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 maximum number of linearly independent 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 N-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.
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 A is a N × N matrix, X a non-zero vector, and λ is a scalar, such that A X = λ X {\displaystyle AX=\lambda X} , then the scalar λ is said to be an eigenvalue of A and the vector X is said to be the eigenvector corresponding to λ. Together with the zero vector, the set of all eigenvectors corresponding to a given eigenvalue λ form a subspace of Cn, which is called the eigenspace of λ. An eigenvalue λ which corresponds to two or more different linearly independent eigenvectors is said to be degenerate, i.e., A X 1 = λ X 1 {\displaystyle AX_{1}=\lambda X_{1}} and A X 2 = λ X 2 {\displaystyle AX_{2}=\lambda X_{2}} , where X 1 {\displaystyle X_{1}} and X 2 {\displaystyle X_{2}} 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 λ is a subspace (being the kernel of the Hamiltonian minus λ 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 H ^ {\displaystyle {\hat {H}}} has a degenerate eigenvalue E n {\displaystyle E_{n}} of degree gn, the eigenstates associated with it form a vector subspace of dimension gn. In such a case, several final states can be possibly associated with the same result E n {\displaystyle E_{n}} , all of which are linear combinations of the gn orthonormal eigenvectors | E n , i ⟩ {\displaystyle |E_{n,i}\rangle } . In this case, the probability that the energy value measured for a system in the state | ψ ⟩ {\displaystyle |\psi \rangle } will yield the value E n {\displaystyle E_{n}} is given by the sum of the probabilities of finding the system in each of the states in this basis, i.e.,
P ( E n ) = ∑ i = 1 g n | ⟨ E n , i | ψ ⟩ | 2 {\displaystyle P(E_{n})=\sum _{i=1}^{g_{n}}|\langle E_{n,i}|\psi \rangle |^{2}}
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 | ψ ⟩ {\displaystyle |\psi \rangle } moving in a one-dimensional potential V ( x ) {\displaystyle V(x)} , the time-independent Schrödinger equation can be written as
− ℏ 2 2 m d 2 ψ d x 2 + V ψ = E ψ {\displaystyle -{\frac {\hbar ^{2}}{2m}}{\frac {d^{2}\psi }{dx^{2}}}+V\psi =E\psi }
Since this is an ordinary differential equation, there are two independent eigenfunctions for a given energy E {\displaystyle E} 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 V {\displaystyle V} and the energy E {\displaystyle E} is the existence of two real numbers M , x 0 {\displaystyle M,x_{0}} with M ≠ 0 {\displaystyle M\neq 0} such that ∀ x > x 0 {\displaystyle \forall x>x_{0}} we have V ( x ) − E ≥ M 2 {\displaystyle V(x)-E\geq M^{2}} . In particular, V {\displaystyle V} is bounded below in this criterion.
Degeneracy in two-dimensional quantum systems Two-dimensional quantum systems exist in all three states of matter and much of the variety seen in three dimensional matter can be created in two dimensions. Real two-dimensional materials are made of monoatomic layers on the surface of solids. Some examples of two-dimensional electron systems achieved experimentally include MOSFET, two-dimensional superlattices of Helium, Neon, Argon, Xenon etc. and surface of liquid Helium. The presence of degenerate energy levels is studied in the cases of Particle in a box and two-dimensional harmonic oscillator, which act as useful mathematical models for several real world systems.
