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Degenerate energy levels

From Wikipedia, the free encyclopedia
Energy level of a quantum system that corresponds to two or more different measurable states
"Quantum degeneracy" redirects here. For the state of matter, seeDegenerate matter. For other uses, seeDegeneracy.

Inquantum mechanics, anenergy level isdegenerate if it corresponds to two or more different measurable states of aquantum 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 thedegree of degeneracy (or simply thedegeneracy) of the level. It is represented mathematically by theHamiltonian for the system having more than onelinearly independenteigenstate with the same energyeigenvalue.[1]: 48  When this is the case, energy alone is not enough to characterize what state the system is in, and otherquantum numbers are needed to characterize the exact state when distinction is desired. Inclassical mechanics, this can be understood in terms of different possible trajectories corresponding to the same energy.

Degeneracy plays a fundamental role inquantum statistical mechanics. For anN-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.

Degenerate states in a quantum system

Mathematics

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The possible states of a quantum mechanical system may be treated mathematically as abstract vectors in a separable, complexHilbert space, while theobservables may be represented bylinearHermitian operators acting upon them. By selecting a suitablebasis, the components of these vectors and the matrix elements of the operators in that basis may be determined. IfA is aN × N matrix,X a non-zerovector, andλ is ascalar, such thatAX=λX{\displaystyle AX=\lambda X}, then the scalarλ is said to be an eigenvalue ofA and the vectorX is said to be the eigenvector corresponding to λ. Together with the zero vector, the set of alleigenvectors corresponding to a given eigenvalueλ form asubspace ofCn, which is called theeigenspace ofλ. An eigenvalueλ which corresponds to two or more different linearly independent eigenvectors is said to bedegenerate, i.e.,AX1=λX1{\displaystyle AX_{1}=\lambda X_{1}} andAX2=λX2{\displaystyle AX_{2}=\lambda X_{2}}, whereX1{\displaystyle X_{1}} andX2{\displaystyle X_{2}} are linearly independent eigenvectors. Thedimension of the eigenspace corresponding to that eigenvalue is known as itsdegree 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 physicalobservables inquantum 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, anylinear 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 eigenvalueλ is a subspace (being thekernel of the Hamiltonian minusλ times the identity), hence is closed under linear combinations.

Proof of the above theorem.[2]: p. 52 

IfH^{\displaystyle {\hat {H}}} represents theHamiltonian operator and|ψ1⟩{\displaystyle |\psi _{1}\rangle } and|ψ2⟩{\displaystyle |\psi _{2}\rangle } are two eigenstates corresponding to the same eigenvalueE, thenH^|ψ1⟩=E|ψ1⟩H^|ψ2⟩=E|ψ2⟩{\displaystyle {\begin{aligned}{\hat {H}}|\psi _{1}\rangle &=E|\psi _{1}\rangle \\{\hat {H}}|\psi _{2}\rangle &=E|\psi _{2}\rangle \end{aligned}}}

Let|ψ⟩=c1|ψ1⟩+c2|ψ2⟩{\displaystyle |\psi \rangle =c_{1}|\psi _{1}\rangle +c_{2}|\psi _{2}\rangle }, wherec1{\displaystyle c_{1}} andc2{\displaystyle c_{2}} are complex(in general) constants, be any linear combination of|ψ1⟩{\displaystyle |\psi _{1}\rangle } and|ψ2⟩{\displaystyle |\psi _{2}\rangle }.Then,H^|ψ⟩=H^(c1|ψ1⟩+c2|ψ2⟩)=c1H^|ψ1⟩+c2H^|ψ2⟩=E(c1|ψ1⟩+c2|ψ2⟩)=E|ψ⟩{\displaystyle {\begin{aligned}{\hat {H}}|\psi \rangle &={\hat {H}}(c_{1}|\psi _{1}\rangle +c_{2}|\psi _{2}\rangle )\\&=c_{1}{\hat {H}}|\psi _{1}\rangle +c_{2}{\hat {H}}|\psi _{2}\rangle \\&=E(c_{1}|\psi _{1}\rangle +c_{2}|\psi _{2}\rangle )\\&=E|\psi \rangle \end{aligned}}}which shows that|ψ⟩{\displaystyle |\psi \rangle } is an eigenstate ofH^{\displaystyle {\hat {H}}} with the same eigenvalueE.

