Operator methods in quantum mechanics

1 Chapter 3 Operator methods inquantum mechanicsWhile the wave mechanical formulation has proved successful in describingthe quantum mechanics of bound and unbound particles, some properties cannot be represented through a wave-like description. For example, the electronspin degree of freedom does not translate to the action of a gradient is therefore useful to reformulate quantum mechanics in a framework thatinvolves only discussing properties of operators, it is helpful to introduce a furthersimplification of notation. One advantage of the Operator algebra is that itdoes not rely upon a particular basis. For example, when one writes H= p22m,where the hat denotes an Operator , we can equally represent the momentumoperator in the spatial coordinate basis, when it is described by the differentialoperator, p= i!

2 X, or in the momentum basis, when it is just a number p=p. Similarly, it would be useful to work with a basis for the wavefunctionwhich is coordinate independent. Such a representation was developed byDirac early in the formulation of quantum the parlons of mathematics, square integrable functions (such as wave-functions) are said form a vector space, much like the familiar three-dimensionalvector spaces. In theDirac notation, a state vector or wavefunction, , isrepresented as a ket ,| . Just as we can express any three-dimensionalvector in terms of the basis vectors,r=x e1+y e2+z e3, so we can expandany wavefunction as a superposition of basis state vectors,| = 1| 1 + 2| 2 + .Alongside the ket, we can define the bra , |. Together, the bra and ketdefine thescalar product | dx (x) (x),from which follows the identity, | = |.

3 In this formulation, the realspace representation of the wavefunction is recovered from the inner prod-uct (x)= x| while the momentum space wavefunction is obtained from (p)= p| . As with a three-dimensional vector space wherea b |a||b|,the magnitude of the scalar product is limited by the magnitude of the vectors, | | | ,a relation known as theSchwartz quantum OperatorsAn Operator Ais a mathematical object that maps one state vector,| ,into another,| , A| =| . If A| =a| ,withareal, then| is said to be aneigenstate(oreigenfunction) of Awitheigenvaluea. For example, the plane wave state p(x)= x| p =Aeipx/!isan eigenstate of themomentum Operator , p= i! x, with a free particle, the plane wave is also an eigenstate of the Hamiltonian, H= p22mwith quantum mechanics , for any observableA, there is an Operator Awhichacts on the wavefunction so that, if a system is in a state described by| ,the expectation value ofAis A = | A| = dx (x) A (x).

4 ( )Every Operator corresponding to an observable is both linear and Hermitian:That is, for any two wavefunctions| and| , and any two complex numbers and ,linearityimplies that A( | + | )= ( A| )+ ( A| ).Moreover, for any linear Operator A, theHermitian conjugateoperator(also known as the adjoint) is defined by the relation | A = dx ( A )= dx ( A ) = A | .( )From the definition, A | = | A , we can prove some useful rela-tions: Taking the complex conjugate, A | = | A = A | , andthen finding the Hermitian conjugate of A , we have | A = ( A ) | = A | , ( A ) = , if we take the Hermitian conjugate twice, we get back to the sameoperator. Its easy to show that ( A) = A and ( A+ B) = A + B justfrom the properties of the dot product. We can also show that ( A B) = B A from the identity, | A B = A | B = B A |.

5 Note that operatorsareassociativebut not (in general)commutative, A B| = A( B| )=( A B)| &= B A| .A physical variable must have real expectation values (and eigenvalues).This implies that the operators representing physical variables have some spe-cial properties. By computing the complex conjugate of the expectation valueof a physical variable, we can easily show that physical operators are their ownHermitian conjugate, | H| =[ (x) H (x)dx] = (x)( H (x)) dx= H | . H | = | H = H | , and H = H. Operators that are theirown Hermitian conjugate are calledHermitian(or self-adjoint).Advanced quantum OPERATORS21' that the momentum Operator p= i! is Hermitian. Fur-ther show that the parity Operator , defined by P (x)= ( x) is also of Hermitian operators H|i =Ei|i form an orthonormal ( i|j = ij) complete basis: For a complete set of states|i , we can expanda state function| as| = i|i i|.