Particle in a rectangular plane Consider a free particle in a plane of dimensions L x {\displaystyle L_{x}} and L y {\displaystyle L_{y}} in a plane of impenetrable walls. The time-independent Schrödinger equation for this system with wave function | ψ ⟩ {\displaystyle |\psi \rangle } can be written as
− ℏ 2 2 m ( ∂ 2 ψ ∂ x 2 + ∂ 2 ψ ∂ y 2 ) = E ψ {\displaystyle -{\frac {\hbar ^{2}}{2m}}\left({\frac {\partial ^{2}\psi }{{\partial x}^{2}}}+{\frac {\partial ^{2}\psi }{{\partial y}^{2}}}\right)=E\psi }
The permitted energy values are
E n x , n y = π 2 ℏ 2 2 m ( n x 2 L x 2 + n y 2 L y 2 ) {\displaystyle E_{n_{x},n_{y}}={\frac {\pi ^{2}\hbar ^{2}}{2m}}\left({\frac {n_{x}^{2}}{L_{x}^{2}}}+{\frac {n_{y}^{2}}{L_{y}^{2}}}\right)}
The normalized wave function is
ψ n x , n y ( x , y ) = 2 L x L y sin ( n x π x L x ) sin ( n y π y L y ) {\displaystyle \psi _{n_{x},n_{y}}(x,y)={\frac {2}{\sqrt {L_{x}L_{y}}}}\sin \left({\frac {n_{x}\pi x}{L_{x}}}\right)\sin \left({\frac {n_{y}\pi y}{L_{y}}}\right)}
where n x , n y = 1 , 2 , 3 , … {\displaystyle n_{x},n_{y}=1,2,3,\dots }
So, quantum numbers n x {\displaystyle n_{x}} and n y {\displaystyle n_{y}} are required to describe the energy eigenvalues and the lowest energy of the system is given by
E 1 , 1 = π 2 ℏ 2 2 m ( 1 L x 2 + 1 L y 2 ) {\displaystyle E_{1,1}=\pi ^{2}{\frac {\hbar ^{2}}{2m}}\left({\frac {1}{L_{x}^{2}}}+{\frac {1}{L_{y}^{2}}}\right)}
For some commensurate ratios of the two lengths L x {\displaystyle L_{x}} and L y {\displaystyle L_{y}} , certain pairs of states are degenerate. If L x / L y = p / q {\displaystyle L_{x}/L_{y}=p/q} , where p and q are integers, the states ( n x , n y ) {\displaystyle (n_{x},n_{y})} and ( p n y / q , q n x / p ) {\displaystyle (pn_{y}/q,qn_{x}/p)} have the same energy and so are degenerate to each other.
Particle in a square box In this case, the dimensions of the box L x = L y = L {\displaystyle L_{x}=L_{y}=L} and the energy eigenvalues are given by
E n x , n y = π 2 ℏ 2 2 m L 2 ( n x 2 + n y 2 ) {\displaystyle E_{n_{x},n_{y}}={\frac {\pi ^{2}\hbar ^{2}}{2mL^{2}}}(n_{x}^{2}+n_{y}^{2})}
Since n x {\displaystyle n_{x}} and n y {\displaystyle n_{y}} can be interchanged without changing the energy, each energy level has a degeneracy of at least two when n x {\displaystyle n_{x}} and n y {\displaystyle n_{y}} are different. Degenerate states are also obtained when the sum of squares of quantum numbers corresponding to different energy levels are the same. For example, the three states (nx = 7, ny = 1), (nx = 1, ny = 7) and (nx = ny = 5) all have E = 50 π 2 ℏ 2 2 m L 2 {\displaystyle E=50{\frac {\pi ^{2}\hbar ^{2}}{2mL^{2}}}} and constitute a degenerate set. Degrees of degeneracy of different energy levels for a particle in a square box:
Particle in a cubic box In this case, the dimensions of the box L x = L y = L z = L {\displaystyle L_{x}=L_{y}=L_{z}=L} and the energy eigenvalues depend on three quantum numbers.
E n x , n y , n z = π 2 ℏ 2 2 m L 2 ( n x 2 + n y 2 + n z 2 ) {\displaystyle E_{n_{x},n_{y},n_{z}}={\frac {\pi ^{2}\hbar ^{2}}{2mL^{2}}}(n_{x}^{2}+n_{y}^{2}+n_{z}^{2})}
Since n x {\displaystyle n_{x}} , n y {\displaystyle n_{y}} and n z {\displaystyle n_{z}} can be interchanged without changing the energy, each energy level has a degeneracy of at least three when the three quantum numbers are not all equal.
