I pedagogically show that the momentum operator in quantum mechanics, in
          the position representation, commonly known to be a derivative with
          respect to a spatial x-coordinate, can be derived by identifying momentum
          as the generator of space translations

Reece, Ryan

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A Derivation of the Quantum MechanicalMomentum Operator in the PositionRepresentationRyan D. ReeceSeptember 23, 2006AbstractI show that the momentum operator in quantum mechanics, inthe position representation, commonly known to be a derivative withrespect to a spacial x-coordinate, can be derived by identifying mo-mentum as the generator of space translations.1 Translation OperatorGiven an eigenstate of position |~x〉, with eigenvalue x, we define a Transla-tion Operator, T (~a), which transforms an eigenstate of position to anothereigenstate of position, with the eigenvalue increased by ~a.T (~a) |~x〉 ≡ |~x+ ~a〉 (1)By the following argument, we note that the adjoint of T (~a) moves a statebackward. It transforms an eigenstate of position to another eigenstate ofposition, with the eigenvalue decreased by ~a.〈~x′ |T (~a) | ~x〉 = 〈~x′ | ~x+ ~a〉 (2)= δ((~x+ ~a)− ~x′) (3)= δ(~x− (~x′ − ~a)) (4)= 〈~x′ − ~a | ~x〉 (5)⇒ 〈~x′|T (~a) = 〈~x′ − ~a| (6)1T †(~a) |~x′〉 = |~x′ − ~a〉 (7)Note that if we translate forwards by some amount, it is the same as trans-lating backwards by negative that amount.T (~a) = T †(−~a) (8)If we translate a state forwards and then backwards by the same amount,the state remains unchanged. This implies that the translation operator isunitary.T †(~a) T (~a) |~x〉 = |~x〉 (9)⇒ T †(~a) = T−1(~a) (10)Any unitary operator can be written asT (~a) = e−i~K·~a (11)1 = T †(~a) T (~a) (12)= ei~K†·~a e−i~K·~a (13)= ei(~K†− ~K)·~a (14)⇒ ~K = ~K† (15)Where evidently, ~K must be hermitian. In general, when writing a unitaryoperator this way, the operators ~K are known as the generators of what everunitary operator one is expressing, in this case: translation.2 Eigenstates of ~KLet us call the eigenstates of ~K, which are also eigenstates of T (~a), |~k〉.~K |~k〉 = ~k |~k〉 and T (~a) |~k〉 = e−i~k·~a |~k〉 (16)Let us consider the position projection of the translation operator actingon an eigenstate of translation. Letting the translation operator, operate tothe right, we have〈~x|T (~a)|~k〉 = e−i~k·~a 〈~x|~k〉 (17)= e−i~k·~a ψ~k(~x) (18)2where we have defined the wavefunction to beψ~k(~x) = 〈~x|~k〉 (19)Now consider the same projection, replacing T (~a) with T †(−~a), and let-ting it operate to the left.〈~x|T (~a)|~k〉 = 〈~x|T †(−~a)|~k〉 (20)= 〈~x− ~a|~k〉 (21)= ψ~k(~x− ~a) (22)Equating the two methods, we haveψ~k(~x− ~a) = e−i~k·~a ψ~k(~x) (23)Letting ~x = 0, and ~a = −~y, we recognize that this gives plane wave solutionsfor the wavefunction.ψ~k(~y) = ψ~k(0) ei~k·~y (24)As hypothesized by de Broglie, and first experimentally verified by elec-tron diffraction, a particle in an eigenstate of momentum has a wavefunctionwith with a wavevector, ~k, related to its momentum ~p by~p = ~ ~k (25)This means that the ~K operator that we have been discussing is indeed thewavevector operator. We can now write the translation operator asT (~a) = e−i~P ·~a/~ (26)Aside from the constant, ~, momentum is the generator of translation.3 Matrix Elements of ~P in the |~x〉 BasisFor simplicity, let us now consider translation in only one dimension.T (a) = e−iPa/~ (27)The following clever manipulation reveals how to write the momentum op-erator in terms of the translation operator.∂∂a∣∣∣∣a=0T (a) = − i~P (28)3P = i~∂∂a∣∣∣∣a=0T (a) (29)We should now ask what the matrix elements are of the momentum operatorin the position basis.〈x′ |P |x〉 = i~ ∂∂a∣∣∣∣a=0〈x′ |T (a) |x〉 (30)= i~∂∂a∣∣∣∣a=0δ(x+ a− x′) (31)= i~ δ′(x− x′) (32)4 ~P Acting on a WavefunctionWe should now take a digression to investigate what is meaning of this deriva-tive of a delta function, δ′(x). We integrate by parts, a δ′(x − y) acting onsome arbitrary function, f(x). Note that the boundary term is zero becauseδ(x− y) is zero on the boundary, provided a boundary of integration is notat position y.∫δ′(x− y) f(x) dx = 0−∫δ(x− y) f ′(x) dx (33)= −f ′(y) (34)Evidently, the derivative of a delta function is sort of a tool for evaluatingthe derivative of some function at a certain point.Now we may ask how we can represent the momentum operator in theposition basis. Because the number of states in the position basis are un-countably infinite, a matrix representation would be awkward. We see bythe following argument that there is a much more elegant way of writing themomentum operator.Consider the momentum operator acting on the wavefunction of some4state state |ψ〉.P ψ(x) = 〈x |P |ψ〉 (35)=∫〈x|P |x′〉 〈x′|ψ〉 dx′ (36)= i~∫δ′(x′ − x) ψ(x′) dx′ (37)= −i~ ∂ψ(x′)∂x′∣∣∣∣x′=x(38)= −i~ ∂ψ(x)∂x(39)∴ P → −i~ ∂∂x(40)5

Derivation of the Quantum Mechanical Momentum Operator in the Position Representation

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Derivation of the Quantum Mechanical Momentum Operator in the Position Representation

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