Quantum harmonic oscillator

In the one-dimensional harmonic oscillator problem, a particle of mass ${\displaystyle \displaystyle m}$  is subject to a potential ${\displaystyle \displaystyle V(x)={\frac {1}{2}}m\omega ^{2}x^{2}}$ . In classical mechanics, ${\displaystyle \displaystyle m\omega ^{2}=k}$  is called the spring stiffness coefficient, force constant or spring constant, and ${\displaystyle \displaystyle \omega }$  the circular frequency.

The Hamiltonian of the particle is:

${\displaystyle H={\frac {p^{2}}{2m}}+{\frac {1}{2}}m\omega ^{2}x^{2}}$ 

where x is the position operator, and p is the momentum operator ${\displaystyle \left(p=-i\hbar {d \over dx}\right)}$ . The first term represents the kinetic energy of the particle, and the second term represents the potential energy in which it resides. In order to find the energy levels and the corresponding energy eigenstates, we must solve the time-independent Schrödinger equation,

${\displaystyle H\left|\psi \right\rangle =E\left|\psi \right\rangle }$ .

We can solve the differential equation in the coordinate basis, using a spectral method. It turns out that there is a family of solutions,

${\displaystyle \left\langle x|\psi _{n}\right\rangle ={\sqrt {\frac {1}{2^{n}\,n!}}}\cdot \left({\frac {m\omega }{\pi \hbar }}\right)^{1/4}\cdot \exp \left(-{\frac {m\omega x^{2}}{2\hbar }}\right)\cdot H_{n}\left({\sqrt {\frac {m\omega }{\hbar }}}x\right)}$ 

${\displaystyle n=0,1,2,\ldots }$ 

The first six solutions (n = 0 to 5) are shown on the right. The functions ${\displaystyle H_{n}}$  are the Hermite polynomials:

${\displaystyle H_{n}(x)=(-1)^{n}e^{x^{2}}{\frac {d^{n}}{dx^{n}}}e^{-x^{2}}}$ 

They should not be confused with the Hamiltonian, which is also denoted by H. The corresponding energy levels are

${\displaystyle E_{n}=\hbar \omega \left(n+{1 \over 2}\right)}$ .

This energy spectrum is noteworthy for three reasons. Firstly, the energies are "quantized", and may only take the discrete values of ${\displaystyle \hbar \omega }$  times 1/2, 3/2, 5/2, and so forth. This is a feature of many quantum mechanical systems. In the following section on ladder operators, we will engage in a more detailed examination of this phenomenon. Secondly, the lowest achievable energy is not zero, but ${\displaystyle \hbar \omega /2}$ , which is called the "ground state energy" or zero-point energy. In the ground state, according to quantum mechanics, an oscillator performs null oscillations and its average kinetic energy is positive. It is not obvious that this is significant, because normally the zero of energy is not a physically meaningful quantity, only differences in energies. Nevertheless, the ground state energy has many implications, particularly in quantum gravity. The final reason is that the energy levels are equally spaced, unlike the Bohr model or the particle in a box.

Note that the ground state probability density is concentrated at the origin. This means the particle spends most of its time at the bottom of the potential well, as we would expect for a state with little energy. As the energy increases, the probability density becomes concentrated at the "classical turning points", where the state's energy coincides with the potential energy. This is consistent with the classical harmonic oscillator, in which the particle spends most of its time (and is therefore most likely to be found) at the turning points, where it is the slowest. The correspondence principle is thus satisfied.

The spectral method solution, though straightforward, is rather tedious. The "ladder operator" method, due to Paul Dirac, allows us to extract the energy eigenvalues without directly solving the differential equation. Furthermore, it is readily generalizable to more complicated problems, notably in quantum field theory. Following this approach, we define the operators a and its adjoint a^†

${\displaystyle {\begin{matrix}a&=&{\sqrt {m\omega \over 2\hbar }}\left(x+{i \over m\omega }p\right)\\a^{\dagger }&=&{\sqrt {m\omega \over 2\hbar }}\left(x-{i \over m\omega }p\right)\end{matrix}}}$ 

The operator a is not Hermitian since it and its adjoint a^† are not equal.

