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=                    Quantum harmonic oscillator                     =
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                            Introduction
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The quantum harmonic oscillator is the quantum-mechanical analog of
the classical harmonic oscillator.   Because an arbitrary potential
can usually be approximated as a harmonic potential at the vicinity of
a stable equilibrium point,  it is one of the most important model
systems in quantum mechanics. Furthermore, it is one of the few
quantum-mechanical systems for which an exact, analytical solution is
known.


Hamiltonian and energy eigenstates
====================================
Wavefunction representations for the first eight bound eigenstates,
'n' = 0 to 7. The horizontal axis shows the position 'x'. Note: The
graphs are not normalized, and the signs of some of the functions
differ from those given in the text.
Corresponding probability densities.

The Hamiltonian of the particle is:
:\hat H = \frac{{\hat p}^2}{2m} + \frac{1}{2} k {\hat x}^2 =
\frac{{\hat p}^2}{2m} + \frac{1}{2} m \omega^2 {\hat x}^2\, ,
where  is the particle's mass,  is the force constant, \omega =
\sqrt{\frac{k}{m}} is the angular frequency of the oscillator, \hat{x}
is the position operator (given by ), and   \hat{p}  is the momentum
operator (given by \hat p = - i \hbar {\partial \over \partial x} \,).
The first term in the Hamiltonian represents the kinetic energy of the
particle, and the second term represents its potential energy, as in
Hooke's law.

One may write the time-independent Schrödinger equation,
:: \hat H \left| \psi \right\rangle = E \left| \psi \right\rangle  ~,
where  denotes a to-be-determined real number that will specify a
time-independent energy level, or eigenvalue, and the solution
denotes that level's energy eigenstate.

One may solve the differential equation representing this eigenvalue
problem in the coordinate basis, for the wave function , using a
spectral method. It turns out that there is a family of solutions. In
this basis, they amount to  Hermite functions,
:  \psi_n(x) = \frac{1}{\sqrt{2^n\,n!}} \cdot \left(\frac{m\omega}{\pi
\hbar}\right)^{1/4} \cdot e^{
- \frac{m\omega x^2}{2 \hbar}} \cdot
H_n\left(\sqrt{\frac{m\omega}{\hbar}} x \right), \qquad n =
0,1,2,\ldots.

The functions 'Hn' are the physicists' Hermite polynomials,
:H_n(z)=(-1)^n~ e^{z^2}\frac{d^n}{dz^n}\left(e^{-z^2}\right).

The corresponding energy levels are
: E_n = \hbar \omega \left(n + {1\over 2}\right)=(2 n + 1) {\hbar
\over 2} \omega~.

This energy spectrum is noteworthy for three reasons.  First, the
energies are quantized, meaning that only discrete energy values
(integer-plus-half multiples of ) are possible; this is a general
feature of quantum-mechanical systems when a particle is confined.
Second, these discrete energy levels are equally spaced, unlike in the
Bohr model of the atom, or the particle in a box.  Third, the lowest
achievable energy (the energy of the  state, called the ground state)
is not equal to the minimum of the potential well, but  above it; this
is called zero-point energy. Because of the zero-point energy, the
position and momentum of the oscillator in the ground state are not
fixed (as they would be in a classical oscillator), but have a small
range of variance, in accordance with the Heisenberg uncertainty
principle.

The ground state probability density is concentrated at the origin,
which means the particle spends most of its time at the bottom of the
potential well, as one would expect for a state with little energy. As
the energy increases, the probability density  peaks at the classical
"turning points", where the state's energy coincides with the
potential energy. (See the discussion below of the highly excited
states.) This is consistent with the classical harmonic oscillator, in
which the particle spends more of its time (and is therefore more
likely to be found) near the turning points, where it is moving the
slowest. The correspondence principle is thus satisfied. Moreover,
special nondispersive wave packets, with minimum uncertainty,  called
coherent states oscillate very much like classical objects, as
illustrated in the figure; they are 'not' eigenstates of the
Hamiltonian.


