\sqrt{n}} a^\dagger \left| n - 1 \right\rangle = \frac{1}{\sqrt{n(n - 1)}} \left(a^\dagger\right)^2 \left| n - 2 \right\rangle = \cdots = \frac{1}{\sqrt{n!}} \left(a^\dagger\right)^n \left|0\right\rangle.
\end{align}</math>}}

====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.