The Position Representation in Quantum Mechanics Sungwook Lee Department of Mathematics University of Southern Mississippi sunglee@usm.edu July 13, 2007 The x-coordinate of a particle is associated with an operator ˆx. The set of numbers that are eigenvalues of an operator is called the spectrum of the operator. The spectrum of ˆx consists of real numbers < x <. x is the state in which the particle is certainly at x and satisfies the eigenvalue equation ˆx x = x x. If the particle is in some state ψ, the amplitude to find the particle at x is the complex number x ψ. The functional dependence of this number on x is what is called the wave function ψ(x): ψ(x) = x ψ. Since the particle has to be somewhere, the sum of the probabilities ψ(x) 2 has to be 1. Since the values of x is continuous, 1 = dx ψ(x) 2 = dx ψ x x ψ. (1) In discrete case, if ψ n, n = 1, 2, denote eigenkets, then ψ = ψ n ψ n ψ. n=1 So, n=1 ψ n ψ n = I. Similarly, from (1) we obtain dx x x = I. (2) By (2), x ψ = dx x x x ψ. (3) 1
Recall that the Dirac delta function is defined by ψ(x ) = dxδ(x x )ψ(x). Compare this with (3), we get If φ and ψ are two states, then 1 by (2) φ ψ = dx φ x x ψ = dxφ (x)ψ(x). x x = δ(x x ). (4) In the position representation, the functional form ψ( ) is identified with the state ψ and the complex number ψ(x) with the amplitude x ψ. Now, let us calculate x ˆx ψ. First we assume that ˆx is hermitian, i.e., (ˆx) = ˆx as usual in quantum mechanics. x ˆx ψ = (ˆx) x ψ = ˆxx ψ = x x ψ = xψ(x). This calculation also can be done without assuming that ˆx is hermitian: x ˆx ψ = x ˆx I ψ = = = = = xψ(x). dx x ˆx x x ψ dx x ˆxx ψ(x ) dxx δ(x x )ψ(x ) dxδ(x x )x ψ(x ) 1 Here, we consider only 1-dimensional case, so φ (x) = φ(x). 2
Hence, the operator ˆx turns ψ that evaluates ψ(x) into the one that evaluates xψ(x). In the position representation, x-momentum p operates on wave functions by differentiation 2 : h x ˆp ψ = x ˆpψ = ˆpψ(x) = i ψ. = 1.055 10 34 J s is the Dirac constant or the reduced Planck constant, where h 6.626 10 34 J s is the Planck constant. Let u p (x) := x p. Then x ˆp p = x ˆpp = i u p = pu p (x), since ˆp p = p p. That is, we obtain the differential equation i u p = pu p. The differential equation has a general solution u p (x) = Ae ipx/. The unknown constant A can be determined as follows: δ(p p ) = p p = dx p x x p = dxu p (x)u p(x) = A 2 dxe i(p p )x/ = A 2 δ(p p ). 1 Here, we used the familiar integral dte ixt = δ(x) in Fourier transforms. Recall that this integral came from the sequence of functions δ n (x) = sin nx πx = 1 n n dte ixt, n = 1, 2, 2 For now, we will take it as a definition, but it will be proved later. 3
which approximates the Dirac delta function δ(x), i.e., δ(x) lim n δ n(x) = 1 Since A > 0, A = 1. Hence, dte ixt. Figure 1: The sequence δ n (x) u p (x) = 1 e ipx/. This is the wave function of a state of momentum. In physics, the hamiltonian H plays an important role in the study of dynamics. It is given by H = T + V, where T and V are the kinetic energy and the potential energy, respectively. The hamiltonian H represents the energy of a physical system. The kinetic energy is given by T = p2 2m where p is the momentum. So, the hamiltonian can be written as H(p, x) = p2 2m + V (x). 4
