A toy model of quantum electrodynamics in (1 + 1) dimensions

Transcription

1 IOP PUBLISHING Eur. J. Phys. 29 (2008) EUROPEAN JOURNAL OF PHYSICS doi: / /29/4/014 A toy odel of quantu electrodynaics in (1 + 1) diensions ADBoozer Departent of Physics, California Institute of Technology, Pasadena, CA 91125, USA E-ail: boozer@caltech.edu Received 31 March 2008, in final for 19 May 2008 Published 23 June 2008 Online at stacks.iop.org/ejp/29/815 Abstract We present a toy odel of quantu electrodynaics (QED) in (1 +1) diensions. The QED odel is uch sipler than QED in (3+1) diensions but exhibits any of the sae physical phenoena, and serves as a pedagogical introduction to both QED and quantu field theory in general. We show how the QED odel can be derived by quantizing a toy odel of classical electrodynaics, and we discuss the connections between the classical and quantu odels. In addition, we use the QED odel to discuss the radiation of atos and the Lab shift. 1. Introduction Quantu electrodynaics (QED) is of fundaental iportance in quantu field theory: it was the first quantu field theory to be constructed, and has provided a odel for the construction of subsequent quantu field theories. For this reason, quantu field theory is usually introduced to students by using QED as an exaple. Fro a pedagogical standpoint, however, QED is not ideal for this purpose, because it is a rather coplicated and technically deanding theory. QED is plagued by infinities of various sorts that require careful treatent; for exaple, a calculation of the Lab shift yields a result that is infinite when expressed in ters of the bare electron ass; to obtain a finite result, one needs to introduce ass renoralization. In addition, the gauge invariance of electrodynaics raises a nuber of subtle issues when the theory is quantized. Finally, calculations in QED are often lengthy and difficult, and for the beginning student the atheatical coplexity of the theory can obscure its physical eaning. To address these issues, we present a siple toy odel of QED in (1+1) diensions. The odel is obtained by quantizing a toy odel of classical electrodynaics that describes a scalar potential coupled to one or ore charged particles in (1 + 1) diensions. The classical toy odel is uch sipler than ordinary electrodynaics, but shares any of the sae physical features: there are fields that correspond to the electric and agnetic fields, these fields both ediate an interaction between the charged particles and support freely propagating radiation, and the coupling of a particle to its own field causes it to radiate and to undergo radiation /08/ $30.00 c 2008 IOP Publishing Ltd Printed in the UK 815

2 816 A D Boozer daping. A full treatent of the classical odel is given in [1], which provides a detailed discussion of these features 1. In the present paper we quantize this odel, and copare the classical and quantu versions of the theory. Quantu electrodynaics is usually taught to students in two different settings: first, it is used in undergraduate quantu echanics classes to treat the radiation of atos, and second, it is used in graduate level quantu field theory classes to describe relativistic collisions. The toy odel presented here is intended to bridge the gap between these two treatents, and could be used in either setting. The paper is organized as follows. In section 2, we describe the classical toy odel, and point out a nuber of features that we will later copare with corresponding features of the quantu toy odel. In section 3, we quantize the classical toy odel using the ethod of canonical quantization. In section 4, we consider a quantu field coupled to a classical source; this allows us to introduce soe features of the quantized odel in a siplified setting. In section 5, we investigate the full quantu theory, which describes a quantized particle coupled to a quantized field. We calculate the spontaneous decay rate and the Lab shift for a haronically bound particle, and discuss the connections between the QED toy odel and ordinary QED. 2. Classical theory The classical toy odel that we will be considering describes one or ore point particles that obey Newtonian dynaics and are coupled to a pair of fields E(t,x) and B(t,x), which correspond to the electric and agnetic fields of ordinary electrodynaics. For siplicity we will assue that there is only one particle; the generalization to ultiple particles is straightforward. The equation of otion for the particle is z = z V(z) 2gE(t, z), (1) where z(t) is the position of the particle at tie t, and g are its ass and charge, and V(z) is an arbitrary external potential. The equations of otion for the E and B fields are t E(t,x) = x B(t,x), (2) t B(t,x) = x E(t,x) 2ρ(t,x), (3) where ρ(t,x), the charge density, is given by ρ(t,x) = gδ(x z(t)). (4) Equation (1) can be thought of as the analogue to the Lorentz force law of ordinary electrodynaics, and equations (2) and (3) can be thought of as the analogues to Maxwell s equations. Together, these equations give a coplete description of the toy odel. In ordinary electrodynaics, it is often useful to express the electric and agnetic fields as derivatives of the vector potential A µ (t, r). We can do soething siilar in the toy odel: fro equation (2), it follows that we can define a scalar potential φ(t,x) such that E(t,x) = x φ(t,x)and B(t,x) = t φ(t,x). By substituting these relations into equation (3), we obtain a field equation for the scalar potential 2 : φ(t,x) = 2ρ(t,x). (5) This is just the inhoogeneous wave equation, where ρ(t,x) acts as the source. 1 In order to ake the present paper self-contained, a brief overview of the classical odel is given in section 2, which suarizes all the results needed to understand the quantu odel. 2 The following notation is used in this paper: = t 2 x 2 is the d Alebertian operator in (1+1) diensions, and ɛ(x) is the sign function, defined such that ɛ(x) = 1forx>0,ɛ(x) = 0forx = 0,ɛ(x) = 1forx<0. We have chosen a syste of units in which the speed at which waves are propagated by the field is equal to 1.

