A Derivation of the Quantum Mechanical Momentum Operator in the Position Representation Ryan D. Reece September 23, 2006 Abstract I show that the momentum operator in quantum mechanics, in the position representation, commonly known to be a derivative with respect to a spacial x-coordinate, can be derived by identifying momentum as the generator of space translations. 1 Translation Operator Given an eigenstate of position |~x〉, with eigenvalue x, we define a Translation Operator, T (~a), which transforms an eigenstate of position to another eigenstate 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 state backward. It transforms an eigenstate of position to another eigenstate of position, 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) 1 T †(~a) |~x′〉 = |~x′ − ~a〉 (7) Note that if we translate forwards by some amount, it is the same as translating 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 is unitary. T †(~a) T (~a) |~x〉 = |~x〉 (9) ⇒ T †(~a) = T−1(~a) (10) Any unitary operator can be written as T (~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 unitary operator this way, the operators ~K are known as the generators of what ever unitary operator one is expressing, in this case: translation. 2 Eigenstates of ~K Let 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 acting on an eigenstate of translation. Letting the translation operator, operate to the right, we have 〈~x|T (~a)|~k〉 = e−i~k*~a 〈~x|~k〉 (17) = e−i ~k*~a ψ~k(~x) (18) 2 where we have defined the wavefunction to be ψ~k(~x) = 〈~x|~k〉 (19) Now consider the same projection, replacing T (~a) with T †(−~a), and letting 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 have ψ~k(~x− ~a) = e −i~k*~a ψ~k(~x) (23) Letting ~x = 0, and ~a = −~y, we recognize that this gives plane wave solutions for the wavefunction. ψ~k(~y) = ψ~k(0) e i~k*~y (24) As hypothesized by de Broglie, and first experimentally verified by electron diffraction, a particle in an eigenstate of momentum has a wavefunction with 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 the wavevector operator. We can now write the translation operator as T (~a) = e−i ~P *~a/~ (26) Aside from the constant, ~, momentum is the generator of translation. 3 Matrix Elements of ~P in the |~x〉 Basis For 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 operator in terms of the translation operator. ∂ ∂a ∣∣∣∣ a=0 T (a) = − i ~ P (28) 3 P = i~ ∂ ∂a ∣∣∣∣ a=0 T (a) (29) We should now ask what the matrix elements are of the momentum operator in the position basis. 〈x′ |P |x〉 = i~ ∂ ∂a ∣∣∣∣ a=0 〈x′ |T (a) |x〉 (30) = i~ ∂ ∂a ∣∣∣∣ a=0 δ(x+ a− x′) (31) = i~ δ′(x− x′) (32) 4 ~P Acting on a Wavefunction We should now take a digression to investigate what is meaning of this derivative of a delta function, δ′(x). We integrate by parts, a δ′(x − y) acting on some arbitrary function, f(x). Note that the boundary term is zero because δ(x− y) is zero on the boundary, provided a boundary of integration is not at 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 evaluating the derivative of some function at a certain point. Now we may ask how we can represent the momentum operator in the position basis. Because the number of states in the position basis are uncountably infinite, a matrix representation would be awkward. We see by the following argument that there is a much more elegant way of writing the momentum operator. Consider the momentum operator acting on the wavefunction of some 4 state 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)