Ladder operator method
========================
2  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  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. It is generalizable to more complicated
problems, notably in quantum field theory.  Following this approach,
we define the operators  and its adjoint ,
:\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}

This leads to the useful representation of \hat{x} and  \hat{p},
:\begin{align}
\hat x &=  \sqrt{\frac{\hbar}{2}\frac{1}{m\omega}}(a^\dagger +
a) \\
\hat p &= i\sqrt{\frac{\hbar}{2}m\omega}(a^\dagger - a) ~.
\end{align}

The operator  is not Hermitian, since itself and its adjoint  are not
equal. The energy eigenstates , when operated on by these ladder
operators, give
:\begin{align}
a^\dagger|n\rangle &= \sqrt{n + 1} | n + 1\rangle \\
a|n\rangle &= \sqrt{n} | n - 1\rangle.
\end{align}

It is then evident that , in essence, appends a single quantum of
energy to the oscillator, while  removes a quantum. For this reason,
they are sometimes referred to as "creation" and "annihilation"
operators.

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

The following commutators can be easily obtained by substituting the
canonical commutation relation,
:[a, a^\dagger] = 1,\qquad[N, a^\dagger] = a^{\dagger},\qquad[N, a] =
-a,

And the Hamilton operator can be expressed as
:\hat H = \hbar\omega\left(N + \frac{1}{2}\right),

so the eigenstate of  is also the eigenstate of energy.

The commutation property yields
:\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}

and similarly,
:Na|n\rangle = (n - 1)a | n \rangle.

This means that  acts on    to produce, up to a multiplicative
constant,   , and  acts on    to produce . For this reason,  is called
a annihilation operator ("lowering operator"), and  a creation
operator ("raising operator"). The two operators together are called
ladder operators. In quantum field theory,  and  are alternatively
called "annihilation" and "creation" operators because they destroy
and create particles, which correspond to our quanta of energy.

Given any energy eigenstate, we can act on it with the lowering
operator, , to produce another eigenstate with  less energy. By
repeated application of the lowering operator, it seems that we can
produce energy eigenstates down to . However, since
:n = \langle n | N | n \rangle = \langle n | a^\dagger a | n \rangle =
\Bigl(a | n \rangle \Bigr)^\dagger a | n \rangle \geqslant 0,

the smallest eigen-number is 0, and
:a \left| 0 \right\rangle = 0.

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

Finally, by acting on  |0� with the raising operator and multiplying
by suitable normalization factors, we can produce an infinite set of
energy eigenstates
:\left\{\left| 0 \right\rangle, \left| 1 \right\rangle, \left| 2
\right\rangle, \ldots , \left| n \right\rangle, \ldots\right\},

such that

:\hat H \left| n \right\rangle = \hbar\omega \left( n + \frac{1}{2}
\right) \left| n \right\rangle,

which matches the energy spectrum given in the preceding section.

Arbitrary eigenstates can be expressed in terms of |0�,

:|n\rangle = \frac{(a^\dagger)^n}{\sqrt{n!}} |0\rangle.
:Proof:
::\begin{align}
\langle n | aa^\dagger | n \rangle &= \langle n|\left([a,
a^\dagger] + a^\dagger a\right)| n \rangle = \langle n|(N +
1)|n\rangle = n + 1 \\
\Rightarrow a^\dagger | n\rangle &= \sqrt{n + 1} | n +
1\rangle \\
\Rightarrow|n\rangle &=
\frac{a^\dagger}{\sqrt{n}} | n - 1 \rangle =
\frac{(a^\dagger)^2}{\sqrt{n(n - 1)}} | n - 2 \rangle = \cdots =
\frac{(a^\dagger)^n}{\sqrt{n!}}|0\rangle.
\end{align}


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 a\psi_0 = 0. In the position representation,
this is the first-order differential equation
:\left(x+\frac{\hbar}{m\omega}\frac{d}{dx}\right)\psi_0=0,
whose solution is easily found to be the Gaussian
The normalization constant is C = \left(\frac{m\omega}{\pi
\hbar}\right)^{\frac{1}{4}}, and satisfies the normalization condition
\int_{-\infty}^{\infty}\psi_0(x)^{*}\psi_0(x)dx = 1.
:\psi_0(x)=Ce^{-\frac{m\omega x^2}{2\hbar}}.
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 energy eigenstates \psi_n
constructed by the ladder method form a 'complete' orthonormal set of
functions.