3 A toy odel of quantu electrodynaics in (1+1) diensions 817 As shown in [1], we can derive expressions for the total energy and oentu of the coupled particle-field syste. In the toy odel the E-field of the particle does not fall off with distance, so to obtain finite values for these quantities we will assue that the theory is defined on a finite region of space [ L/2,L/2], where L is an arbitrary length 3. The total energy of the syste is then given by U tot = U p + U f + U i, where U p = 1 2 ż2 + V(z) (6) is the energy of the particle, L/2 U f = 1 (E 2 (t, x) + B 2 (t, x)) dx (7) 2 L/2 is the energy of the E and B fields, and U i = 2 L/2 L/2 φ(t,x)ρ(t,x)dx = 2gφ(t, z) (8) is the energy due to the coupling of the particle to the field. The total oentu of the syste is given by P tot = P f + ż, where P f = L/2 L/2 E(t,x)B(t,x)dx (9) is the oentu of the field. Using the equations of otion (1), (2) and (3), it is straightforward to show that U tot = 0 and Ṗ tot = z V(z). Thus, the total energy is conserved, and in the absence of an external potential the total oentu is also conserved. In section 3 we will quantize the odel using the ethod of canonical quantization, which requires that we first express the classical theory in ters of a Hailtonian. We will take the generalized coordinates of the syste to be the particle position z and the scalar potential φ(t,x), and we will define p and π(t,x) to be the generalized oenta conjugate to these coordinates. The Hailtonian is given by H = H p + H f + H i, where H p = p2 + V(z) (10) 2 is the Hailtonian for the particle, L/2 H f = 1 (( x φ(t,x)) 2 + π 2 (t, x)) dx (11) 2 L/2 is the Hailtonian for the field, and H i = 2 L/2 L/2 φ(t,x)ρ(t,x)dx = 2gφ(t, z) (12) describes the coupling of the particle to the field. Using Hailton s equations, we find that the equations of otion for the particle variables are ż = p/, (13) ṗ = z V(z) 2g x φ(t,x) x=z(t), (14) and the equations of otion for the field variables are t φ(t,x) = π(t,x), (15) t π(t,x) = x 2 φ(t,x) 2ρ(t,x). (16) It is straightforward to verify that these equations of otion are equivalent to equations (1), (2) and (3). Note that when the equations of otion are satisfied, p = ż is the echanical oentu of the particle and π(t,x) = B(t,x) is the B-field. 3 We will assue that L is large enough that the particle is always confined to this region.

4 818 A D Boozer 2.1. Solutions to the field equation Now that we have described the odel, let us consider soe solutions to the field equation (5). Although we are priarily interested in the situation in which the charge distribution ρ(t,x) is given by equation (4), it is ore instructive to solve the field equation for the case of an arbitrary static charge distribution ρ(x). We can express the solutions to the field equation as φ(t,x) = φ s (x) + φ r (t, x), where and φ s (x) = 2 x φ s(x) = 2ρ(x) (17) φ r (t, x) = 0. (18) We can think of φ s (x) as describing the static potential generated by the charge distribution, and φ r (t, x) as describing any radiation that ight be present. Fro equation (17), we find that the static potential is given by 4 φ s (x) = φ 0 + x y ρ(y)dy, (19) where φ 0 is an arbitrary constant. The corresponding E and B fields are E s (x) = x φ s (x) = ɛ(x y)ρ(y)dy, (20) B s (x) = t φ s (x) = 0. (21) Let g denote the total quantity of charge: g = ρ(x)dx. (22) If we substitute equations (19), (20) and (21) into equations (7) and (8), we find that the total energy is U tot = U f + U i = U b + U c, where U b = x y ρ(x)ρ(y)dx dy (23) is the binding energy of the charge distribution, and U c = g 2 L/2 +2gφ 0 is a constant that depends only on the total quantity of charge g, not on how the charge is spatially distributed. As an exaple, let us consider a charge distribution that consists of two stationary point charges: ρ(x) = g 1 δ(x z 1 ) + g 2 δ(x z 2 ). (24) Fro equation (23), we find that the binding energy of the two charges is U b = 2g 1 g 2 z 1 z 2. (25) If the two charges have the sae sign, then U b describes an attractive interaction potential between the charges. Thus, in contrast to electrodynaics, charges of the sae sign attract. This can also be shown by solving for the field produced by one of the charges, and then using equation (1) to calculate the force that this field exerts on the other charge. 4 The ost general solution includes an additional ter E 0 x that describes a constant background E-field E 0,butfor siplicity we will assue that E 0 = 0. Also, we have assued that ρ(x) is nonzero only in the region [ L/2,L/2].

