In quantum mechanics, an energy level is degenerate if it corresponds to two or more different measurable states of a quantum system. Conversely, two or more different states of a quantum mechanical system are said to be degenerate if they give the same value of energy upon measurement. The number of different states corresponding to a particular energy level is known as the degree of degeneracy (or simply the degeneracy) of the level. It is represented mathematically by the Hamiltonian for the system having more than one linearly independent eigenstate with the same energy eigenvalue. When this is the case, energy alone is not enough to characterize what state the system is in, and other quantum numbers are needed to characterize the exact state when distinction is desired. In classical mechanics, this can be understood in terms of different possible trajectories corresponding to the same energy.
Degeneracy plays a fundamental role in quantum statistical mechanics. For an -particle system in three dimensions, a single energy level may correspond to several different wave functions or energy states. These degenerate states at the same level all have an equal probability of being filled. The number of such states gives the degeneracy of a particular energy level.
upright=1.5|thumb|Degenerate states in a quantum system
Mathematics
The possible states of a quantum mechanical system may be treated mathematically as abstract vectors in a separable, complex Hilbert space, while the observables may be represented by linear Hermitian operators acting upon them. By selecting a suitable basis, the components of these vectors and the matrix elements of the operators in that basis may be determined.
If is a matrix, a non-zero vector, and is a scalar, such that <math>AX = \lambda X</math>, then the scalar is said to be an eigenvalue of and the vector is said to be the eigenvector corresponding to . Together with the zero vector, the set of all eigenvectors corresponding to a given eigenvalue form a subspace of , which is called the eigenspace of . An eigenvalue which corresponds to two or more different linearly independent eigenvectors is said to be degenerate, i.e., <math>A X_1 = \lambda X_1</math> and <math> A X_2 = \lambda X_2</math>, where <math> X_1 </math> and <math> X_2 </math> are linearly independent eigenvectors. The dimension of the eigenspace corresponding to that eigenvalue is known as its degree of degeneracy, which can be finite or infinite. An eigenvalue is said to be non-degenerate if its eigenspace is one-dimensional.
The eigenvalues of the matrices representing physical observables in quantum mechanics give the measurable values of these observables while the eigenstates corresponding to these eigenvalues give the possible states in which the system may be found, upon measurement. The measurable values of the energy of a quantum system are given by the eigenvalues of the Hamiltonian operator, while its eigenstates give the possible energy states of the system. A value of energy is said to be degenerate if there exist at least two linearly independent energy states associated with it. Moreover, any linear combination of two or more degenerate eigenstates is also an eigenstate of the Hamiltonian operator corresponding to the same energy eigenvalue. This clearly follows from the fact that the eigenspace of the energy value eigenvalue is a subspace (being the kernel of the Hamiltonian minus times the identity), hence is closed under linear combinations.
Effect of degeneracy on the measurement of energy
In the absence of degeneracy, if a measured value of energy of a quantum system is determined, the corresponding state of the system is assumed to be known, since only one eigenstate corresponds to each energy eigenvalue. However, if the Hamiltonian <math>\hat{H}</math> has a degenerate eigenvalue <math>E_n</math> of degree g<sub>n</sub>, the eigenstates associated with it form a vector subspace of dimension g<sub>n</sub>. In such a case, several final states can be possibly associated with the same result <math>E_n</math>, all of which are linear combinations of the g<sub>n</sub> orthonormal eigenvectors <math>|E_{n,i}\rangle</math>.
In this case, the probability that the energy value measured for a system in the state <math>|\psi\rangle</math> will yield the value <math>E_n</math> is given by the sum of the probabilities of finding the system in each of the states in this basis, i.e.,
<math display="block">P(E_n)=\sum_{i=1}^{g_n}|\langle E_{n,i}|\psi\rangle|^2</math>
Degeneracy in different dimensions
This section intends to illustrate the existence of degenerate energy levels in quantum systems studied in different dimensions. The study of one and two-dimensional systems aids the conceptual understanding of more complex systems.