Explicitly connecting with the previous section, the ground state  |0�
in the position representation is determined by  a| 0\rangle =0,
:  \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
:                                \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  ~,
hence
: \langle x \mid a^\dagger \mid 0 \rangle = \psi_1 (x) ~,
so that \psi_1(x,t)=\langle x \mid e^{-3i\omega t/2} a^\dagger \mid 0
\rangle , and so on.


Natural length and energy scales
==================================
The quantum harmonic oscillator possesses natural scales for length
and energy, which can be used to simplify the problem. These can be
found by nondimensionalization.

The result is that, if   'energy' is measured in units of   and
'distance' in units of , then the Hamiltonian  simplifies to
:: H = -\frac{1}{2} {d^2 \over dx^2} +\frac{1}{2}  x^2 ,
while the energy eigenfunctions and eigenvalues simplify to Hermite
functions and integers offset by a half,
::\psi_n(x)= \left\langle x \mid n \right\rangle = {1 \over \sqrt{2^n
n!}}~ \pi^{-1/4} \exp(-x^2 / 2)~ H_n(x),
::E_n = n + \tfrac{1}{2} ~,
where  are the Hermite polynomials.

To avoid confusion,  these "natural units"   will mostly not be
adopted in this article. However, they frequently come in handy when
performing calculations, by bypassing clutter.

For example, the fundamental solution (propagator) of , the
time-dependent Schrödinger operator for this oscillator, simply boils
down to the Mehler kernel,
::\langle x \mid \exp (-itH) \mid y \rangle \equiv K(x,y;t)=
\frac{1}{\sqrt{2\pi i \sin t}} \exp \left(\frac{i}{2\sin t}\left
((x^2+y^2)\cos t - 2xy\right )\right )~,
where . The most general solution for a given initial configuration
then is simply
::\psi(x,t)=\int dy~ K(x,y;t) \psi(y,0) ~.


Coherent states
=================
The coherent states of the harmonic oscillator are special
nondispersive wave packets, with minimum uncertainty , whose
observables' expectation values evolve like a classical system.  They
are eigenvectors of the annihilation operator, 'not' the Hamiltonian,
and form an overcomplete basis which consequentially lacks
orthogonality.

The coherent states are indexed by  and expressed in the  basis as

:|\alpha\rangle = \sum_{n=0}^\infty |n\rangle \langle n | \alpha
\rangle = e^{-\frac{1}{2} |\alpha|^2}
\sum_{n=0}^\infty\frac{\alpha^n}{\sqrt{n!}} |n\rangle =
e^{-\frac{1}{2} |\alpha|^2} e^{\alpha a^\dagger} |0\rangle.

Because a \left| 0 \right\rangle = 0  and via the Kermack-McCrae
identity, the last form is equivalent to a unitary displacement
operator acting on the ground state: |\alpha\rangle=e^{\alpha \hat
a^\dagger - \alpha^*\hat a}|0\rangle  = D(\alpha)|0\rangle.  The
position space wave functions are

:\psi_\alpha(x')= \left(\frac{m\omega}{\pi\hbar}\right)^{\frac{1}{4}}
e^{\frac{i}{\hbar} \langle\hat{p}\rangle_\alpha x' -
\frac{m\omega}{2\hbar}(x' - \langle\hat{x}\rangle_\alpha)^2} .


Highly excited states
=======================
When  is large, the eigenstates are  localized into the classical
allowed region, that is, the region in which a classical particle with
energy  can move. The eigenstates are peaked near the turning points:
the points at the ends of the classically allowed region where the
classical particle changes direction. This phenomenon can be verified
through asymptotics of the Hermite polynomials,  and also through the
WKB approximation.