Effect of degeneracy on the measurement of energy

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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 HamiltonianH^{\displaystyle {\hat {H}}} has a degenerate eigenvalueEn{\displaystyle E_{n}} of degreegn, the eigenstates associated with it form avector subspace ofdimensiongn. In such a case, several final states can be possibly associated with the same resultEn{\displaystyle E_{n}}, all of which are linear combinations of thegnorthonormal eigenvectors|En,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 valueEn{\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(En)=∑i=1gn|⟨En,i|ψ⟩|2{\displaystyle P(E_{n})=\sum _{i=1}^{g_{n}}|\langle E_{n,i}|\psi \rangle |^{2}}

Degeneracy in different dimensions

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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

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In several cases,analytic results can be obtained more easily in the study of one-dimensional systems. For a quantum particle with awave function|ψ⟩{\displaystyle |\psi \rangle } moving in a one-dimensional potentialV(x){\displaystyle V(x)}, thetime-independent Schrödinger equation can be written as−ℏ22md2ψdx2+Vψ=Eψ{\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 energyE{\displaystyle E} at most, so that the degree of degeneracy never exceeds two. It can be proven that in one dimension, there are no degeneratebound states fornormalizable wave functions. A sufficient condition on apiecewise continuous potentialV{\displaystyle V} and the energyE{\displaystyle E} is the existence of two real numbersM,x0{\displaystyle M,x_{0}} withM≠0{\displaystyle M\neq 0} such that∀x>x0{\displaystyle \forall x>x_{0}} we haveV(x)−E≥M2{\displaystyle V(x)-E\geq M^{2}}.[3] In particular,V{\displaystyle V} is bounded below in this criterion.

Proof of the above theorem.
Considering a one-dimensional quantum system in a potentialV(x){\displaystyle V(x)} with degenerate states|ψ1⟩{\displaystyle |\psi _{1}\rangle } and|ψ2⟩{\displaystyle |\psi _{2}\rangle } corresponding to the same energy eigenvalueE{\displaystyle E}, writing the time-independent Schrödinger equation for the system:

−ℏ22md2ψ1dx2+Vψ1=Eψ1−ℏ22md2ψ2dx2+Vψ2=Eψ2{\displaystyle {\begin{aligned}-{\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{aligned}}}Multiplying the first equation byψ2{\displaystyle \psi _{2}} and the second byψ1{\displaystyle \psi _{1}} and subtracting one from the other, we get:ψ1d2dx2ψ2−ψ2d2dx2ψ1=0{\displaystyle \psi _{1}{\frac {d^{2}}{dx^{2}}}\psi _{2}-\psi _{2}{\frac {d^{2}}{dx^{2}}}\psi _{1}=0}Integrating both sidesψ1dψ2dx−ψ2dψ1dx=constant{\displaystyle \psi _{1}{\frac {d\psi _{2}}{dx}}-\psi _{2}{\frac {d\psi _{1}}{dx}}={\mbox{constant}}}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:ψ1(x)=cψ2(x){\displaystyle \psi _{1}(x)=c\psi _{2}(x)}wherec{\displaystyle c} is, in general, a complex constant. For bound stateeigenfunctions (which tend to zero asx→∞{\displaystyle x\to \infty }), and assumingV{\displaystyle V} andE{\displaystyle E} satisfy the condition given above, it can be shown[3] that also the first derivative of the wave function approaches zero in the limitx→∞{\displaystyle x\to \infty }, so that the above constant is zero and we have no degeneracy.

Degeneracy in two-dimensional quantum systems

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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 includeMOSFET, two-dimensionalsuperlattices ofHelium,Neon,Argon,Xenon etc. and surface ofliquid Helium. The presence of degenerate energy levels is studied in the cases ofParticle in a box and two-dimensionalharmonic oscillator, which act as usefulmathematical models for several real world systems.

Particle in a rectangular plane

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Consider afree particle in a plane of dimensionsLx{\displaystyle L_{x}} andLy{\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−ℏ22m(∂2ψ∂x2+∂2ψ∂y2)=Eψ{\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 areEnx,ny=π2ℏ22m(nx2Lx2+ny2Ly2){\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 isψnx,ny(x,y)=2LxLysin⁡(nxπxLx)sin⁡(nyπyLy){\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)}wherenx,ny=1,2,3,…{\displaystyle n_{x},n_{y}=1,2,3,\dots }

So,quantum numbersnx{\displaystyle n_{x}} andny{\displaystyle n_{y}} are required to describe the energy eigenvalues and the lowest energy of the system is given byE1,1=π2ℏ22m(1Lx2+1Ly2){\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 lengthsLx{\displaystyle L_{x}} andLy{\displaystyle L_{y}}, certain pairs of states are degenerate. IfLx/Ly=p/q{\displaystyle L_{x}/L_{y}=p/q}, where p and q are integers, the states(nx,ny){\displaystyle (n_{x},n_{y})} and(pny/q,qnx/p){\displaystyle (pn_{y}/q,qn_{x}/p)} have the same energy and so are degenerate to each other.