A typical classical example is the hamiltonian for a harmonic oscillator: H(p, x) = p2 2m + 1 2 mω2 x 2. In the position representation, the hamiltonian for a particle acts by a combination of differentiation and multiplications: x H ψ = x Hψ = Hψ(x) ( = 2 ) 2m 2 + V (x) ψ(x). For a single particle moving along a real line R, there are two important observables: position and momentum. As we discussed earlier, in the quantum mechanical description of such a particle, the position operator ˆx and momentum operator ˆp are respectively given by x ˆx ψ = xψ(x), (5) x ˆp ψ = i ψ(x). (6) Now, we prove (6). This is what Larry Mead taught me 3. Let H be a Hilbert space of states. Suppose that O is a hermitian operator 4 on H such that the unitary operator g(a) = e ioa/, < a < satisfies the property: g(a)ψ(x) = ψ(x + a). (7) That is, g(a) gives rise to a translation of ψ(x) along the real line. unitary operator g(a) = e ioa/ acts on ψ(x) as 5 e ioa/ ψ(x) = (1 + = ψ(x) + ia ) ia a2 O 2! 2 O2 + ψ(x) Oψ(x) a2 2! 2 O2 ψ(x) +. The 3 If there is any misleading in the following argument, it is solely due to my misunderstanding on the material. 4 Let G be the group of unitary operators on a Hilbert space H and O a hermitian operator on H. Then {e ioa/ : a R} is a one-parameter (sub)group of G. 5 Here O 2 means the function composition O O. 5
Let x denote the position operator and δa denote infinitesimal change of a, so that (δa) n can be neglected for n 2. For any wave function ψ(x), Thus, On the other hand, e ioδa/ xe ioδa/ ψ(x) = e ioδa/ xψ(x δa) e ioδa/ xe ioδa/ = = (x + δa)ψ(x). e ioδa/ xe ioδa/ = x + δa. (8) ( 1 + iδa ) ( O x 1 iδa ) O = x iδa [x, O], where [x, O] = xo Ox is the canonical commutator. From (8) and (9), we obtain x + δa = x iδa [x, O]. This implies [x, O] = i. (10) Let a R. From (7) one obtains (9) ψ(x) 2 = ψ(x + a) 2. (11) So, one may consider the complex-valued function ψ(x) itself as a periodic function, and a simple candidate for ψ(x) satisfying (11) can be ψ(x) = e ikx. Note that k is the wave number k = λ, where λ is the wave length. Using the de Broglie s formula p = h λ, one obtains p = k or k = p. Hence, ψ(x) = e ipx/. (12) The wave function in (12) may be regarded as x p, i.e., ψ(x) = x p and ψ (x) = p x. Since ˆp p = p p, On the other hand, x ˆp p = p x p = pψ(x). (13) x ˆp p = xˆp I p = dx x ˆp x x p = dx ˆp x ψ(x ). (14) 6
From (13) and (14), one obtains dx x ˆp x e ipx / = pe ipx/. By Fourier transform 6, x ˆp x = This implies that = i dp / e ipx pe ipx/ dp eip(x x )/. = i δ(x x ). ψ ˆp ψ = ψ I ˆp I ψ = dx dx ψ x x ˆp x x ψ [ = dx dx ψ (x) i ] δ(x x ) ψ(x ) ( = dxψ (x) i ) dx δ(x x )ψ(x ) ( = dxψ (x) i ) ψ(x). Since ψ ˆp ψ = dxψ (x)ˆpψ(x), we see that ˆp = i. (15) Let us write the momentum operator ˆp simply as p. Then for any wave then 6 Recall that if φ(k) = 1 ψ(x) = 1 dxe ikx ψ(x), dke ikx φ(k) and vice versa. 7
function ψ(x), Hence, [x, p]ψ(x) = xpψ(x) pxψ(x) ( = x i ψ(x) ) + i ψ(x) = i ψ(x). [x, p] = i. (16) That is, the momentum operator p satisfies the commutation relation (10). Clearly, e ipa/ ψ(x) = e ipa/ e ipx/ = e ip(x+a)/ = ψ(x + a). Therefore, the momentum operator p = i is the desired operator O. 8