The operator a and a^† have properties as below:

${\displaystyle {\begin{matrix}a\left|\phi _{n}\right\rangle &=&{\sqrt {n}}\left|\phi _{n-1}\right\rangle \\a^{\dagger }\left|\phi _{n}\right\rangle &=&{\sqrt {n+1}}\left|\phi _{n+1}\right\rangle \end{matrix}}}$ 

We can also define a number operator N which has the following property:

${\displaystyle N=a^{\dagger }a}$ 

${\displaystyle N\left|\phi _{n}\right\rangle =n\left|\phi _{n}\right\rangle }$ 

In deriving the form of a^†, we have used the fact that the operators x and p, which represent observables, are Hermitian. These observable operators can be expressed as a linear combination of the ladder operators as

${\displaystyle {\begin{matrix}x&=&{\sqrt {\hbar \over 2m\omega }}\left(a^{\dagger }+a\right)\\p&=&i{\sqrt {{\hbar }m\omega \over 2}}\left(a^{\dagger }-a\right)\end{matrix}}}$ 

The x and p operators obey the following identity, known as the canonical commutation relation:

${\displaystyle \left[x,p\right]=i\hbar }$ .

The square brackets in this equation are a commonly-used notational device, known as the commutator, defined as

${\displaystyle \left[A,B\right]\ {\stackrel {\mathrm {def} }{=}}\ AB-BA}$ .

Using the above, we can prove the identities

${\displaystyle H=\hbar \omega \left(a^{\dagger }a+1/2\right)}$ 

${\displaystyle \left[a,a^{\dagger }\right]=1}$ .

Now, let ${\displaystyle \left|\psi _{E}\right\rangle }$  denote an energy eigenstate with energy E. The inner product of any ket with itself must be non-negative, so

${\displaystyle \left(a\left|\psi _{E}\right\rangle ,a\left|\psi _{E}\right\rangle \right)=\left\langle \psi _{E}\right|a^{\dagger }a\left|\psi _{E}\right\rangle \geq 0}$ .

Expressing a^†a in terms of the Hamiltonian:

${\displaystyle \left\langle \psi _{E}\right|{H \over \hbar \omega }-{1 \over 2}\left|\psi _{E}\right\rangle =\left({E \over \hbar \omega }-{1 \over 2}\right)\geq 0}$ ,

so that ${\displaystyle E\geq \hbar \omega /2}$ . Note that when (${\displaystyle a\left|\psi _{E}\right\rangle }$ ) is the zero ket (i.e. a ket with length zero), the inequality is saturated, so that ${\displaystyle E=\hbar \omega /2}$ . It is straightforward to check that there exists a state satisfying this condition; it is the ground (n = 0) state given in the preceding section.

Using the above identities, we can now show that the commutation relations of a and a^† with H are:

${\displaystyle {\begin{matrix}\left[H,a\right]&=&-\hbar \omega a\\\left[H,a^{\dagger }\right]&=&\hbar \omega a^{\dagger }\end{matrix}}}$ .

Thus, provided (${\displaystyle a\left|\psi _{E}\right\rangle }$ ) is not the zero ket,

${\displaystyle {\begin{matrix}H(a\left|\psi _{E}\right\rangle )&=&(\left[H,a\right]+aH)\left|\psi _{E}\right\rangle \\&=&(-\hbar \omega a+aE)\left|\psi _{E}\right\rangle \\&=&(E-\hbar \omega )(a\left|\psi _{E}\right\rangle )\end{matrix}}}$ .

Similarly, we can show that

${\displaystyle H(a^{\dagger }\left|\psi _{E}\right\rangle )=(E+\hbar \omega )(a^{\dagger }\left|\psi _{E}\right\rangle )}$ .

In other words, a acts on an eigenstate of energy E to produce, up to a multiplicative constant, another eigenstate of energy ${\displaystyle E-\hbar \omega }$ , and a^† acts on an eigenstate of energy E to produce an eigenstate of energy ${\displaystyle E+\hbar \omega }$ . For this reason, a is called a "lowering operator", and a^† a "raising operator". The two operators together are called "ladder operators". In quantum field theory, a and a^† 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, a, to produce another eigenstate with ${\displaystyle \hbar \omega }$ -less energy. By repeated application of the lowering operator, it seems that we can produce energy eigenstates down to E = −∞. However, this would contradict our earlier requirement that ${\displaystyle E\geq \hbar \omega /2}$ . Therefore, there must be a ground-state energy eigenstate, which we label ${\displaystyle \left|0\right\rangle }$  (not to be confused with the zero ket), such that

${\displaystyle a\left|0\right\rangle =0{\hbox{(zero ket)}}}$ .