Particle in a square box

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In this case, the dimensions of the boxLx=Ly=L{\displaystyle L_{x}=L_{y}=L} and the energy eigenvalues are given byEnx,ny=π2ℏ22mL2(nx2+ny2){\displaystyle E_{n_{x},n_{y}}={\frac {\pi ^{2}\hbar ^{2}}{2mL^{2}}}(n_{x}^{2}+n_{y}^{2})}

Sincenx{\displaystyle n_{x}} andny{\displaystyle n_{y}} can be interchanged without changing the energy, each energy level has a degeneracy of at least two whennx{\displaystyle n_{x}} andny{\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 haveE=50π2ℏ22mL2{\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:

nx{\displaystyle n_{x}}ny{\displaystyle n_{y}}E(ℏ2π22mL2){\displaystyle E\left({\frac {\hbar ^{2}\pi ^{2}}{2mL^{2}}}\right)}Degeneracy
1121
2
1		1
2		5
5		2
2281
3
1		1
3		10
10		2
3
2		2
3		13
13		2
4
1		1
4		17
17		2
33181
............
7
5
1		1
5
7		50
50
50		3
............
8
7
4
1		1
4
7
8		65
65
65
65		4
............
9
7
6
2		2
6
7
9		85
85
85
85		4
............
11
10
5
2		2
5
10
11		125
125
125
125		4
............
14
10
2		2
10
14		200
200
200		3
............
17
13
7		7
13
17		338
338
338		3

Particle in a cubic box

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In this case, the dimensions of the boxLx=Ly=Lz=L{\displaystyle L_{x}=L_{y}=L_{z}=L} and the energy eigenvalues depend on three quantum numbers.Enx,ny,nz=π2ℏ22mL2(nx2+ny2+nz2){\displaystyle E_{n_{x},n_{y},n_{z}}={\frac {\pi ^{2}\hbar ^{2}}{2mL^{2}}}(n_{x}^{2}+n_{y}^{2}+n_{z}^{2})}

Sincenx{\displaystyle n_{x}},ny{\displaystyle n_{y}} andnz{\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

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If twooperatorsA^{\displaystyle {\hat {A}}} andB^{\displaystyle {\hat {B}}} commute, i.e.,[A^,B^]=0{\displaystyle [{\hat {A}},{\hat {B}}]=0}, then for every eigenvector|ψ⟩{\displaystyle |\psi \rangle } ofA^{\displaystyle {\hat {A}}},B^|ψ⟩{\displaystyle {\hat {B}}|\psi \rangle } is also an eigenvector ofA^{\displaystyle {\hat {A}}} with the same eigenvalue. However, if this eigenvalue, sayλ{\displaystyle \lambda }, is degenerate, it can be said thatB^|ψ⟩{\displaystyle {\hat {B}}|\psi \rangle } belongs to the eigenspaceEλ{\displaystyle E_{\lambda }} ofA^{\displaystyle {\hat {A}}}, which is said to be globally invariant under the action ofB^{\displaystyle {\hat {B}}}.

For two commuting observablesA andB, one can construct anorthonormal basis of the state space with eigenvectors common to the two operators. However,λ{\displaystyle \lambda } is a degenerate eigenvalue ofA^{\displaystyle {\hat {A}}}, then it is an eigensubspace ofA^{\displaystyle {\hat {A}}} that is invariant under the action ofB^{\displaystyle {\hat {B}}}, so therepresentation ofB^{\displaystyle {\hat {B}}} in the eigenbasis ofA^{\displaystyle {\hat {A}}} is not a diagonal but ablock diagonal matrix, i.e. the degenerate eigenvectors ofA^{\displaystyle {\hat {A}}} are not, in general, eigenvectors ofB^{\displaystyle {\hat {B}}}. However, it is always possible to choose, in every degenerate eigensubspace ofA^{\displaystyle {\hat {A}}}, a basis of eigenvectors common toA^{\displaystyle {\hat {A}}} andB^{\displaystyle {\hat {B}}}.

Choosing a complete set of commuting observables

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If a given observableA is non-degenerate, there exists a unique basis formed by its eigenvectors. On the other hand, if one or several eigenvalues ofA^{\displaystyle {\hat {A}}} are degenerate, specifying an eigenvalue is not sufficient to characterize a basis vector. If, by choosing an observableB^{\displaystyle {\hat {B}}}, which commutes withA^{\displaystyle {\hat {A}}}, it is possible to construct an orthonormal basis of eigenvectors common toA^{\displaystyle {\hat {A}}} andB^{\displaystyle {\hat {B}}}, which is unique, for each of the possible pairs of eigenvalues {a,b}, thenA^{\displaystyle {\hat {A}}} andB^{\displaystyle {\hat {B}}} are said to form acomplete 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 observableC^{\displaystyle {\hat {C}}}, which commutes with bothA^{\displaystyle {\hat {A}}} andB^{\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