Finding a unique eigenbasis in case of degeneracy If two operators A ^ {\displaystyle {\hat {A}}} and B ^ {\displaystyle {\hat {B}}} commute, i.e., [ A ^ , B ^ ] = 0 {\displaystyle [{\hat {A}},{\hat {B}}]=0} , then for every eigenvector | ψ ⟩ {\displaystyle |\psi \rangle } of A ^ {\displaystyle {\hat {A}}} , B ^ | ψ ⟩ {\displaystyle {\hat {B}}|\psi \rangle } is also an eigenvector of A ^ {\displaystyle {\hat {A}}} with the same eigenvalue. However, if this eigenvalue, say λ {\displaystyle \lambda } , is degenerate, it can be said that B ^ | ψ ⟩ {\displaystyle {\hat {B}}|\psi \rangle } belongs to the eigenspace E λ {\displaystyle E_{\lambda }} of A ^ {\displaystyle {\hat {A}}} , which is said to be globally invariant under the action of B ^ {\displaystyle {\hat {B}}} . For two commuting observables A and B, one can construct an orthonormal basis of the state space with eigenvectors common to the two operators. However, λ {\displaystyle \lambda } is a degenerate eigenvalue of A ^ {\displaystyle {\hat {A}}} , then it is an eigensubspace of A ^ {\displaystyle {\hat {A}}} that is invariant under the action of B ^ {\displaystyle {\hat {B}}} , so the representation of B ^ {\displaystyle {\hat {B}}} in the eigenbasis of A ^ {\displaystyle {\hat {A}}} is not a diagonal but a block diagonal matrix, i.e. the degenerate eigenvectors of A ^ {\displaystyle {\hat {A}}} are not, in general, eigenvectors of B ^ {\displaystyle {\hat {B}}} . However, it is always possible to choose, in every degenerate eigensubspace of A ^ {\displaystyle {\hat {A}}} , a basis of eigenvectors common to A ^ {\displaystyle {\hat {A}}} and B ^ {\displaystyle {\hat {B}}} .
Choosing a complete set of commuting observables If a given observable A is non-degenerate, there exists a unique basis formed by its eigenvectors. On the other hand, if one or several eigenvalues of A ^ {\displaystyle {\hat {A}}} are degenerate, specifying an eigenvalue is not sufficient to characterize a basis vector. If, by choosing an observable B ^ {\displaystyle {\hat {B}}} , which commutes with A ^ {\displaystyle {\hat {A}}} , it is possible to construct an orthonormal basis of eigenvectors common to A ^ {\displaystyle {\hat {A}}} and B ^ {\displaystyle {\hat {B}}} , which is unique, for each of the possible pairs of eigenvalues {a,b}, then A ^ {\displaystyle {\hat {A}}} and B ^ {\displaystyle {\hat {B}}} are said to form a complete set of commuting observables. However, if a unique set of eigenvectors can still not be specified, for at least one of the pairs of eigenvalues, a third observable C ^ {\displaystyle {\hat {C}}} , which commutes with both A ^ {\displaystyle {\hat {A}}} and B ^ {\displaystyle {\hat {B}}} can be found such that the three form a complete set of commuting observables. It follows that the eigenfunctions of the Hamiltonian of a quantum system with a common energy value must be labelled by giving some additional information, which can be done by choosing an operator that commutes with the Hamiltonian. These additional labels required naming of a unique energy eigenfunction and are usually related to the constants of motion of the system.
Degenerate energy eigenstates and the parity operator The parity operator is defined by its action in the | r ⟩ {\displaystyle |r\rangle } representation of changing r to −r, i.e.