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

${\displaystyle H\left|0\right\rangle =(\hbar \omega /2)\left|0\right\rangle }$ 

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

${\displaystyle H\left|n\right\rangle =\hbar \omega (n+1/2)\left|n\right\rangle }$ 

which matches the energy spectrum which we gave in the preceding section.

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 we measure energy in units of ${\displaystyle \hbar \omega }$  and distance in units of ${\displaystyle \left(\hbar /\left(m\omega \right)\right)^{1/2}}$ , then the Schrödinger equation becomes:

${\displaystyle H=-{1 \over 2}{d^{2} \over du^{2}}+{1 \over 2}u^{2}}$ ,

and the energy eigenfunctions and eigenvalues become

${\displaystyle \left\langle x|\psi _{n}\right\rangle ={1 \over {\sqrt {2^{n}n!}}}\pi ^{-1/4}{\hbox{exp}}(-u^{2}/2)H_{n}(u)}$ 

${\displaystyle E_{n}=n+{1 \over 2}}$ 

where ${\displaystyle H_{n}(u)}$  are the Hermite polynomials.

To avoid confusion, we will not adopt these natural units in this article. However, they frequently come in handy when performing calculations.

In diatomic molecules, the natural frequency can be found by: [1]

${\displaystyle \omega ={\sqrt {\frac {k}{m_{r}}}}}$ 

where

${\displaystyle \omega =2\pi f}$    is the angular frequency,

k is the bond force constant, and

${\displaystyle m_{r}}$  is the reduced mass.

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 x₁, ..., x_N. Corresponding to each position coordinate is a momentum; we label these p₁, ..., p_N. The canonical commutation relations between these operators are

${\displaystyle {\begin{matrix}\left[x_{i},p_{j}\right]&=&i\hbar \delta _{i,j}\\\left[x_{i},x_{j}\right]&=&0\\\left[p_{i},p_{j}\right]&=&0\end{matrix}}}$ .

The Hamiltonian for this system is

${\displaystyle H=\sum _{i=1}^{N}\left({p_{i}^{2} \over 2m}+{1 \over 2}m\omega ^{2}x_{i}^{2}\right)}$ .

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 x₁, ..., x_N would refer to the positions of each of the N particles. This is a happy property of the r² 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:

${\displaystyle \langle \mathbf {x} |\psi _{\{n\}}\rangle =\prod _{i=1}^{N}\langle x_{i}|\psi _{n_{i}}\rangle }$ 

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

${\displaystyle {\begin{matrix}a_{i}&=&{\sqrt {m\omega \over 2\hbar }}\left(x_{i}+{i \over m\omega }p_{i}\right)\\a_{i}^{\dagger }&=&{\sqrt {m\omega \over 2\hbar }}\left(x_{i}-{i \over m\omega }p_{i}\right)\end{matrix}}}$ .

By a procedure analogous 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 ℏω respectively. The Hamiltonian is

${\displaystyle H=\hbar \omega \,\sum _{i=1}^{N}\left(a_{i}^{\dagger }\,a_{i}+{\frac {1}{2}}\right).}$ 

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

${\displaystyle U\,a_{i}^{\dagger }\,U^{\dagger }=\sum _{j=1}^{N}a_{j}^{\dagger }\,U_{ji}\quad {\hbox{for all}}\quad U\in U(N),}$ 

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

The energy levels of the system are

${\displaystyle E=\hbar \omega \left[(n_{1}+\cdots +n_{N})+{N \over 2}\right]}$ .

