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# Quantum harmonic oscillator

The quantum harmonic oscillator is a quantum mechanical analogue of the classical harmonic oscillator. It is one of the most important problems in quantum mechanics, because (i) a simple exact solution exists, and (ii) a wide variety of physical situations can be reduced to this. In particular, a system near an equilibrium configuration can often be described in terms of one or more harmonic oscillators.

The following discussion of the quantum harmonic oscillator relies on the article Mathematical formulation of quantum mechanics.

 Table of contents

### Hamiltonian and energy eigenstates

In the one-dimensional harmonic oscillator problem, a particle of mass m is subject to a potential V(x) = (1/2)mω2 x2. The Hamiltonian of the particle is:

$\mathbf{H} = {p^2 \over 2m} + {1\over 2} m \omega^2 x^2$

where x is the position operator, and p is the momentum operator (p = - iℏ ∂ /∂x). In order to find energy eigenstates |ψE⟩, and the corresponding energy levels E, we must solve the time-independent Schrödinger equation,

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

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

$\left\langle x | \psi_n \right\rangle = \frac{1}{\sqrt{2^n n!}} \left(\frac{m\omega}{\pi \hbar}\right)^{1/4} \hbox{exp} \left(- \frac{m\omega x^2}{2 \hbar} \right) H_n\left(\sqrt{\frac{m\omega}{\hbar}} x \right)$
$n = 0, 1, 2, ...$

The functions Hn(θ) are the Hermite polynomials. The corresponding energy levels are

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

This energy spectrum is noteworthy for two reasons. Firstly, the energies are "quantized", and may only take the discrete values of ℏω 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 ℏω/2, which is called the "ground state energy". 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 probability densities of the energy eigenstates are shown below, beginning with the ground state (n = 0) at the bottom of the picture and increasing in energy toward the top of the picture. The horizontal axis corresponds to the position x, and brighter colors represent higher probability densities.

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.

### Ladder operator method

The power series 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 operator

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

where a is the Hermitian conjugate of a. Note that a is not Hermitian, since a and a are not equal. In deriving the form of a, we have used the fact that the operators x and p, which represent observables, are Hermitian.

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

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

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

$\left[A , B \right] \equiv AB - BA$.

Using the above, we can prove the identities

$\mathbf{H} = \hbar \omega \left(a^{\dagger}a + 1/2\right)$
$\left[a , a^{\dagger} \right] = 1$.

Now, let |ψE⟩ denote an energy eigenstate with energy E. The inner product of any ket with itself must be non-negative, so

$\left(a \left|\psi_E \right\rangle, a \left|\psi_E \right\rangle\right) = \left\langle\psi_E \right| a^\dagger a \left| \psi_E \right\rangle \ge 0$.

Expressing aa in terms of the Hamiltonian:

$\left\langle\psi_E \right| {H \over \hbar \omega} - {1 \over 2} \left|\psi_E\right\rangle = \left({E \over \hbar \omega} - {1 \over 2} \right) \ge 0$,

so that E ≥ (ℏω / 2). Note that when (a|ψE⟩) is the zero ket (i.e. a ket with length zero), the inequality is saturated, so that E = (ℏω/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:

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

Thus, provided (a|ψE⟩) is not the zero ket,

$\begin{matrix} \mathbf{H} (a \left| \psi_E \right\rangle)  &=& (\left[\mathbf{H},a\right] + a \mathbf{H}) \left|\psi_E\right\rangle \\ &=& (- \hbar\omega a + a E) \left|\psi_E\right\rangle \\ &=& (E - \hbar\omega) (a\left|\psi_E\right\rangle)  \end{matrix}$.

Similarly, we can show that

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

In other words, a acts on an eigenstate of energy E to produce, up to a multiplicative constant, another eigenstate of energy (E - ℏω), and a acts on an eigenstate of energy E to produce an eigenstate of energy (E + ℏω.) 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 ℏω 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 E ≥ (ℏω / 2). Therefore, there must be a ground-state energy eigenstate, which we label |0⟩ (not to be confused with the zero ket), such that

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

In this case, subsequent applications of the lowering operator will just produce zero kets, instead of additional energy eigenstate. Furthermore, we have shown above that

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

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

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

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

### Natural length and energy scales

The quantum harmonic oscillator possesses natural scales for length and energy, which can be used to simplify the problem. If we measure energy in units of ℏω and distance in units of (ℏ/(mω))1/2, then the Schrödinger equation becomes:

$\mathbf{H} = - {1\over2} {\partial^2 \over \partial u^2 } + {1 \over 2} u^2$,

and the energy eigenfunctions and eigenvalues become

$\left\langle x | \psi_n \right\rangle = {1 \over \sqrt{2^n n!}} \pi^{-1/4} \hbox{exp} (-u^2 / 2) H_n(u)$
$E_n = n + {1\over 2}$.

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

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

$\begin{matrix} \left[x_i , p_j \right] &=& i\hbar\delta_{i,j} \\ \left[x_i , x_j \right] &=& 0 \\ \left[p_i , p_j \right] &=& 0 \end{matrix}$.

The Hamiltonian for this system is

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

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

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

By a procedure analogous to the one-dimensional case, we can then show that each of the ai and ai operators lower and raise the energy by ℏω respectively. The energy levels of the system are

$E = \hbar \omega \left[(n_1 + ... + n_N) + {N\over 2}\right]$.
$n_i = 0, 1, 2, ...$

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 quantum harmonic oscillator can be extended in many interesting ways. We will briefly discuss two of the more important extensions, the anharmonic oscillator and coupled harmonic oscillators.

### Anharmonic oscillator

As mentioned in the introduction, a system residing "near" the minimum of some potential may be treated as a harmonic oscillator. In this approximation, we Taylor expand the potential energy around the minimum and discard terms of third or higher order, resulting in an approximate quadratic potential. Once we have studied the system in this approximation, we may wish to investigate the corrections due to the discarded higher-order terms, particularly the third-order term.

The anharmonic oscillator Hamiltonian is the harmonic oscillator Hamiltonian with an additional x3 potential:

$\mathbf{H} = {p^2 \over 2m} + {1\over 2} m \omega^2 x^2 + \lambda x^3$

If the harmonic approximation is valid, the coefficient λ is small compared to the quadratic term. We may therefore use perturbation theory to determine the corrections to the states and energy levels imposed by the anharmonic term. This task may be simplified by using the ladder operators to rewrite the anharmonic term as

$\lambda \left({\hbar \over 2m\omega}\right)^{3\over2} (a + a^\dagger)^3$.

It turns out that the correction to the energies vanish to first-order in λ. The second-order corrections are given by the usual formula in perturbation theory:

$\Delta E^{(2)} = \lambda^2 \left\langle \psi_E \right| x^3 {1 \over E - H_0} x^3 \left| \psi_E \right\rangle$.

This is straightforward, though tedious, to evaluate.

### Coupled Harmonic Oscillators

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. xk = 0 if particle k is at its equilibrium position.) In two or more dimensions, the xs are are vector quantities. The Hamiltonian of the total system is

$\mathbf{H} = \sum_{i=1}^N {p_i^2 \over 2m} + {1\over 2} m \omega^2 \sum_{\{ij\} (nn)} (x_i - x_j)^2$

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.

All Wikipedia text is available under the terms of the GNU Free Documentation License

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