Degeneracy in one dimension
In several cases, analytic results can be obtained more easily in the study of one-dimensional systems. For a quantum particle with a wave function <math>|\psi\rangle</math> moving in a one-dimensional potential <math>V(x)</math>, the time-independent Schrödinger equation can be written as
<math display="block"> -\frac{\hbar^2}{2m}\frac{d^2\psi}{dx^2} + V\psi = E\psi</math>
Since this is an ordinary differential equation, there are two independent eigenfunctions for a given energy <math>E</math> at most, so that the degree of degeneracy never exceeds two. It can be proven that in one dimension, there are no degenerate bound states for normalizable wave functions. A sufficient condition on a piecewise continuous potential <math>V</math> and the energy <math>E</math> is the existence of two real numbers <math>M,x_0</math> with <math>M \neq 0</math> such that <math>\forall x > x_0</math> we have <math>V(x) - E \geq M^2</math>. In particular, <math>V</math> is bounded below in this criterion.
:{| class="toccolours collapsible collapsed" width="60%" style="text-align:left"
!Proof of the above theorem.
|-
|Considering a one-dimensional quantum system in a potential <math>V(x)</math> with degenerate states <math>|\psi_1\rangle</math> and <math>|\psi_2\rangle</math> corresponding to the same energy eigenvalue <math>E</math>, writing the time-independent Schrödinger equation for the system:
<math display="block">\begin{align}
- \frac{\hbar^2}{2m}\frac{d^2\psi_1}{dx^2} + V\psi_1 &= E\psi_1 \\
- \frac{\hbar^2}{2m}\frac{d^2\psi_2}{dx^2} + V\psi_2 &= E\psi_2
\end{align}</math>
Multiplying the first equation by <math> \psi_2 </math> and the second by <math>\psi_1</math> and subtracting one from the other, we get:
<math display="block">\psi_1\frac{d^2}{dx^2}\psi_2-\psi_2\frac{d^2}{d x^2}\psi_1 = 0</math>
Integrating both sides
<math display="block">\psi_1\frac{d\psi_2}{dx} - \psi_2\frac{d\psi_1}{dx} = \mbox{constant}</math>
In case of well-defined and normalizable wave functions, the above constant vanishes, provided both the wave functions vanish at at least one point, and we find:
<math display="block">\psi_1(x) = c\psi_2(x)</math>
where <math>c</math> is, in general, a complex constant. For bound state eigenfunctions (which tend to zero as <math>x \to \infty</math>), and assuming <math>V</math> and <math>E</math> satisfy the condition given above, it can be shown It also results in conserved quantities, which are often not easy to identify. Accidental symmetries lead to these additional degeneracies in the discrete energy spectrum. An accidental degeneracy can be due to the fact that the group of the Hamiltonian is not complete. These degeneracies are connected to the existence of bound orbits in classical physics.
Examples: Coulomb and Harmonic Oscillator potentials
For a particle in a central potential, the Laplace–Runge–Lenz vector is a conserved quantity resulting from an accidental degeneracy, in addition to the conservation of angular momentum due to rotational invariance.
For a particle moving on a cone under the influence of and potentials, centred at the tip of the cone, the conserved quantities corresponding to accidental symmetry will be two components of an equivalent of the Runge-Lenz vector, in addition to one component of the angular momentum vector. These quantities generate SU(2) symmetry for both potentials.
Example: Particle in a constant magnetic field
A particle moving under the influence of a constant magnetic field, undergoing cyclotron motion on a circular orbit is another important example of an accidental symmetry. The symmetry multiplets in this case are the Landau levels which are infinitely degenerate.
Examples
The hydrogen atom
In atomic physics, the bound states of an electron in a hydrogen atom show us useful examples of degeneracy. In this case, the Hamiltonian commutes with the total orbital angular momentum <math>\hat{L}^2</math>, its component along the z-direction, <math>\hat{L}_z</math>, total spin angular momentum <math>\hat{S}^2</math> and its z-component <math>\hat{S}_z</math>. The quantum numbers corresponding to these operators are <math>\ell</math>, <math>m_\ell</math>, <math>s</math> (always 1/2 for an electron) and <math>m_s</math> respectively.
The energy levels in the hydrogen atom depend only on the principal quantum number . For a given , all the states corresponding to <math>\ell = 0, \ldots, n-1</math> have the same energy and are degenerate. Similarly for given values of and , the <math>(2\ell+1)</math>, states with <math>m_\ell = -\ell, \ldots, \ell</math> are degenerate. The degree of degeneracy of the energy level E<sub>n</sub> is therefore <math display="block">\sum_{\ell \mathop = 0}^{n-1} (2\ell+1) = n^2,</math> which is doubled if the spin degeneracy is included.