5 A toy odel of quantu electrodynaics in (1+1) diensions Solutions to the particle equation of otion Now that we have considered soe solutions to the field equation, let us turn our attention to equation (1), the equation of otion for the particle. One can show that the field E(t,z) that appears on the right hand side of this equation can be divided into two coponents: a coponent E r (t, z) = gż that describes the retarded E-field generated by the particle itself, and a coponent E ext (t, z) that describes the E-field due to other charges and to incoing radiation 5. By dividing the E-field in this way, we can express the equation of otion as z + γ ż = z V(z) 2gE ext (t, z), (26) where γ 2g 2 /. Thus, the coupling of the particle to its own field results in a daping ter with daping constant γ. As the particle oves it radiates energy to the field, and this daping ter describes the reaction force that the radiation exerts on the particle. Let us solve equation (26) for the special case of a haronically bound particle in the absence of an external field (E ext (t, z) = 0). The potential is V(z)= 1 2 ω2 A z2, (27) where ω A is the haronic frequency, so the equation of otion is z + γ ż + ωa 2 z = 0. (28) The solution to this equation is z(t) = A e γt/2 cos(ωt + θ), (29) ( where ω ω / A 1 γ 2 1/2, 4ωA) 2 and A and θ are set by the initial conditions. Note that because of the daping ter, the oscillation frequency is shifted by δω = ω ω A ;thisisthe classical analogue of the quantu Lab shift, which we will discuss in section 5.2. Let us assue that the daping constant is sall copared to the haronic frequency (γ ω A ).In this liit, the frequency shift is given by δω = γ 2 /8ω A. Also, if we substitute the particle trajectory given by equation (29) into equation (6) for the particle energy, we find that U p = 1 2 ω2 A A2 e γt. (30) Thus, the energy of the particle decays exponentially at rate γ. 3. Quantizing the toy odel We can quantize the toy odel by replacing the dynaical variables z, p, φ(t, x), and π(t,x) with quantu operators. For the particle variables z and p, this procedure is failiar fro eleentary quantu echanics: in the position basis, the operators corresponding to these variables are ẑ = z, ˆp = i z. (31) Fro equation (10), we find that the quantized particle Hailtonian is Ĥ p = ˆp2 1 + V(ẑ) = z + V(z)= E n n n, (32) n where n and E n, the eigenstates and eigenvalues of Ĥ p, are obtained by solving the tieindependent Schrödinger equation. For the special case of a haronically bound particle, it is convenient to express ẑ and ˆp in ters of creation and annihilation operators b and b: ẑ = (2ω A ) 1/2 (b + b ), ˆp = i(ω A /2) 1/2 (b b ). (33) 5 This is shown in [1]. For siplicity, we have assued that the speed of the particle is uch less than the speed at which waves are propagated by the field ( ż 1); in general, E r (t, z) = gż(1 ż 2 ) 1.

6 820 A D Boozer The quantized particle Hailtonian can then be expressed as Ĥ p = ω A (b b +1/2). (34) The eigenstates of Ĥ p are Fock states, and the eigenvalues are E n = ω A (n +1/2). In what follows, it will soeties be useful to write the position operator in the for ẑ = (η/ω A )(b + b ), (35) where η (2ω A ) 1/2 ω A is a diensionless paraeter, called the Lab Dicke paraeter, that characterizes the ratio of the size of the ground state wavepacket to the wavelength of light with frequency ω A. To quantize the field variables φ(t,x)and π(t,x), we first use the field Hailtonian given in equation (11) to derive the classical equations of otion for a free field: t φ(t,x) = π(t,x), t π(t,x) = x 2 φ(t,x). (36) We then find the noral odes of the syste 6, and quantize each ode by introducing creation and annihilation operators. The details of this procedure are described in a nuber of textbooks (see [2 4]), so we will oit the interediate steps and siply quote the end result: ˆφ(x) = (2ω L) 1/2( a e ik x + a e ik x ), (37) ˆπ(x) = i (ω /2L) 1/2( a e ik x a e ik x ). (38) Here is the ode nuber, which ranges over the nonzero integers, a and a are the creation and annihilation operators associated with ode, and k 2π/L and ω k are the wavenuber and frequency of ode. The quantized field operators are given by Ê(x) = x ˆφ(x), ˆB(x) = ˆπ(x). (39) If we substitute the quantized field operators into the field Hailtonian H f equation (11), and note that L/2 L/2 given in e i(k 1 +k 2 )x dx = Lδ 1 + 2, (40) we find that the quantized field Hailtonian is Ĥ f = ω a a, (41) where for siplicity we have dropped a constant ter that describes the zero-point energy. We will denote the ground state of Ĥ f by 0, and we will denote states in which one photon is present in ode by 1 a 0. Fro equation (9), we find that the quantized field oentu is ˆP f = k a a. (42) Note that ˆP f generates displaceents of the field operators; for exaple, [ ˆφ(x), ˆP f ] = i x ˆφ(x), [ˆπ(x), ˆP f ] = i x ˆπ(x). (43) 6 The noral odes depend on the choice of boundary conditions. We will assue that periodic boundary conditions are iposed on the region [ L/2,L/2], so φ(t,l/2) = φ(t, L/2) and π(t,l/2) = π(t, L/2).