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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 operatorA^{\displaystyle {\hat {A}}} is one that satisfies,A~=PA^P{\displaystyle {\tilde {A}}=P{\hat {A}}P}[P,A^]=0{\displaystyle [P,{\hat {A}}]=0}while an odd operatorB^{\displaystyle {\hat {B}}} is one that satisfiesPB^+B^P=0{\displaystyle P{\hat {B}}+{\hat {B}}P=0}Since the square of the momentum operatorp^2{\displaystyle {\hat {p}}^{2}} is even, if the potential V(r) is even, the HamiltonianH^{\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 ofH^{\displaystyle {\hat {H}}} among even and odd states. However, if one of the energy eigenstates has no definiteparity, it can be asserted that the corresponding eigenvalue is degenerate, andP|ψ⟩{\displaystyle P|\psi \rangle } is an eigenvector ofH^{\displaystyle {\hat {H}}} with the same eigenvalue as|ψ⟩{\displaystyle |\psi \rangle }.

Degeneracy and symmetry

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The physical origin of degeneracy in a quantum-mechanical system is often the presence of somesymmetry 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 asymmetry operation associated with aunitary operatorS. Under such an operation, the new Hamiltonian is related to the original Hamiltonian by asimilarity transformation generated by the operatorS, such thatH′=SHS−1=SHS†{\displaystyle H'=SHS^{-1}=SHS^{\dagger }}, sinceS is unitary. If the Hamiltonian remains unchanged under the transformation operationS, we haveSHS†=HSHS−1=HSH=HS[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 }whereE is the corresponding energy eigenvalue.HS|α⟩=SH|α⟩=SE|α⟩=ES|α⟩{\displaystyle HS|\alpha \rangle =SH|\alpha \rangle =SE|\alpha \rangle =ES|\alpha \rangle }which means thatS|α⟩{\displaystyle S|\alpha \rangle } is also an energy eigenstate with the same eigenvalueE. If the two states|α⟩{\displaystyle |\alpha \rangle } andS|α⟩{\displaystyle S|\alpha \rangle } are linearly independent (i.e. physically distinct), they are therefore degenerate.

In cases whereS is characterized by a continuousparameterϵ{\displaystyle \epsilon }, all states of the formS(ϵ)|α⟩{\displaystyle S(\epsilon )|\alpha \rangle } have the same energy eigenvalue.

Symmetry group of the Hamiltonian

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The set of all operators which commute with the Hamiltonian of a quantum system are said to form thesymmetry group of the Hamiltonian. Thecommutators of thegenerators of this group determine thealgebra of the group. An n-dimensional representation of the Symmetry group preserves themultiplication table of the symmetry operators. The possible degeneracies of the Hamiltonian with a particular symmetry group are given by the dimensionalities of theirreducible 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

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Degeneracies in a quantum system can be systematic or accidental in nature.

Systematic or essential degeneracy

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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

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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.[4] 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

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For a particle in a central1/r potential, theLaplace–Runge–Lenz vector is a conserved quantity resulting from an accidental degeneracy, in addition to the conservation ofangular momentum due torotational invariance.

For a particle moving on a cone under the influence of1/r andr2 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 generateSU(2) symmetry for both potentials.

Example: Particle in a constant magnetic field

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A particle moving under the influence of a constant magnetic field, undergoingcyclotron motion on a circular orbit is another important example of an accidental symmetry. The symmetrymultiplets in this case are theLandau levels which are infinitely degenerate.

Examples

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The hydrogen atom

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Main article:Hydrogen Atom

Inatomic physics, the bound states of an electron in ahydrogen atom show us useful examples of degeneracy. In this case, the Hamiltonian commutes with the totalorbital angular momentumL^2{\displaystyle {\hat {L}}^{2}}, its component along the z-direction,L^z{\displaystyle {\hat {L}}_{z}}, totalspin angular momentumS^2{\displaystyle {\hat {S}}^{2}} and its z-componentS^z{\displaystyle {\hat {S}}_{z}}. The quantum numbers corresponding to these operators areℓ{\displaystyle \ell },mℓ{\displaystyle m_{\ell }},s{\displaystyle s} (always 1/2 for an electron) andms{\displaystyle m_{s}} respectively.