6 Equivalently, in a coordinate rep-resentation, we have (x)= x| = i x|i i| = i i| i(x), where i(x)= x|i .' operators and completeness:A ket state vector fol-lowed by a bra state vector is an example of an Operator . The Operator whichprojects a vector onto thejth eigenstate is given by|j j|. First the bra vector dotsinto the state, giving the coefficient of|j in the state, then its multiplied by the unitvector|j , turning it back into a vector, with the right length to be a projection. Anoperator maps one vector into another vector, so this is an Operator . If we sum overa complete set of states, like the eigenstates of a Hermitian Operator , we obtain the(useful)resolution of identity i|i i|= , in coordinate form, we can write i i(x) i(x )= (x x ). Indeed, we canform a projection Operator into a subspace, P= subspace|i i|.

7 As in a three-dimensional vector space, the expansion of the vectors| and| , as| = ibi|i and| = ici|i , allows the dot product to be takenby multiplying the components, | = ib ici.'Example:The basis states can be formed from any complete set of orthogonalstates. In particular, they can be formed from the basis states of the position orthe momentum Operator , dx|x x|= dp|p p|=I. If we apply thesedefinitions, we can then recover the familiar Fourier representation, (x) x| = dp x|p eipx/!/ 2 ! p| =1 2 ! dp eipx/! (p),where x|p denotes the plane wave state|p expressed in the real space Time-evolution operatorThe ability to develop an eigenfunction expansion provides the means to ex-plore the time evolution of a general wave packet,| under the action ofa Hamiltonian.

8 Formally, we can evolve a wavefunction forward in timeby applying the time-evolution Operator . For a Hamiltonian which is time-indepenent, we have| (t) = U| (0) , where U=e i Ht/!,denotes the time-evolution inserting the resolution of identity,I= i|i i|, where the states|i are eigenstates of the Hamiltonian witheigenvalueEi, we find that| (t) =e i Ht/! i|i i| (0) = i|i i| (0) e iEit/!.1 This equation follows from integrating the time-dependent Schr odinger equation, H| =i! t| .Advanced quantum OPERATORS22'Example:Consider the harmonic oscillator Hamiltonian H= p22m+12m in this chapter, we will see that the eigenstates,|n , have equally-spaced eigen-values,En=! (n+1/2), forn=0,1,2, . Let us then consider the time-evolutionof a general wavepacket,| (0) , under the action of the Hamiltonian.

9 From the equa-tion above, we find that| (t) = n|n n| (0) e iEnt/!. Since the eigenvalues areequally spaced, let us consider what happens whent=tr 2 r/ , this case, sincee2 inr= 1, we have| (tr) = n|n n| (0) e i tr/2=( 1)r| (0) .From this result, we can see that, up to an overall phase, the wave packet is perfectlyreconstructed at these times. This recurrence or echo is not generic, but is amanifestation of the equal separation of eigenvalues in the harmonic oscillator.' the symmetry of the harmonic oscillator wavefunctions underparity show that, at timestr= (2r+ 1) / , x| (tr) =e i tr/2 x| (0) . Explainthe origin of this time-evolution Operator is an example of aunitary Operator . Thelatter are defined as transformations which preserve the scalar product, | = U | U = | U U !

10 = | , U U= Uncertainty principle for non-commuting operatorsFor non-commuting Hermitian operators, [ A, B]&= 0, it is straightforward toestablish a bound on the uncertainty in their expectation values. Given a state| , the mean square uncertainty is defined as( A)2= |( A A )2 = | U2 ( B)2= |( B B )2 = | V2 ,where we have defined the operators U= A | A and V= B | B .Since A and B are just constants, [ U, V] = [ A, B]. Now let us take thescalar product of U| +i V| with itself to develop some information aboutthe uncertainties. As a modulus, the scalar product must be greater than orequal to zero, expanding, we have | U2 + 2 | V2 +i U | V i V | U 0. Reorganising this equation in terms of the uncertainties, wethus find( A)2+ 2( B)2+i |[ U, V]| we minimise this expression with respect to , we can determine whenthe inequality becomes strongest.