7 A toy odel of quantu electrodynaics in (1+1) diensions Quantu field coupled to a classical charge distribution In section 2.1, we described solutions to the classical field equation when we replaced the point particle with an arbitrary static charge distribution ρ(x). Before turning to the full quantu theory, it is instructive to consider the quantu version of this proble. We will take the Hailtonian for the syste to be Ĥ = Ĥ f + Ĥ i, where Ĥ f, the field Hailtonian, is given by equation (41), and Ĥ i, the interaction Hailtonian, is given by Ĥ i = 2 ˆφ(x)ρ(x)dx. (44) 4.1. Perturbation theory In section 2.1, we found that for the classical theory we could divide the total potential φ into two coponents: a coponent φ s that described the static potential of the charge distribution, and a coponent φ r that described the dynaical radiation field. We expect that the ground state of the quantu theory will correspond to the situation in which the radiation coponent vanishes and only the static coponent is present. Thus, the properties of the quantu ground state should tell us soething about the static field generated by the charge distribution. We can deterine the ground state wavefunction by using perturbation theory; we will take Ĥ f to be the unperturbed Hailtonian and Ĥ i to be the perturbation. To first order in ρ(x), the ground state wavefunction is given by = 0 (1/ω ) 1 Ĥ i 0 1. (45) Fro equations (37) and (44), we find that 1 Ĥ i 0 =2(2ω L) 1/2 e ik x ρ(x)dx. (46) Thus, the vacuu expectation value of the scalar potential ˆφ(x) is ˆφ(x) = (2/L) (1 / ω 2 ) e ik (x y) ρ(y)dy. (47) We show how to evaluate the su in appendix B; fro equation (B.6), we find that ˆφ(x) =φ 0 + x y ρ(y)dy = φ s (x), (48) where φ 0 = gl/6 and φ s (x) is the classical solution given by equation (19). Siilarly, one can show that the vacuu expectation values of the field operators Ê(x) and ˆB(x) are Ê(x) =E s (x), ˆB(x) =B s (x), (49) where E s (x) and B s (x) are the classical solutions given by equations (20) and (21). Thus, in the quantu theory the charge distribution generates a static field that agrees with the predictions of the classical theory. We can also use perturbation theory to calculate the energy shift to the ground state; to second order in ρ(x), it is given by E = (1/ω ) 1 Ĥ i 0 2. (50) If we substitute for 1 Ĥ i 0 using equation (46), and evaluate the su using equation (B.6), we find that E = E c + x y ρ(x)ρ(y)dx dy = E c + U b, (51)

8 822 A D Boozer where E c = g 2 L/6 and U b is the classical binding energy given in equation (23). As we showed in section 2.1, we can take the charge distribution to consist of two point charges; the binding energy U b can then be thought of as describing an interaction potential between the charges. Since this binding energy has the sae for in both the classical and quantu theories, the interaction potential is likewise the sae, and for both theories charges of the sae sign feel an attractive force Unitary transforation Although we only calculated the vacuu expectation values to first order and the energy shift to second order, the results we obtained are actually exact to all orders. To see this, first note that we can express the interaction Hailtonian as Ĥ i = ω ( λ a + λ a ), (52) where we have defined λ = (2/ω )(2ω L) 1/2 e ik x ρ(x)dx. (53) Thus, the total Hailtonian is Ĥ f + Ĥ i = ω a a + ω ( λ a + λ a ). (54) We can diagonalize this Hailtonian by defining a unitary transforation ( ( U = exp λ a λ a ) ), (55) which has the property that 7 Ua U = a λ. (56) Fro equations (54) and (56), it follows that the transfored Hailtonian is U(Ĥ f + Ĥ i )U = Ĥ f ω λ λ. (57) If we substitute for λ, and use the results of appendix B to perfor the su over, we find that U(Ĥ f + Ĥ i )U = Ĥ f + E c + U b, (58) where as before E c = g 2 L/6 and U b is the classical binding energy given in equation (23). We can understand the physical eaning of the unitary transforation U by considering its effect on the field operators. Using equation (56), it is straightforward to show that U ˆφ(x)U = ˆφ(x) + φ s (x), (59) where φ s (x) is the classical static potential given by equation (19), and UÊ(x)U = Ê(x) + E s (x), U ˆB(x)U = ˆB(x) + B s (x), (60) where E s (x) and B s (x) are the classical static fields given by equations (20) and (21). Thus, we can think of U as reoving the static coponent of the field, leaving only the quantized radiation field (in the classical theory, this would correspond to subtracting the static potential φ s fro the total potential φ to obtain the radiation potential φ r ). In equation (58) forthe 7 For each ode, U acts as a displaceent operator D(λ ); see appendix A, equation (A.6).