The energy levels in the hydrogen atom depend only on theprincipal quantum numbern. For a givenn, all the states corresponding toℓ=0,…,n−1{\displaystyle \ell =0,\ldots ,n-1} have the same energy and are degenerate. Similarly for given values ofn andℓ, the(2ℓ+1){\displaystyle (2\ell +1)}, states withmℓ=−ℓ,…,ℓ{\displaystyle m_{\ell }=-\ell ,\ldots ,\ell } are degenerate. The degree of degeneracy of the energy levelEn is therefore∑ℓ=⁡0n−1(2ℓ+1)=n2,{\displaystyle \sum _{\ell \mathop {=} 0}^{n-1}(2\ell +1)=n^{2},} which is doubled if the spin degeneracy is included.[1]: 267f 

The degeneracy with respect tomℓ{\displaystyle m_{\ell }} is an essential degeneracy which is present for anycentral potential, and arises from the absence of a preferred spatial direction. The degeneracy with respect toℓ{\displaystyle \ell } is often described as an accidental degeneracy, but it can be explained in terms of special symmetries of the Schrödinger equation which are only valid for the hydrogen atom in which the potential energy is given byCoulomb's law.[1]: 267f 

Isotropic three-dimensional harmonic oscillator

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It is a spinlessparticle of mass m moving inthree-dimensional space, subject to acentral force whose absolute value is proportional to the distance of the particle from the centre of force.F=−kr{\displaystyle F=-kr}It is said to be isotropic since the potentialV(r){\displaystyle V(r)} acting on it is rotationally invariant, i.e.,V(r)=12mω2r2{\displaystyle V(r)={\tfrac {1}{2}}m\omega ^{2}r^{2}} whereω{\displaystyle \omega } is theangular frequency given byk/m{\textstyle {\sqrt {k/m}}}.

Since the state space of such a particle is thetensor product of the state spaces associated with the individual one-dimensional wave functions, the time-independent Schrödinger equation for such a system is given by-−ℏ22m(∂2ψ∂x2+∂2ψ∂y2+∂2ψ∂z2)+12mω2(x2+y2+z2)ψ=Eψ{\displaystyle -{\frac {\hbar ^{2}}{2m}}\left({\frac {\partial ^{2}\psi }{\partial x^{2}}}+{\frac {\partial ^{2}\psi }{\partial y^{2}}}+{\frac {\partial ^{2}\psi }{\partial z^{2}}}\right)+{\frac {1}{2}}{m\omega ^{2}\left(x^{2}+y^{2}+z^{2}\right)\psi }=E\psi }

So, the energy eigenvalues areEnx,ny,nz=(nx+ny+nz+32)ℏω{\displaystyle E_{n_{x},n_{y},n_{z}}=\left(n_{x}+n_{y}+n_{z}+{\tfrac {3}{2}}\right)\hbar \omega } or,En=(n+32)ℏω{\displaystyle E_{n}=\left(n+{\tfrac {3}{2}}\right)\hbar \omega } wheren is a non-negative integer.So, the energy levels are degenerate and the degree of degeneracy is equal to the number of different sets{nx,ny,nz}{\displaystyle \{n_{x},n_{y},n_{z}\}} satisfyingnx+ny+nz=n{\displaystyle n_{x}+n_{y}+n_{z}=n}The degeneracy of then{\displaystyle n}-th state can be found by considering the distribution ofn{\displaystyle n} quanta acrossnx{\displaystyle n_{x}},ny{\displaystyle n_{y}} andnz{\displaystyle n_{z}}. Having 0 innx{\displaystyle n_{x}} givesn+1{\displaystyle n+1} possibilities for distribution acrossny{\displaystyle n_{y}} andnz{\displaystyle n_{z}}. Having 1 quanta innx{\displaystyle n_{x}} givesn{\displaystyle n} possibilities acrossny{\displaystyle n_{y}} andnz{\displaystyle n_{z}} and so on. This leads to the general result ofn−nx+1{\displaystyle n-n_{x}+1} and summing over alln{\displaystyle n} leads to the degeneracy of then{\displaystyle n}-th state,∑nx=0n(n−nx+1)=(n+1)(n+2)2{\displaystyle \sum _{n_{x}=0}^{n}(n-n_{x}+1)={\frac {(n+1)(n+2)}{2}}}For theground staten=0{\displaystyle n=0}, the degeneracy is1{\displaystyle 1} so the state is non-degenerate. For all higher states, the degeneracy is greater than 1 so the state is degenerate.

Removing degeneracy

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The degeneracy in a quantum mechanical system may be removed if the underlying symmetry is broken by an externalperturbation. This causes splitting in the degenerate energy levels. This is essentially a splitting of the original irreducible representations into lower-dimensional such representations of the perturbed system.

Mathematically, the splitting due to the application of a small perturbation potential can be calculated using time-independent degenerateperturbation theory. This is an approximation scheme that can be applied to find the solution to the eigenvalue equation for the Hamiltonian H of a quantum system with an applied perturbation, given the solution for the Hamiltonian H0 for the unperturbed system. It involves expanding the eigenvalues and eigenkets of the Hamiltonian H in a perturbation series. The degenerate eigenstates with a given energy eigenvalue form a vector subspace, but not every basis of eigenstates of this space is a good starting point for perturbation theory, because typically there would not be any eigenstates of the perturbed system near them. The correct basis to choose is one that diagonalizes the perturbation Hamiltonian within the degenerate subspace.

