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Wednesday, February 3, 2016

A Curious Algebraic Structure Underlying Pairs of Operators

The Quantum Mechanical Harmonic Oscillator

In my research, I've been working on a generalization of the quantum mechanical harmonic oscillator (QMHO). Much of the algebraic structure of the QMHO comes from the canonical commutation relation:

[x,p]=i.

The time-independent QMHO is governed by the eigenvalue equation (in dimensionless units)

Hψ=12d2dx2ψ+12x2ψ=Eψ.

The difference-of-squares-like expression suggests a factorization for the QMHO a la

12(ddx+x)(ddx+x).

However since these are now operators, it is not guaranteed that the product gives exactly (2). Indeed, expanding the product, we instead get

12d2dx2+12x212.

This is not exactly H, but is H12, that is to say that

H=12(ddx+x)(ddx+x)+12.

Defining a=ddx+x, we see that the adjoint of a, denoted a, which satisfies af,g=f,ag, is given by ddx+x (which can be verified by integration by parts). Noting that [x,ddx]=1 (which is directly related to (1)), we have the following commutation relation:

[a,a]=[ddx+x,ddx+x]=[ddx,ddx]+[x,ddx]+[ddx,x]+[x,x]=0+1+1+0=2.

Therefore aa=aa2. Moreover we also have

[H,a]=[12aa+12,a]=12(aaaaaa)=12a[a,a]=a.

Thus Ha=aH+a. What this in turn gives is that if ψ is an eigenfunction of H, i.e. it solves (2), with eigenvalue E, then aψ also solves (2), but it has an eigenvalue of E+1:

Haψ=(aH+a)ψ=a(Hψ)+aψ=a(Eψ)+aψ=(E+1)aψ.

Analogously, aψ has an eigenvalue of E1 (assuming aψ0). Since a increases the eigenvalue, it is called a raising operator; similarly since a decreases the eigenvalue, it is called a lowering operator. Collectively, they are called ladder operators. What we conclude is that the algebraic structure of the QMHO encodes deep information about the analytic nature of the QMHO.

The ladder structure is inherently an algebraic structure about the operators. The structure concerning the QMHO is at its most fundamental level the Heisenberg-Weyl Lie group/Lie algebra. Other Lie algebras have raising and lowering operators, such as su(2).


A More General Setting

Without divulging the nature of my current project (which is as of posting unpublished), I stumbled upon the following result concerning pairs of operators a and b (and their adjoints):

Theorem If a and b are operators satisfying aa=bb+γ and aa=bb+δ for γ,δR, where γ<δ, then the operator ab (resp. ba) is a raising (resp. lowering) operator for aa (and bb) and the operator ba (resp. ab) is a raising (resp. lowering) operator for aa (and bb).

Proof. Without loss of generality, we assume that γ<δ since otherwise we can just switch the roles of a and b. Following the argument for the QMHO, we need only to show that [aa,ab] is a multiple of ab and likewise for ba with aa. To do that, we show a couple of simpler identities first.

aaa=abb+γaaaa=bba+δa

Likewise we have

baa=bbb+γbaab=bbb+δb

Subtracting these equations from each other gives the following "intertwining" relations:

bbaabb=(γδ)aaabbaa=(γδ)b

From this we get the identities

[bb,ba]=b(bbaabb)=(γδ)ba[bb,ab]=(bbaabb)b=(γδ)ab[aa,ab]=a(aababa)=(δγ)ab[aa,ba]=(aabbaa)a=(δγ)ba

Note the importance of the assumption that γδ. Otherwise the mixed operators ab, etc. would not resemble ladder operators in the truest sense. In the above commutation relations, aa can be replaced with bb and vice versa; likewise for the aa and bb.

The relations aa=bb+γ and aa=bb+δ seem rather strange at first, but in physical terms, it means that the "Hamiltonians" aa and bb are really the same, but with an energy level shift of γ. They capture the same fundamental dynamics, but they have a different zero point. Likewise for aa and bb. We have a further result:

Theorem The set {aaγ2,ab,ba} forms a Lie algebra isomorphic to su(1,1).