9 A toy odel of quantu electrodynaics in (1+1) diensions 823 transfored Hailtonian, Ĥ f is the Hailtonian for the radiation field and E c + U b is the energy of the static field. Note that the ground state of the transfored Hailtonian is 0 ; as expected, this corresponds to the situation in which there is no radiation (Ĥ f 0 =0). Also, note that the energy of the ground state is E c + U b, which agrees with the value we obtained by using second-order perturbation theory. By inverting the unitary transforation, we can obtain an exact expression for the ground state of the original Hailtonian: ( =U 0 =exp ( λ a λ a ) ) 0 = α, (61) where α is a coherent state of ode with aplitude α = λ (see appendix A). Fro equation (59), it follows that the vacuu expectation value of the scalar potential ˆφ(x) is ˆφ(x) = 0 U ˆφ(x)U 0 = 0 ˆφ(x) 0 + φ s (x) = φ s (x), (62) which agrees with the value we obtained by using first-order perturbation theory. 5. Quantu theory Let us now consider the full quantu theory, which describes a quantized particle coupled to a quantized field. The Hailtonian for the syste is Ĥ = Ĥ p + Ĥ f + Ĥ i, where Ĥ p and Ĥ f, the Hailtonians for the particle and field, are given by equations (32) and (41), and Ĥ i,the interaction Hailtonian, is obtained by quantizing the Hailtonian given in equation (12): Ĥ i = 2g ˆφ(ẑ) = 2g (2ω L) 1/2( a e ikẑ + a e ik ẑ ). (63) It is straightforward to check that [ ˆP f + ˆp, Ĥ ] = i z V(z), (64) so in the absence of an external potential the total oentu of the syste is conserved, just as for the classical theory. We can study the effects of the coupling between the particle and field by using perturbation theory: we will take Ĥ 0 Ĥ p + Ĥ f to be the unperturbed Hailtonian and Ĥ i to be the perturbation. The eigenstates of Ĥ 0 are tensor products of the particle and field eigenstates; for exaple, n, 0 n 0 is an eigenstate of Ĥ 0 in which the particle is in state n and the field is in state 0, and n, 1 n 1 is an eigenstate in which the particle is in state n and the field is in state Spontaneous decay The interaction Hailtonian allows the particle to exchange energy with the field, and this eans that excited states of the particle can spontaneously decay. A state with n>0can spontaneously decay to any state r<nby eitting a photon, where the decay rate is given by Feri s golden rule: Ɣ n r = 2π r, 1 Ĥ i n, 0 2 δ(e n E r ω ). (65) Fro equation (63) for the interaction Hailtonian, we find that r, 1 Ĥ i n, 0 =2g(2ω L) 1/2 r e ikẑ n. (66) If we substitute this atrix eleent into equation (65), and convert the su over into an integral over k (see equation (B.4)), we find that Ɣ n r = 2g 2 (1/ω k ) r e ikẑ n 2 δ(e n E r ω k ) dk, (67)