Lifting of degeneracy by first-order degenerate perturbation theory.
Consider an unperturbed HamiltonianH0^{\displaystyle {\hat {H_{0}}}} and perturbationV^{\displaystyle {\hat {V}}}, so that the perturbed Hamiltonian

H^=H0^+V^{\displaystyle {\hat {H}}={\hat {H_{0}}}+{\hat {V}}}The perturbed eigenstate, for no degeneracy, is given by-|ψ0⟩=|n0⟩+∑k≠0Vk0/(E0−Ek)|nk⟩{\displaystyle |\psi _{0}\rangle =|n_{0}\rangle +\sum _{k\neq 0}V_{k0}/(E_{0}-E_{k})|n_{k}\rangle }The perturbed energy eigenket as well as higher order energy shifts diverge whenE0=Ek{\displaystyle E_{0}=E_{k}}, i.e., in the presence of degeneracy in energy levels. AssumingH0^{\displaystyle {\hat {H_{0}}}} possesses N degenerate eigenstates|m⟩{\displaystyle |m\rangle } with the same energy eigenvalue E, and also in general some non-degenerate eigenstates. A perturbed eigenstate|ψj⟩{\displaystyle |\psi _{j}\rangle } can be written as a linear expansion in the unperturbed degenerate eigenstates as-|ψj⟩=∑i|mi⟩⟨mi|ψj⟩=∑icji|mi⟩{\displaystyle |\psi _{j}\rangle =\sum _{i}|m_{i}\rangle \langle m_{i}|\psi _{j}\rangle =\sum _{i}c_{ji}|m_{i}\rangle }[H0^+V^]ψj⟩=[H0^+V^]∑icji|mi⟩=Ej∑icji|mi⟩{\displaystyle [{\hat {H_{0}}}+{\hat {V}}]\psi _{j}\rangle =[{\hat {H_{0}}}+{\hat {V}}]\sum _{i}c_{ji}|m_{i}\rangle =E_{j}\sum _{i}c_{ji}|m_{i}\rangle }whereEj{\displaystyle E_{j}} refer to the perturbed energy eigenvalues. SinceE{\displaystyle E} is a degenerate eigenvalue ofH0^{\displaystyle {\hat {H_{0}}}},∑icjiV^|mi⟩=(Ej−E)∑icji|mi⟩=ΔEj∑icji|mi⟩{\displaystyle \sum _{i}c_{ji}{\hat {V}}|m_{i}\rangle =(E_{j}-E)\sum _{i}c_{ji}|m_{i}\rangle =\Delta E_{j}\sum _{i}c_{ji}|m_{i}\rangle }Premultiplying by another unperturbed degenerate eigenket⟨mk|{\displaystyle \langle m_{k}|} gives-∑icji[⟨mk|V^|mi⟩−δik(Ej−E)]=0{\displaystyle \sum _{i}c_{ji}[\langle m_{k}|{\hat {V}}|m_{i}\rangle -\delta _{ik}(E_{j}-E)]=0}This is an eigenvalue problem, and writingVik=⟨mi|V^|mk⟩{\displaystyle V_{ik}=\langle m_{i}|{\hat {V}}|m_{k}\rangle }, we have-|V11−ΔEjV12…V1NV21V22−ΔEj…V2N⋮⋮⋱⋮VN1VN2…VNN−ΔEj|.{\displaystyle {\begin{vmatrix}V_{11}-\Delta E_{j}&V_{12}&\dots &V_{1N}\\V_{21}&V_{22}-\Delta E_{j}&\dots &V_{2N}\\\vdots &\vdots &\ddots &\vdots \\V_{N1}&V_{N2}&\dots &V_{NN}-\Delta E_{j}\end{vmatrix}}.}

TheN eigenvalues obtained by solving this equation give the shifts in the degenerate energy level due to the applied perturbation, while the eigenvectors give the perturbed states in the unperturbed degenerate basis|m⟩{\displaystyle |m\rangle }. To choose the good eigenstates from the beginning, it is useful to find an operatorV^{\displaystyle {\hat {V}}} which commutes with the original HamiltonianH0^{\displaystyle {\hat {H_{0}}}} and has simultaneous eigenstates with it.

Physical examples of removal of degeneracy by a perturbation

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Some important examples of physical situations where degenerate energy levels of a quantum system are split by the application of an external perturbation are given below.

Symmetry breaking in two-level systems

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See also:Avoided crossing § In two state systems

Atwo-level system essentially refers to a physical system having two states whose energies are close together and very different from those of the other states of the system. All calculations for such a system are performed on a two-dimensionalsubspace of the state space.

If the ground state of a physical system is two-fold degenerate, any coupling between the two corresponding states lowers the energy of the ground state of the system, and makes it more stable.