The frequency of oscillation at  is proportional to the momentum  of a
classical particle of energy   and position .  Furthermore, the square
of the amplitude (determining the probability density) is 'inversely'
proportional to  ,  reflecting the  length of time the classical
particle spends near .   The system behavior in a small neighborhood
of the turning point does not have a simple classical explanation, but
can be modeled using an Airy function. Using properties of the Airy
function,  one may estimate the probability of finding the particle
outside the classically allowed region,  to be approximately
:\frac{2}{n^{1/3}3^{2/3}\Gamma^2(\tfrac{1}{3})}=\frac{1}{n^{1/3}\cdot
7.46408092658...}
This is also given, asymptotically, by the integral
:\frac{1}{2\pi}\int_{0}^{\infty}e^{(2n+1)\left
(x-\tfrac{1}{2}\sinh(2x)  \right )}dx ~.


Phase space solutions
=======================
In the phase space formulation of quantum mechanics, solutions to the
quantum harmonic oscillator in several different representations of
the quasiprobability distribution can be written in closed form.  The
most widely used of these is for the Wigner quasiprobability
distribution, which has the solution
:F_n(u) = \frac{(-1)^n}{\pi \hbar} L_n\left(4\frac{u}{\hbar
\omega}\right) e^{-2u/\hbar \omega} ~,
where
:u=\frac{1}{2} m \omega^2 x^2 + \frac{p^2}{2m},
and 'Ln' are the Laguerre polynomials.

This example illustrates how the Hermite  and Laguerre polynomials are
linked through the Wigner map.

Meanwhile, the Husimi Q function of the harmonic oscillator
eigenstates have an even simpler form. If we work in the natural units
described above, we have
:Q(\psi_n)(x,p)=\frac{(x^2+p^2)^n}{n!}\frac{e^{-(x^2+p^2)}}{\pi}
This claim can be verified using the Segal-Bargmann transform.
Specifically, since the raising operator in the Segal-Bargmann
representation is simply multiplication by z=x+ip and the ground state
is the constant function 1, the normalized harmonic oscillator states
in this representation are simply z^n/\sqrt{n!} . At this point, we
can appeal to the formula for the Husimi Q function in terms of the
Segal-Bargmann transform.


               ''N''-dimensional harmonic oscillator
======================================================================
The one-dimensional harmonic oscillator is readily generalizable to
'N' dimensions, where 'N' = 1, 2, 3, ... . In one dimension, the
position of the particle was specified by a single coordinate, 'x'. In
'N' dimensions, this is replaced by 'N' position coordinates, which we
label 'x'1, ..., 'x''N'. Corresponding to each position coordinate is
a momentum; we label these 'p'1, ..., 'p''N'. The canonical
commutation relations between these operators are

:
\begin{align}
{[}x_i , p_j{]} &= i\hbar\delta_{i,j} \\
{[}x_i , x_j{]} &= 0                  \\
{[}p_i , p_j{]} &= 0
\end{align}


The Hamiltonian for this system is

: H = \sum_{i=1}^N \left( {p_i^2 \over 2m} + {1\over 2} m \omega^2
x_i^2 \right).

As the form of this Hamiltonian makes clear, the 'N'-dimensional
harmonic oscillator is exactly analogous to 'N' independent
one-dimensional harmonic oscillators with the same mass and spring
constant. In this case, the quantities 'x'1, ..., 'x''N' would refer
to the positions of each of the 'N' particles. This is a convenient
property of the r^2 potential, which allows the potential energy to be
separated into terms depending on one coordinate each.