10 824 A D Boozer where ω k k. The delta function in the integral tells us that the frequency of the eitted photon is equal to the energy difference between the initial and final states: ω k = E n E r. We will assue that the wavelength associated with this frequency is large copared to the spatial extent of the particle wavefunction, so we can ake the dipole approxiation: r e ikẑ n 2 r 1 ikẑ n 2 = ωk 2 r ẑ n 2. (68) If we substitute equation (68) into equation (67) and perfor the integral over k, we find that Ɣ n r = 4g 2 (E n E r ) r ẑ n 2. (69) Note that the spatial extent of state n is given by n ẑ 2 n 1/2, so the dipole approxiation is justified when ω k n ẑ 2 n 1/2 1. (70) Let us now specialize to the case of a haronically bound particle. If we substitute equation (35) forẑ into equation (70), and use that ω k = E n E r = (n r)ω A, we find that the dipole approxiation is justified when η(n r) 2n Thus, we will assue that η 1 and that n is sall enough that this condition is et. Since the atrix eleent r ẑ n vanishes unless r = n ± 1, in the dipole approxiation state n only decays to state n 1. Fro equation (69), we find that the decay rate is Ɣ n n 1 = 4g 2 ω A n 1 ẑ n 2 = γn, (71) where γ = 2g 2 / is the classical daping constant. If we start the particle in an arbitrary initial state, it will eventually decay to the ground state by spontaneously eitting photons. To describe this process, let p n (t) denote the probability that the particle occupies Fock state n at tie t. These probabilities evolve in tie according to a set of rate equations: ṗ n (t) = Ɣ n+1 n p n+1 (t) Ɣ n n 1 p n (t). (72) The first ter describes the rate at which the population in n increases due to spontaneous decay fro n +1ton, and the second ter describes the rate at which the population in n decreases due to spontaneous decay fro n to n 1. Equation (72) can be thought of as a quantu analogue 8 to equation (28), the classical equation of otion for a haronically bound particle undergoing radiation daping, and it is instructive to copare the classical and quantu theories. The closest analogue in the quantu theory to a classical state of the oscillator is a coherent state (see appendix A). Suppose that at t = 0 we start the particle in a coherent state α 0. Using equation (A.2) to expand the coherent state in a basis of Fock states, we find that the initial probabilities are p n (0) = n α 0 2 = (1/n!) e n0 n n 0, (73) where n 0 = α 0 2 is the ean vibrational quantu nuber. Using equations (71) and (72), it is straightforward to check that the probabilities at tie t are p n (t) = (1/n!) e n(t) n n (t), (74) where n(t), the ean vibrational quantu nuber at tie t, is given by n(t) = n 0 e γt. Since the ean energy of the particle is Ē(t) = ω A n(t), we see that the ean energy of the particle decays exponentially at rate γ, just as we found for the classical theory. 8 The probabilities p n are the diagonal eleents of the density atrix for the quantu particle, and one can derive an equation of otion that describes the evolution of the entire density atrix. This equation, called a aster equation, is an even closer analogue to the classical equation of otion, but a discussion of these atters is beyond the scope of this paper (see [5]).

11 A toy odel of quantu electrodynaics in (1+1) diensions Lab shift In addition to causing spontaneous decay, the interaction Hailtonian shifts the energy levels of the particle fro E n to E n + δ n ; for ordinary QED, this effect is known as the Lab shift 9. We can calculate the shift to state n by using second-order perturbation theory: δ n = r, 1 Ĥ i n, 0 2 (E n E r ω ) 1 = n,r, (75) r r where n,r = (2g) 2 (2ω L) 1 r e ikẑ n 2 (E n E r ω ) 1. (76) We can use the results of appendix B to evaluate the n,n ter explicitly: n,n = (2g 2 /L) ( 1 / ω 2 ) n e ik ẑ n 2 (77) = E c + g 2 x y ψ n (x) 2 ψ n (y) 2 dx dy, (78) where E c = g 2 L/6. Here we have substituted n e ikẑ n = n z e ikz z n dz = e ik z ψ n (z) 2 dz, (79) where z is a position eigenstate with eigenvalue z, and ψ n (z) = z n is the position-space wavefunction for the state n. If we use equation (78) to substitute for n,n in equation (75), we find that δ n = E c + g 2 x y ψ n (x) 2 ψ n (y) 2 dx dy + n,r. (80) r n We can interpret this result as follows. The first ter is a constant that is the sae for all energy levels, and siply redefines the zero of energy. Since it is only the energy differences between levels that are physically eaningful, we can neglect this constant ter and consider only the n-dependent portion of the level shift that is given by δ n δ n E c. The second ter is just the binding energy for a classical charge distribution with charge density ρ(x) = g ψ n (x) 2 ; we will call this the seiclassical binding energy. The third ter describes corrections to the seiclassical binding energy due to the contributions of other levels. Note that if the energy spacings E n E r are large, then these corrections are suppressed and δ n is doinated by the seiclassical binding energy. Let us now specialize to the case of a haronic oscillator potential. Fro equation (76), we find that n,r = (γ /2πη)I n (η), (81) r n where we have defined a diensionless quantity I n (η) = r e iu(b+b ) n 2 (u 2 + (r n)ηu) 1 du. (82) r n 0 Here we have substituted for ẑ using equation (35), converted the su on to an integral over k using equation (B.4), and defined a diensionless variable u = (2ω A ) 1/2 k. Note that 9 Discussions of the Lab shift in hydrogen that are accessible to undergraduates are given in [6 8].

