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CONTINUOUS SYMMETRY OPERATORS   85

where we used Equation (3.12) for the first equality, Equation (3.9) for thesecond equality, Equation (3.8) for the fourth equality, and Equation (3.9)for the last equality. This relation must hold for any ket

x+ ax

 so we have

the operator equationT xXT 

−1x  = X − ax1 (3.14)

We define exp(A), the exponentiation of the operator  A, by its Taylor seriesexpansion

eA = 1 +A+   12A2+

  16A

3+ · · ·   (3.15)

It follows from this definition that the inverse operator of exp(A) is exp(−A).Without loss of generality we can write the operator  T x   in the form of an

exponential T x  = e−iaxPx (3.16)

where  Px  is an operator yet to be determined. The usefulness of writing  T xin the form Equation (3.16) will become clear soon. For the moment wenote that Equation (3.16) gives  T x  = 1 for   ax  = 0, as expected because if ax  = 0 the kets

x+ ax and

x should be equal. The imaginary unit i in the

exponent in Equation (3.16) together with the fact that  T x   is unitary, seeEquation (3.11), means that the operator  Px   is Hermitian (show this) andthat its eigenvalues, if any, will be real and observable. This remedies thefact that the eigenvalues of  T x  are not observable.

To identify the operator   Px   in Equation (3.16) we consider the case of an infinitely small   ax. We expand Equation (3.16) in powers of   ax   andsubstitute the result in Equation (3.14). Keeping terms to order   a1 we get(1 − iaxPx)X (1 + iaxP

†) = X − ax1 or   iPxX − iXPx  = 1 o r [Px,X ] = 1/i.We conclude that the unknown operator  Px  introduced in Equation (3.16)is the x-component of the momentum operator P. Had we not introduced theminus sign in Equation (3.16) we would have concluded that the unknownoperator  Px  was the negative of the momentum operator and we would be

led to introduce the minus sign.If   T x   is a symmetry operator it must commute with the Hamiltonian,

see Equation (3.5). It is easy to show using Equation (3.15) that if   T xcommutes with   H   that   Px   also commutes with   H . Therefore   Px   and   H share the same eigenkets (or in the case of degeneracy, shared eigenketscan be constructed). Thus the eigenvalues of   Px  and the energy  E  of thephysical system are simultaneously observable and the eigenvalues of Px areconserved. Thus we have derived conservation of the   x  component of themomentum  P from the symmetry of the physical system under translations

in the  x direction.In general we may write a continuous unitary operator  U  as

U = e−iεG (3.17)