This observation makes the solution straightforward. For a particular
set of quantum numbers {'n'} the energy eigenfunctions for the
'N'-dimensional oscillator are expressed in terms of the 1-dimensional
eigenfunctions as:

:
\langle \mathbf{x}|\psi_{\{n\}}\rangle
=\prod_{i=1}^N\langle x_i\mid \psi_{n_i}\rangle


In the ladder operator method, we define 'N' sets of ladder operators,

:\begin{align}
a_i &= \sqrt{m\omega \over 2\hbar} \left(x_i + {i \over m \omega}
p_i \right), \\
a^{\dagger}_i &= \sqrt{m \omega \over 2\hbar} \left( x_i - {i
\over m \omega} p_i \right).
\end{align}

By an analogous procedure to the one-dimensional case, we can then
show that each of the 'a''i' and 'a'� 'i' operators lower and raise the
energy by �� respectively. The Hamiltonian is
:
H =  \hbar \omega \, \sum_{i=1}^N \left(a_i^\dagger \,a_i +
\frac{1}{2}\right).

This Hamiltonian is invariant under the dynamic symmetry group
'U'('N') (the unitary group in 'N' dimensions), defined by

:
U\, a_i^\dagger \,U^\dagger = \sum_{j=1}^N
a_j^\dagger\,U_{ji}\quad\text{for all}\quad
U \in U(N),

where U_{ji} is an element in the defining matrix representation of
'U'('N').

The energy levels of the system are

: E = \hbar \omega \left[(n_1 + \cdots + n_N) + {N\over 2}\right].
:n_i = 0, 1, 2, \dots \quad (\text{the energy level in dimension } i).


As in the one-dimensional case, the energy is quantized. The ground
state energy is 'N' times the one-dimensional ground energy, as we
would expect using the analogy to 'N' independent one-dimensional
oscillators. There is one further difference: in the one-dimensional
case, each energy level corresponds to a unique quantum state. In
'N'-dimensions, except for the ground state, the energy levels are
'degenerate', meaning there are several states with the same energy.

The degeneracy can be calculated relatively easily.  As an example,
consider the 3-dimensional case: Define 'n' = 'n'1 + 'n'2 + 'n'3. All
states with the same 'n' will have the same energy.  For a given 'n',
we choose a particular 'n'1. Then 'n'2 + 'n'3 = 'n' � 'n'1.   There
are 'n' � 'n'1 + 1 possible pairs {'n'2, 'n'3}.  'n'2 can take on the
values 0 to 'n' � 'n'1, and for each 'n'2 the value of 'n'3 is fixed.
The degree of degeneracy therefore is:

:
g_n = \sum_{n_1=0}^n n - n_1 + 1 = \frac{(n+1)(n+2)}{2}

Formula for general 'N' and 'n' ['g''n' being the dimension of the
symmetric irreducible 'n'th power representation of the unitary group
'U'('N')]:
:
g_n = \binom{N+n-1}{n}

The special case 'N' = 3, given above, follows directly from this
general equation.  This is however, only true for distinguishable
particles, or one particle in N dimensions (as dimensions are
distinguishable). For the case of 'N' bosons in a one-dimension
harmonic trap, the degeneracy scales as the number of ways to
partition an integer 'n' using integers less than or equal to 'N'.

:
g_n = p(N_{-},n)


This arises due to the constraint of putting 'N' quanta into a state
ket where \sum_{k=0}^\infty k n_k = n     and  \sum_{k=0}^\infty  n_k
= N , which are the same constraints as in integer partition.


Example: 3D isotropic harmonic oscillator
===========================================
The Schrödinger equation of a spherically-symmetric three-dimensional
harmonic oscillator can be solved explicitly by separation of
variables; see this article for the present case. This procedure is
analogous to the separation performed in the hydrogen-like atom
problem, but with the spherically symmetric potential
:V(r) = {1\over 2} \mu \omega^2 r^2,
where  is the mass of the problem. Because  will be used below for the
magnetic quantum number, mass is indicated by  , instead of ,   as
earlier in this article.