12 826 A D Boozer when r<nthere are poles in the integral; these are to be evaluated by taking the principal part. If we substitute the copleteness relation n e ikẑ n 2 = 1 r e ikẑ n 2 (83) r n into equation (77), we find that n,n = E c + (γ /2πη)I n (0). (84) Thus, the level shift is given by δ n = δ n E c = (γ /2πη)(I n (0) I n (η)). (85) The I n (0) ter describes the seiclassical binding energy, and the I n (η) ter describes the corrections due to the contributions of other levels. As an exaple, let us calculate the Lab shift to the ground state. Using the displaceent operator D(α) that is defined in appendix A, we find that r e iu(b+b ) 0 2 = r D(α) 0 2 = r α 2 = 1 r! e u2 u 2r, (86) where α is a coherent state with aplitude α = iu. Thus, 1 I 0 (η) = e u2 u 2r 1 (u + rη) 1 du. (87) r! r=1 0 Note that 1 I 0 (0) = e u2 u 2r 2 du = (1/u 2 ) ( ) 1 e u2 du = π. (88) r! r=1 0 0 This result could also be obtained by substituting the ground state wavefunction ψ 0 (x) into equation (78) for n,n. Fro equation (87) we see that 0 I 0 (η) I 0 (0), soδ 0 is always positive. In figure 1, we nuerically evaluate 10 I 0 (η), and plot I 0 (η)/i 0 (0) and δ 0 /γ as a function of η. Note that as η,i 0 (η)/i 0 (0) 0, so in the liit of large η the level shift is doinated by the seiclassical binding energy. It is interesting that although the quantu and classical versions of the theory predict the sae spontaneous decay rate γ, the quantu level shift δ n is copletely different fro the classical frequency shift δω that we calculated in section 2.2. The situation is the sae in ordinary QED: the classical and quantu theories agree regarding radiative decay rates, but disagree regarding radiative frequency shifts (this is discussed in [9]) Unitary transforation In section 4.2, we considered a quantu field coupled to a classical charge distribution, and showed that one could reove the static field of the distribution by defining a suitable unitary transforation. For the full quantu theory, we can define a siilar unitary transforation that reoves the static field generated by the particle. Following the procedure described in section 4.2, we write the interaction Hailtonian in the for Ĥ i = 2g ˆφ(ẑ) = ω ( λ a + λ a ), (89) 10 As a consistency check, we nuerically evaluate I 0 (0) and copare the result with the exact value I 0 (0) = π; we find that that the nuerical and exact values differ by

13 A toy odel of quantu electrodynaics in (1+1) diensions 827 (a) (b) Figure 1. Lab shift to the ground state. (a) I 0 (η)/i 0 (0) versus η; (b)δ 0 /γ versus η. For coparison, the dashed curve is U b /γ,whereu b is the seiclassical binding energy. where λ = (2g/ω )(2ω L) 1/2 e ikẑ, (90) and we define a unitary transforation ( ( U = exp λ a λ a ) ). (91) Since Ua U = a λ, we find that U(Ĥ f + Ĥ i )U = Ĥ f ω λ λ = Ĥ f + E c, (92) where E c = g 2 L/6. We can also transfor the field operators; for exaple, the transfored scalar potential is U ˆφ(x)U = ˆφ(x) + ˆφ s (x), (93) where ˆφ s (x) = φ 0 + g x z z z dz (94) and φ 0 = gl/6. The operator ˆφ s (x) describes the instantaneous static potential generated by the particle: note that if the state of the particle is ψ, then the expectation value of ˆφ s (x) is ψ ˆφ s (x) ψ =φ 0 + g x z ψ(z) 2 dz. (95) This is just the static potential for a classical charge distribution with charge density ρ(x) = g ψ(x) 2, where ψ(z) = z ψ is the position-space wavefunction of the particle. Equation (92) describes how the field and interaction Hailtonians transfor under U;let us now consider the transforation of the particle Hailtonian Ĥ p. It is convenient to work in the position basis, in which ẑ = z, and to express the transforation as U = exp( 2igŜ(z)), where we have defined a new quantu field Ŝ(x) = i (1/ω )(2ω L) 1/2( a e ik x a e ik x ). (96) Note that U ˆpU = U(U ˆp +[ˆp, U ]) = ˆp iu z U = ˆp +2g z Ŝ(z), (97)