⟨ r | P | ψ ⟩ = ψ ( − r ) {\displaystyle \langle r|P|\psi \rangle =\psi (-r)}
The eigenvalues of P can be shown to be limited to ± 1 {\displaystyle \pm 1} , which are both degenerate eigenvalues in an infinite-dimensional state space. An eigenvector of P with eigenvalue +1 is said to be even, while that with eigenvalue −1 is said to be odd. Now, an even operator A ^ {\displaystyle {\hat {A}}} is one that satisfies,
A ~ = P A ^ P {\displaystyle {\tilde {A}}=P{\hat {A}}P}
[ P , A ^ ] = 0 {\displaystyle [P,{\hat {A}}]=0}
while an odd operator B ^ {\displaystyle {\hat {B}}} is one that satisfies
P B ^ + B ^ P = 0 {\displaystyle P{\hat {B}}+{\hat {B}}P=0}
Since the square of the momentum operator p ^ 2 {\displaystyle {\hat {p}}^{2}} is even, if the potential V(r) is even, the Hamiltonian H ^ {\displaystyle {\hat {H}}} is said to be an even operator. In that case, if each of its eigenvalues are non-degenerate, each eigenvector is necessarily an eigenstate of P, and therefore it is possible to look for the eigenstates of H ^ {\displaystyle {\hat {H}}} among even and odd states. However, if one of the energy eigenstates has no definite parity, it can be asserted that the corresponding eigenvalue is degenerate, and P | ψ ⟩ {\displaystyle P|\psi \rangle } is an eigenvector of H ^ {\displaystyle {\hat {H}}} with the same eigenvalue as | ψ ⟩ {\displaystyle |\psi \rangle } .
Degeneracy and symmetry The physical origin of degeneracy in a quantum-mechanical system is often the presence of some symmetry in the system. Studying the symmetry of a quantum system can, in some cases, enable us to find the energy levels and degeneracies without solving the Schrödinger equation, hence reducing effort. Mathematically, the relation of degeneracy with symmetry can be clarified as follows. Consider a symmetry operation associated with a unitary operator S. Under such an operation, the new Hamiltonian is related to the original Hamiltonian by a similarity transformation generated by the operator S, such that H ′ = S H S − 1 = S H S † {\displaystyle H'=SHS^{-1}=SHS^{\dagger }} , since S is unitary. If the Hamiltonian remains unchanged under the transformation operation S, we have
S H S † = H S H S − 1 = H S H = H S [ S , H ] = 0 {\displaystyle {\begin{aligned}SHS^{\dagger }&=H\\[1ex]SHS^{-1}&=H\\[1ex]SH&=HS\\[1ex][S,H]&=0\end{aligned}}}
Now, if | α ⟩ {\displaystyle |\alpha \rangle } is an energy eigenstate,
H | α ⟩ = E | α ⟩ {\displaystyle H|\alpha \rangle =E|\alpha \rangle }
where E is the corresponding energy eigenvalue.
H S | α ⟩ = S H | α ⟩ = S E | α ⟩ = E S | α ⟩ {\displaystyle HS|\alpha \rangle =SH|\alpha \rangle =SE|\alpha \rangle =ES|\alpha \rangle }
which means that S | α ⟩ {\displaystyle S|\alpha \rangle } is also an energy eigenstate with the same eigenvalue E. If the two states | α ⟩ {\displaystyle |\alpha \rangle } and S | α ⟩ {\displaystyle S|\alpha \rangle } are linearly independent (i.e. physically distinct), they are therefore degenerate. In cases where S is characterized by a continuous parameter ϵ {\displaystyle \epsilon } , all states of the form S ( ϵ ) | α ⟩ {\displaystyle S(\epsilon )|\alpha \rangle } have the same energy eigenvalue.
Symmetry group of the Hamiltonian The set of all operators which commute with the Hamiltonian of a quantum system are said to form the symmetry group of the Hamiltonian. The commutators of the generators of this group determine the algebra of the group. An n-dimensional representation of the Symmetry group preserves the multiplication table of the symmetry operators. The possible degeneracies of the Hamiltonian with a particular symmetry group are given by the dimensionalities of the irreducible representations of the group. The eigenfunctions corresponding to a n-fold degenerate eigenvalue form a basis for a n-dimensional irreducible representation of the Symmetry group of the Hamiltonian.
Types of degeneracy Degeneracies in a quantum system can be systematic or accidental in nature.
Systematic or essential degeneracy This is also called a geometrical or normal degeneracy and arises due to the presence of some kind of symmetry in the system under consideration, i.e. the invariance of the Hamiltonian under a certain operation, as described above. The representation obtained from a normal degeneracy is irreducible and the corresponding eigenfunctions form a basis for this representation.
Accidental degeneracy It is a type of degeneracy resulting from some special features of the system or the functional form of the potential under consideration, and is related possibly to a hidden dynamical symmetry in the system. It also results in conserved quantities, which are often not easy to identify.