The solution reads
:\psi_{klm}(r,\theta,\phi) = N_{kl} r^{l}e^{-\nu r^2}L_k^{(l+{1\over
2})}(2\nu r^2) Y_{lm}(\theta,\phi)
where
:N_{kl}=\sqrt{\sqrt{\frac{2\nu^3}{\pi }}\frac{2^{k+2l+3}\;k!\;\nu^l}{
(2k+2l+1)!!}}~~ is a normalization constant; \nu \equiv {\mu \omega
\over 2 \hbar}~;
:{L_k}^{(l+{1\over 2})}(2\nu r^2)
are generalized Laguerre polynomials; The order   of the polynomial is
a non-negative integer;
:Y_{lm}(\theta,\phi)\, is a spherical harmonic function;
:  is the reduced Planck constant:   \hbar\equiv\frac{h}{2\pi}~.

The energy eigenvalue is
:E=\hbar \omega \left(2k+l+\frac{3}{2}\right) ~.
The energy is usually described by the single quantum number
:n\equiv 2k+l  ~.

Because  is a non-negative integer, for every even  we have  and for
every odd   we have  . The magnetic quantum number  is an integer
satisfying ,    so for every  and �  there are 2'�' + 1 different
quantum states, labeled by  . Thus, the degeneracy at level    is
:\sum_{l=\ldots,n-2,n} (2l+1) = {(n+1)(n+2)\over 2} ~,
where the sum starts from 0 or 1, according to whether  is even or
odd.
This result is in accordance with the dimension formula above, and
amounts to the dimensionality of a symmetric representation of , the
relevant degeneracy group.


Harmonic oscillators lattice: phonons
=======================================
We can extend the notion of a harmonic oscillator to a one-dimensional
lattice of many particles. Consider a one-dimensional quantum
mechanical 'harmonic chain' of 'N' identical atoms. This is the
simplest quantum mechanical model of a lattice, and we will see how
phonons arise from it. The formalism that we will develop for this
model is readily generalizable to two and three dimensions.

As in the previous section, we denote the positions of the masses by
, as measured from their equilibrium positions (i.e.    = 0  if the
particle  is at its equilibrium position). In two or more dimensions,
the  are vector quantities. The Hamiltonian for this system is

:\mathbf{H} = \sum_{i=1}^N {p_i^2 \over 2m} + {1\over 2} m \omega^2
\sum_{\{ij\} (nn)} (x_i - x_j)^2 ~,
where  is the (assumed uniform) mass of each atom, and   and   are the
position and momentum operators for the 'i' th atom and the sum is
made over the nearest neighbors (nn).  However, it is customary to
rewrite the Hamiltonian in terms of the normal modes of the wavevector
rather than in terms of the particle coordinates so that one can work
in the more convenient Fourier space.

We introduce, then, a set of  "normal coordinates" , defined as the
discrete Fourier transforms of the s, and  "conjugate momenta"
defined as the Fourier transforms of the s,
:
Q_k = {1\over\sqrt{N}} \sum_{l} e^{ikal} x_l
:
\Pi_{k} = {1\over\sqrt{N}} \sum_{l}  e^{-ikal} p_l    ~.


The quantity  will turn out to be the wave number of the phonon, i.e.
2'�'  divided by the wavelength. It takes on quantized values, because
the number of atoms is finite.

This preserves the desired commutation relations in either real space
or wave vector space
:  \begin{align}
\left[x_l , p_m \right]&=i\hbar\delta_{l,m} \\
\left[ Q_k , \Pi_{k'} \right] &={1\over N} \sum_{l,m} e^{ikal}
e^{-ik'am}  [x_l , p_m ] \\
&= {i \hbar\over N} \sum_{m} e^{iam(k-k')} = i\hbar\delta_{k,k'}
\\
\left[ Q_k , Q_{k'} \right] &= \left[ \Pi_k , \Pi_{k'} \right] = 0
~.
\end{align}

From the general result
:  \begin{align}
\sum_{l}x_l x_{l+m}&={1\over N}\sum_{kk'}Q_k Q_{k'}\sum_{l}
e^{ial\left(k+k'\right)}e^{iamk'}= \sum_{k}Q_k Q_{-k}e^{iamk} \\
\sum_{l}{p_l}^2 &= \sum_{k}\Pi_k \Pi_{-k}   ~,
\end{align}
it is easy to show, through elementary trigonometry, that the
potential energy term is
:
{1\over 2} m \omega^2 \sum_{j} (x_j - x_{j+1})^2= {1\over 2} m
\omega^2\sum_{k}Q_k Q_{-k}(2-e^{ika}-e^{-ika})= {1\over 2} m
\sum_{k}{\omega_k}^2Q_k Q_{-k} ~ ,
where
:\omega_k = \sqrt{2 \omega^2 (1 - \cos(ka))} ~.