IfE1{\displaystyle E_{1}} andE2{\displaystyle E_{2}} are the energy levels of the system, such thatE1=E2=E{\displaystyle E_{1}=E_{2}=E}, and the perturbationW{\displaystyle W} is represented in the two-dimensional subspace as the following 2×2 matrixW=[0W12W12∗0].{\displaystyle \mathbf {W} ={\begin{bmatrix}0&W_{12}\\[1ex]W_{12}^{*}&0\end{bmatrix}}.}then the perturbed energies areE+=E+|W12|E−=E−|W12|{\displaystyle {\begin{aligned}E_{+}&=E+|W_{12}|\\E_{-}&=E-|W_{12}|\end{aligned}}}

Examples of two-state systems in which the degeneracy in energy states is broken by the presence of off-diagonal terms in the Hamiltonian resulting from an internal interaction due to an inherent property of the system include:

  • Benzene, with two possible dispositions of the three double bonds between neighbouringCarbon atoms.
  • Ammonia molecule, where the Nitrogen atom can be either above or below the plane defined by the threeHydrogen atoms.
  • H+
    2     molecule, in which the electron may be localized around either of the two nuclei.

Fine-structure splitting

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Main article:Fine structure

The corrections to the Coulomb interaction between the electron and the proton in a Hydrogen atom due to relativistic motion andspin–orbit coupling result in breaking the degeneracy in energy levels for different values ofl corresponding to a single principal quantum numbern.

The perturbation Hamiltonian due to relativistic correction is given byHr=−p4/8m3c2{\displaystyle H_{r}=-p^{4}/8m^{3}c^{2}}wherep{\displaystyle p} is the momentum operator andm{\displaystyle m} is the mass of the electron. The first-order relativistic energy correction in the|nlm⟩{\displaystyle |nlm\rangle } basis is given byEr=(−1/8m3c2)⟨nℓm|p4|nℓm⟩{\displaystyle E_{r}=\left(-1/8m^{3}c^{2}\right)\left\langle n\ell m\right|p^{4}\left|n\ell m\right\rangle }

Nowp4=4m2(H0+e2/r)2{\displaystyle p^{4}=4m^{2}(H^{0}+e^{2}/r)^{2}}Er=−12mc2[En2+2Ene2⟨1r⟩+e4⟨1r2⟩]=−12mc2α4[−3/(4n4)+1/n3(ℓ+1/2)]{\displaystyle {\begin{aligned}E_{r}&=-{\frac {1}{2mc^{2}}}\left[E_{n}^{2}+2E_{n}e^{2}\left\langle {\frac {1}{r}}\right\rangle +e^{4}\left\langle {\frac {1}{r^{2}}}\right\rangle \right]\\&=-{\frac {1}{2}}mc^{2}\alpha ^{4}\left[-3/(4n^{4})+1/{n^{3}(\ell +1/2)}\right]\end{aligned}}}whereα{\displaystyle \alpha } is thefine structure constant.

The spin–orbit interaction refers to the interaction between the intrinsicmagnetic moment of the electron with the magnetic field experienced by it due to the relative motion with the proton. The interaction Hamiltonian isHso=−emcm⋅Lr3=e2m2c2r3S⋅L{\displaystyle H_{so}=-{\frac {e}{mc}}{\frac {\mathbf {m} \cdot \mathbf {L} }{r^{3}}}={\frac {e^{2}}{m^{2}c^{2}r^{3}}}\mathbf {S} \cdot \mathbf {L} }which may be written asHso=e24m2c2r3[J2−L2−S2]{\displaystyle H_{so}={\frac {e^{2}}{4m^{2}c^{2}r^{3}}}\left[J^{2}-L^{2}-S^{2}\right]}

The first order energy correction in the|j,m,ℓ,1/2⟩{\displaystyle |j,m,\ell ,1/2\rangle } basis where the perturbation Hamiltonian is diagonal, is given byEso=ℏ2e24m2c2j(j+1)−ℓ(ℓ+1)−34a03n3ℓ(ℓ+12)(ℓ+1){\displaystyle E_{so}={\frac {\hbar ^{2}e^{2}}{4m^{2}c^{2}}}{\frac {j(j+1)-\ell (\ell +1)-{\frac {3}{4}}}{a_{0}^{3}n^{3}\ell (\ell +{\frac {1}{2}})(\ell +1)}}}wherea0{\displaystyle a_{0}} is theBohr radius.The total fine-structure energy shift is given byEfs=−mc2α42n3[1/(j+1/2)−3/4n]{\displaystyle E_{fs}=-{\frac {mc^{2}\alpha ^{4}}{2n^{3}}}\left[1/(j+1/2)-3/4n\right]}forj=ℓ±12{\textstyle j=\ell \pm {\tfrac {1}{2}}}.