${\displaystyle n_{i}=0,1,2,\dots \quad ({\hbox{the number of bosons in mode }}i).}$ 

As in the one-dimensional case, the energy is quantized. The ground state energy is N times the one-dimensional 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 = n₁ + n₂ + n₃. All states with the same n will have the same energy. For a given n, we choose a particular n₁. Then n₂ + n₃ = n − n₁. There are n − n₁ + 1 possible groups {n₂, n₃}. n₂ can take on the values 0 to n − n₁, and for each n₂ the value of n₃ is fixed. The degree of degeneracy therefore is:

${\displaystyle g_{n}=\sum _{n_{1}=0}^{n}n-n_{1}+1={\frac {(n+1)(n+2)}{2}}}$ 

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)]:

${\displaystyle g_{n}={\binom {N+n-1}{n}}}$ 

The special case N = 3, given above, follows directly from this general equation.

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

${\displaystyle V(r)={1 \over 2}\mu \omega ^{2}r^{2},}$ 

where ${\displaystyle \mu }$  is the mass of the problem. Because m will be used below for the magnetic quantum number, mass is indicated by ${\displaystyle \mu }$ , instead of m, as earlier in this article.

The solution reads

${\displaystyle \psi _{klm}(r,\theta ,\phi )=N_{kl}r^{l}e^{-\nu r^{2}}{L_{k}}^{(l+{1 \over 2})}(2\nu r^{2})Y_{lm}(\theta ,\phi )}$ 

where

${\displaystyle N_{kl}={\sqrt {{\sqrt {\frac {2\nu ^{3}}{\pi }}}{\frac {2^{k+2l+3}\;k!\;\nu ^{l}}{(2k+2l+1)!!}}}}\,}$  is a normalization constant.

${\displaystyle \nu \equiv {\mu \omega \over 2\hbar }}$ 

${\displaystyle {L_{k}}^{(l+{1 \over 2})}(2\nu r^{2})}$  are generalized Laguerre polynomials. The order k of the polynomial is a non-negative integer.

${\displaystyle Y_{lm}(\theta ,\phi )\,}$  is a spherical harmonic function.

${\displaystyle \hbar }$  is the reduced Planck constant: ${\displaystyle \hbar \equiv {\frac {h}{2\pi }}}$ .

The energy eigenvalue is

${\displaystyle E=\hbar \omega (2k+l+{3 \over 2})}$ 

The energy is usually described by the single quantum number

${\displaystyle n\equiv 2k+l\,}$ 

Because k is a non-negative integer, for every even n we have ${\displaystyle l=0,2...,n-2,n}$  and for every odd n we have ${\displaystyle l=1,3...,n-2,n}$ . The magnetic quantum number m is an integer satisfying ${\displaystyle -l\leq m\leq l}$ , so for every n and l there are 2l+1 different quantum states, labeled by m. Thus, the degeneracy at level n is

${\displaystyle \sum _{l=\ldots ,n-2,n}(2l+1)={(n+1)(n+2) \over 2}}$ 

where the sum starts from 0 or 1, according to whether n is even or odd. This result is in accordance with the dimension formula above.

In this problem, we consider N equal masses which are connected to their neighbors by springs, in the limit of large N. The masses form a linear chain in one dimension, or a regular lattice in two or three dimensions.

As in the previous section, we denote the positions of the masses by x₁, x₂, ..., as measured from their equilibrium positions (i.e. x_k = 0 if particle k is at its equilibrium position.) In two or more dimensions, the xs are vector quantities. The Hamiltonian of the total system is

${\displaystyle H=\sum _{i=1}^{N}{p_{i}^{2} \over 2m}+{1 \over 2}m\omega ^{2}\sum _{\{ij\}(nn)}(x_{i}-x_{j})^{2}.}$ 

The potential energy is summed over "nearest-neighbor" pairs, so there is one term for each spring.

Remarkably, there exists a coordinate transformation to turn this problem into a set of independent harmonic oscillators, each of which corresponds to a particular collective distortion of the lattice. These distortions display some particle-like properties, and are called phonons. Phonons occur in the ionic lattices of many solids, and are extremely important for understanding many of the phenomena studied in solid state physics.

• Gas in a harmonic trap
• Creation and annihilation operators
• Coherent state
• Morse potential

• Griffiths, David J. (2004). Introduction to Quantum Mechanics (2nd ed.). Prentice Hall. ISBN 0-13-805326-X.
• Liboff, Richard L. (2002). Introductory Quantum Mechanics. Addison-Wesley. ISBN 0-8053-8714-5.

• Quantum Harmonic Oscillator
• Calculation using a noncommutative free monoid: (mathematical version) / (abbreviated version)