The Hamiltonian may be written in wave vector space as
:\mathbf{H} = {1\over {2m}}\sum_k \left(
{ \Pi_k\Pi_{-k} } + m^2 \omega_k^2 Q_k Q_{-k}
\right) ~.

Note that the couplings between the position variables have been
transformed away; if the s and s were hermitian(which they are not),
the transformed Hamiltonian would describe  'uncoupled' harmonic
oscillators.

The form of the quantization depends on the choice of boundary
conditions; for simplicity, we impose 'periodic' boundary conditions,
defining the th atom as equivalent to the first atom. Physically, this
corresponds to joining the chain at its ends. The resulting
quantization is

:k=k_n = {2n\pi \over Na}
\quad \hbox{for}\ n = 0, \pm1, \pm2, \ldots , \pm {N \over 2}.\

The upper bound to  comes from the minimum wavelength, which is twice
the lattice spacing , as discussed above.

The  harmonic oscillator eigenvalues or energy levels for the mode
are
::E_n = \left({1\over2}+n\right)\hbar\omega_k   \quad\hbox{for}\quad
n=0,1,2,3,\ldots

If we ignore the zero-point energy then the levels are evenly spaced
at
::\hbar\omega,\, 2\hbar\omega,\, 3\hbar\omega,\, \ldots

So an exact amount of energy ,  must be supplied to the harmonic
oscillator lattice to push it to the next energy level. In comparison
to the photon case when the electromagnetic field is quantised, the
quantum of vibrational energy is called a phonon.

All quantum systems show wave-like and particle-like properties. The
particle-like properties of the phonon are best understood using the
methods of second quantization and operator techniques described
later.

In the continuum limit, 'a'�0, 'N'��, while 'Na' is held fixed. The
canonical coordinates 'Qk' devolve to the decoupled momentum modes of
a scalar field, \phi_k, whilst the location index  ('not the
displacement dynamical  variable') becomes the 'parameter  argument of
the scalar field, \phi (x,t).


Molecular vibrations
======================
* The vibrations of a diatomic molecule are an example of a two-body
version of the quantum harmonic oscillator. In this case, the angular
frequency is given by
::\omega = \sqrt{\frac{k}{\mu}}
:where \mu = \frac{m_1 m_2}{m_1 + m_2} is the reduced mass and m_1 and
m_2 are the masses of the two atoms.
* The Hooke's atom is a simple model of the helium atom using the
quantum harmonic oscillator.
* Modelling phonons, as discussed above.
* A charge q, with mass m,  in a uniform magnetic field \mathbf{B}, is
an example of a one-dimensional quantum harmonic oscillator: the
Landau quantization.


                              See also
======================================================================
*Quantum machine
*Gas in a harmonic trap
*Creation and annihilation operators
*Coherent state
*Morse potential
*Bertrand's theorem
*Mehler kernel
*Molecular vibration


                           External links
======================================================================
*[http://hyperphysics.phy-astr.gsu.edu/hbase/quantum/hosc.html Quantum
Harmonic Oscillator]
*[http://behindtheguesses.blogspot.com/2009/03/quantum-harmonic-oscillator-ladde
r.html
Rationale for choosing the ladder operators]
*[http://www.brummerblogs.com/curvature/3d-harmonic-oscillator-eigenfunctions/
Live 3D intensity plots of quantum harmonic oscillator]
*[http://ltl.tkk.fi/~ethuneb/courses/monqo.pdf  Driven and damped
quantum harmonic oscillator (lecture notes of course "quantum optics
in electric circuits")]


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