Zeeman effect

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Main article:Zeeman effect

The splitting of the energy levels of an atom when placed in an external magnetic field because of the interaction of themagnetic momentm→{\displaystyle {\vec {m}}} of the atom with the applied field is known as theZeeman effect.

Taking into consideration the orbital and spin angular momenta,L{\displaystyle \mathbf {L} } andS{\displaystyle \mathbf {S} }, respectively, of a single electron in the Hydrogen atom, the perturbation Hamiltonian is given byV^=−(mℓ+ms)⋅B{\displaystyle {\hat {V}}=-(\mathbf {m} _{\ell }+\mathbf {m} _{s})\cdot \mathbf {B} }wheremℓ=−eL/2m{\displaystyle \mathbf {m} _{\ell }=-e\mathbf {L} /2m} andms=−eS/m{\displaystyle \mathbf {m} _{s}=-e\mathbf {S} /m}.Thus,V^=e2m(L+2S)⋅B{\displaystyle {\hat {V}}={\frac {e}{2m}}(\mathbf {L} +2\mathbf {S} )\cdot \mathbf {B} }Now, in case of the weak-field Zeeman effect, when the applied field is weak compared to the internal field, thespin–orbit coupling dominates andL{\textstyle \mathbf {L} } andS{\textstyle \mathbf {S} } are not separately conserved. Thegood quantum numbers aren,ℓ,j andmj, and in this basis, the first order energy correction can be shown to be given byEz=−μBgjBmj,{\displaystyle E_{z}=-\mu _{B}g_{j}Bm_{j},} whereμB=eℏ/2m{\displaystyle \mu _{B}={e\hbar }/2m} is called theBohr Magneton. Thus, depending on the value ofmj{\displaystyle m_{j}}, each degenerate energy level splits into several levels.

Lifting of degeneracy by an external magnetic field

In case of the strong-field Zeeman effect, when the applied field is strong enough, so that the orbital and spin angular momenta decouple, the good quantum numbers are nown,l,ml, andms. Here,Lz andSz are conserved, so the perturbation Hamiltonian is given by-V^=eB(Lz+2Sz)/2m{\displaystyle {\hat {V}}=eB(L_{z}+2S_{z})/2m}assuming the magnetic field to be along thez-direction. So,V^=eB(mℓ+2ms)/2m{\displaystyle {\hat {V}}=eB(m_{\ell }+2m_{s})/2m}For each value ofmℓ, there are two possible values ofms,±1/2{\displaystyle \pm 1/2}.

Stark effect

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Main article:Stark effect

The splitting of the energy levels of an atom or molecule when subjected to an external electric field is known as theStark effect.

For the hydrogen atom, the perturbation Hamiltonian isH^s=−|e|Ez{\displaystyle {\hat {H}}_{s}=-|e|Ez}if the electric field is chosen along thez-direction.

The energy corrections due to the applied field are given by the expectation value ofH^s{\displaystyle {\hat {H}}_{s}} in the|nℓm⟩{\displaystyle |n\ell m\rangle } basis. It can be shown by the selection rules that⟨nℓmℓ|z|n1ℓ1mℓ1⟩≠0{\displaystyle \langle n\ell m_{\ell }|z|n_{1}\ell _{1}m_{\ell 1}\rangle \neq 0} whenℓ=ℓ1±1{\displaystyle \ell =\ell _{1}\pm 1} andmℓ=mℓ1{\displaystyle m_{\ell }=m_{\ell 1}}.

The degeneracy is lifted only for certain states obeying the selection rules, in the first order. The first-order splitting in the energy levels for the degenerate states|2,0,0⟩{\displaystyle |2,0,0\rangle } and|2,1,0⟩{\displaystyle |2,1,0\rangle }, both corresponding ton = 2, is given byΔE2,1,mℓ=±|e|ℏ2/(mee2)E{\displaystyle \Delta E_{2,1,m_{\ell }}=\pm |e|\hbar ^{2}/(m_{e}e^{2})E}.

See also

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References

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  1. ^abcMerzbacher, Eugen (1998).Quantum Mechanics (3rd ed.). New York: John Wiley.ISBN 0-471-88702-1.
  2. ^Levine, Ira N. (1991).Quantum Chemistry (4th ed.). Prentice Hall. p. 52.ISBN 0-205-12770-3.
  3. ^abMessiah, Albert (1967).Quantum mechanics (3rd ed.). Amsterdam, NLD: North-Holland. pp. 98–106.ISBN 0-471-88702-1.
  4. ^McIntosh, Harold V. (1959)."On Accidental Degeneracy in Classical and Quantum Mechanics"(PDF).American Journal of Physics.27 (9). American Association of Physics Teachers (AAPT):620–625.Bibcode:1959AmJPh..27..620M.doi:10.1119/1.1934944.ISSN 0002-9505.

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