Proof. Based on the work above, we need only to consider [ab,ba].

[ab,ba]=abbabaab=a(aaδ)ab(bb+δ)b=(aa)2(bb)2δ(aa+bb)=(aa)2(aaγ)2δ(aa+aaγ)=2(γδ)aaγ(γδ)=2(δγ)(aaγ2)

The su(1,1) Lie algebra is generated by elements K3,K+,K satisfying the commutation relations

[K3,K±]=±K±,[K+,K]=2K3.

Letting aaγ2=(δγ)K3, ab=δγK+, and ba=δγK, our commutation relations exactly mimic those of su(1,1).

A similar analysis holds for aa and its ladder operators. The Lie algebra su(1,1) does not have any finite dimensional unitary irreducible representations, however we can always make a finite dimensional representation of su(2) into a finite dimensional representation of su(1,1) by complexifying the operators in the proper way. The resulting representation will not be unitary. Finite dimensional representations correspond to matrices, so it is natural to ask if any pair of matrices satisfies the conditions for the theorem. We give an answer in the next result.

Theorem No pair of matrices a, b can satisfy the conditions for the theorem.

Proof. Suppose such a pair of, say, n×n matrices existed. Taking the trace of aa, we have

tr(aa)=tr(bb+γI)=tr(bb)+tr(γI)=tr(bb)+γn.

However taking the trace of aa gives

tr(aa)=tr(bb+δI)=tr(bb)+tr(δI)=tr(bb)+δn.

Noting then that the trace is cyclic, we have tr(aa)=tr(aa) and likewise tr(bb)=tr(bb), which gives the following system of equations:

tr(aa)=tr(bb)+γntr(aa)=tr(bb)+δn

This system can only be solved if γ=δ; however by supposition, γδ and so we must conclude that no such matrices exist.

Thus every representation of the system must be infinite dimensional in nature. The above works in more generality. If one is working in an operator system with a tracial state, the above will show that again no such a or b exist.


Schwinger's "Spinification" of the QMHO

The above methodology is philosophically similar to that of Schwinger's "spinification" of the two-particle QMHO, but casts a much wider net than Schwinger's approach. Schwinger's approach to the two-particle QMHO is as follows:

Let a,a be the ladder operators for particle A and b,b be the ladder operators for particle B. Form the composite system by tensoring, i.e. a1 and a1 act only on the A element and similarly for 1b and 1b. Define then the following operators

J+=(a1)(1b),J=(1b)(a1),J3=(a1)(a1)(1b)(1b).

Then J+ and J act as ladder operators for J3. The use of the letter J suggests a connection to spin. Indeed, after a reparameterization, the above operators exactly satisfy the commutation relations for the spin operators, i.e. form the Lie algebra su(2).

A seeming oversight of Schwinger's approach is that ultimately it is hinged on the QMHO and the first order commutation relations it obeys. These commutation relations are so deeply ingrained in the theory of the QMHO that it is difficult to see how to generalize. While my approach above is similar in effect, the idea was developed in a much different way. After some of my work is published, I will expand upon the ideas presented here and how I came to the result above.

I am unaware if this result has yet appeared anywhere in the literature. Given the odd nature of the set-up, I would not be surprised that it has not. In effect, what it is assuming is that two related Hamiltonians (via supersymmetric arguments) have two distinct factorizations which are governed by the same pairs of operators. (The distinctness comes from the fact that γδ - otherwise one may take b=a.)

In the case of the QMHO, one may take b=a and the above analysis gives second order ladder operators (which amount to (a)2 and a2) as opposed to the usual ladder operators.

Bonus: It is natural to ask if the condition aa=bb+δ is necessary or not in the theorem (taking aa=bb+γ to be true) or if there is some weakening of this that will allow the theorem to remain true. As of posting this, I am unsure, however I am inclined to believe that it is indeed necessary.