14 828 A D Boozer so the transfored particle Hailtonian is UĤ p U = (1/2)( ˆp +2g z Ŝ(z)) 2 + V(z). (98) Cobining these results, we find that the transfored total Hailtonian is UĤU = U(Ĥ p + Ĥ f + Ĥ i )U = Ĥ pi + Ĥ f + E c, (99) where we have defined Ĥ pi UĤ p U. In the transfored representation, the static field of the particle has been copletely eliinated fro our description of the syste: Ĥ f describes the pure radiation field, and Ĥ pi describes the particle and its coupling to this radiation field. The transfored Hailtonian is directly analogous to the Hailtonian for the nonrelativistic liit of ordinary QED in the Coulob gauge 11 : Ĥ = (1/2)(ˆp + qâ(r)) 2 + Ĥ rad. (100) Here ˆp is the oentu of the particle, and q are its ass and charge, and Â(r) and Ĥ rad are the vector potential and Hailtonian that describe the quantized radiation field. In perforing calculations, it is useful to express the transfored Hailtonian as UĤU = Ĥ p + Ĥ f + Ĥ i, (101) where Ĥ i Ĥ pi Ĥ p + E c. This has the sae for as the original Hailtonian, only the interaction Hailtonian is now given by Ĥ i instead of Ĥ i. Note that 12 Ĥ pi = UĤ p U = Ĥ p i[a, Ĥ p ] (1/2)[A, [A, Ĥ p ]], (102) where A 2gŜ(ẑ). Thus, we can express the new interaction Hailtonian as Ĥ i = Ĥ pi Ĥ p + E c = E c i[a, Ĥ p ] (1/2)[A, [A, Ĥ p ]]. (103) It is instructive to use the new for of the transfored Hailtonian to repeat our previous calculations of the spontaneous decay rate and the Lab shift. We will take Ĥ p + Ĥ f to be the unperturbed Hailtonian and Ĥ i to be the perturbation. To second order in g, we find that the Lab shift to state n is given by δ n = E c (1/2) n, 0 [A, [A, Ĥ p ]] n, 0 + r, 1 [A, Ĥ p ] n, 0 2 (E n E r ω ) 1. (104) r Note that r, 1 [A, Ĥ p ] n, 0 =(E n E r ) r, 1 A n, 0 (105) and r, 1 A n, 0 =(i/ω ) r, 1 Ĥ i n, 0. (106) Also, by inserting a coplete set of states, we find that (1/2) n, 0 [A, [A, Ĥ p ]] n, 0 = (E n E r ) r, 1 A n, 0 2. (107) r If we substitute these results into equation (104), it is straightforward to check that it agrees with the expression for the Lab shift given in equation (75). We can also calculate the spontaneous decay rate; to lowest order in g,itisgivenby Ɣ n r = 2π r, 1 [A, Ĥ p ] n, 0 2 δ(e n E r ω ). (108) Using equations (105) and (106), it is straightforward to check that this agrees with the expression for the spontaneous decay rate given in equation (65). 11 This is the for of the QED Hailtonian that is usually presented in undergraduate quantu echanics courses, and should be failiar to students; see [6]. Our unitary transforation U is analogous to a siilar transforation used in ordinary QED; see [10]. 12 For arbitrary operators X and Y, one can show that e X Ye X = Y +[X, Y ]+(1/2)[X, [X, Y ]] +. If we substitute X = ia and Y = Ĥ p, we find that [X, [X, Y ]] is a c-nuber, so the series truncates after the third ter.

15 A toy odel of quantu electrodynaics in (1+1) diensions 829 Appendix A. Coherent states Consider a quantu haronic oscillator that is described in ters of creation and annihilation operators a and a. For any coplex nuber α, one can define a state α that is an eigenstate of a with eigenvalue α: a α =α α. (A.1) Such states are called coherent states, and have a nuber of applications in quantu field theory. Here we suarize soe iportant properties of coherent states that we ake use of in the paper; a ore coplete treatent can be found in [5]. A coherent state α can be expanded in a basis of Fock states as follows: α =e α 2 /2 (1/n!) 1/2 α n n. (A.2) n The ean vibrational quantu nuber of the coherent state α is n = α a a α = α 2. One can define a unitary displaceent operator D(α) = e αa α a, which acts on the vacuu to produce a coherent state α : D(α) 0 = α. The displaceent operator also has the property that D(α)aD (α) = a α. (A.3) (A.4) (A.5) (A.6) Appendix B. Su over odes We will often want to evaluate the su S(x) = (2/L) ( / ) 1 ω 2 e ik x = (2/L) ( 1 / ω 2 ) cos k x. (B.1) Note that S(0) = (2/L) 1 / ω 2 = (L/2π 2 ) 1/ 2 = L/6 (B.2) and ( 1 / ω 2 ) (1 cos k x). (B.3) S(0) S(x) = (2/L) Recall that k = 2π/L,soasLincreases the spacing between adjacent odes decreases. Thus, for large L we can approxiate sus over by integrals over k: (L/2π) dk. (B.4) If we approxiate equation (B.3) in this way, we find that S(0) S(x) = (1/π) (1/k 2 )(1 cos kx) dk = x. (B.5) Cobining equations (B.2) and (B.5), we find that the su is given by (2/L) ( / ) 1 ω 2 e ik x = L/6 x. (B.6)

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