Quantization of electromagnetic eld in a circular cylindrical cavity K. Kakazu Department of Phyic, Univerity of the Ryukyu, Okinawa 903-0, Japan Y. S. Kim Department of Phyic, Univerity of Maryland, College Park, Maryland 20742 We preent a quantization procedure for the electromagnetic eld in a circular cylindrical cavity with perfectly conducting wall, which i baed on the decompoition of the eld. A new decompoition procedure i propoed; all vector mode function atifying the boundary condition are obtained with the help of thi decompoition. After expanding the quantized eld in term of the vector mode function, it i poible to derive the Hamiltonian for thi quantized ytem. PACS:42.50.-p, 03.70.+k
Introduction The behavior of an atom depend on the mode tructure of the electromagnetic eld urrounding it []. The mode tructure in a cavity i determined by it boundary condition which are quite dierent from the periodic boundary condition ued in Lorentz-covariant electrodynamic. There are many intereting eect on the interaction of the atom with the eld with variou boundary condition. It i poible to calculate them and compare them with experimental data [2]. Thank to many new \high-tech" experimental intrument, cavity electrodynamic ha become one of the mot active branche of phyic in recent year. In quantum electrodynamic, we ue the plane wave atifying the periodic boundary condition whoe quantization volume eventually become innite. A a reult, the orientation of the eld i not eential. On the other hand, the eld in a cavitymut atify the boundary condition and the orientation of the eld with repect to cavity wall become an important factor. For thi reaon, we have to ue vector mode function to deal with thi problem. Thee function atify the boundary condition and the boundary eect in turn i contained in the function. Thu, for a given cavity, the mot important tep in the quantization proce i to contruct the vector mode function. Indeed, everal author have invetigated the quantization procedure for the eld in cavitie [3]. They have tudied the eld quantization, the emiion rate, and the atomic level hift in the cavitie with one or two innite plane mirror. Cavity electrodynamic with rectangular coordinate ytem doe not preent further ignicant mathematical problem. However, the cae i quite dierent for cavitie with curved boundary condition. Even for circular cylindrical or pherical cavitie, the quantization procedure ha not yet been rmly etablihed. The purpoe of the preent paper i to dicu in detail the mathematical problem in contructing the vector mode function with a circular cylindrical boundary condition. We hope to dicu the pherical cavity in a later paper. Recently, wehave carried out the eld quantization for everal dierent rectangular boundary condition including a rectangular tube uing the vector mode function [4]. Thee function have been derived with the help of an orthogonal matrix. However, the procedure developed there i not applicable to other cavitie in a traightforward manner. In thi paper, we preent a more general method of obtaining the vector mode function and apply it to the eld quantization in a circular cylindrical cavity with perfectly conducting wall. Thi will be ummarized in two theorem. With thi in mind, we rt decompoe the eld into three part in the circular cylindrical coordinate in the following ection. Then, in Sec. 3, we apply it to the circular cylindrical cavity. After obtaining all vector mode function by uing the decompoition, we arrive at the quantized eld and the quantized electromagnetic Hamiltonian in Sec. 4. 2 Decompoition of Electromagnetic eld Let u now derive the decompoition formula for the electromagnetic eld from Maxwell' equation. Uing the cylindrical coordinate ytem, we hall decompoe the olution of Maxwell' equation into three term, in preparation for the eld quantization within the 2
circular cylindrical cavity in Sec. 3. We emphaize here again that thi decompoition proce i the key to the quantization. Maxwell' equation for the electric eld E and the magnetic eld B in free pace are given by re=0; (2.) rb=0; (2.2) re+@ t B=0; (2.3) rb c 2@ te=0; (2.4) where c i the velocity of light in free pace and @ t = @=@t. Aiwell known, it follow from Maxwell' equation that the eld E and B atify the wave equation: 4 c 2@2 t E=0; 4 c 2@2 t B=0; (2.5) where 4 i the Laplacian operator. The electromagnetic eld E and B can be written in the cylindrical coordinate (r;';z) a E = E T + E z ; B = B T + B z ; (2.6) where E T = e r E r + e ' E ' i the tranvere component of the eld and E z = e z E z. Here e r, e ', and e z are the unit vector in the r, ', and z direction, repectively. Similarly, re, re, and r ( a caler function) can alo be divided into two part. For example, re=r T E+r z E, where the rt term contain the derivative with repect to r and ', while the econd term contain the derivative with repect to z. For implicity, the derivative with repect to r;', and z are decribed a @=@r = @ r, @=@' = @ ' and @=@z = @ z, repectively. Since r z E z = 0, Eq. (2.3) give Similarly, from Eq. (2.4), we have @ t B z = r T E T ; @ t B T = r T E z r z E T : (2.7) c 2@ te z =r T B T ; Equation (2.7) and (2.8) give which lead to c 2@ te T =r T B z +r z B T : (2.8) c 2 @2 t E T = r T @ t B z r z (r T E z +r z E T ); c 2@2 t B T = c 2r T @ t E z r z (r T B z +r z B T ); (2.9) c 2 @2 t E T + r z r z E T r T r T E z =r@ t B z rre z ; c 2@2 t B T +r z r z B T r T r T B z = c 2r@ te z rrb z ;(2.0) 3
where we have ued r T E z = re z. Let u note that r z r z E T = 4 z E T ; r T r T E z = 4 T E z ; c 2 @2 t E T =(4E) T =4E T ; (2.) where (4E) T i the tranvere part of 4E; the lat equation hold in the preent coordinate, although it i not correct in general. (For example, it i not correct in the pherical coordinate.) Then, from Eq. (2.0), we arrive at 4 T E= r r E z + r@ t B z ; 4 T B= c 2r@ te z rrb z : (2.2) To rewrite Eq. (2.2), we mut obtain an expreion for the component E z and B z. Suppoe that the eld i in a nite region and expand E z and B z in term of a certain complete ytem of function with mode : E z (r;t) = B z (r;t) = (E z (r;t)+c:c:); (B z (r;t)+c:c:); (2.3) where E z (r;t)= ~ E z (r)e i! t ; B z (r;t)= ~ B z (r)e i! 2t : (2.4) Here! (! 0; =;2) i determined by uing given boundary condition. Since E z and B z atify the wave equation, their component atify the Helmholtz equation: 4E z = k 2 E z; 4B z = k 2 2 B z; (2.5) where We aume that the component atify k 2 =!2 c 2 : (2.6) @ 2 z E z = h 2 E z; @ 2 z B z = h 2 2 B z; (2.7) where h 2 i alo determined by the boundary condition. Then we havetwo dimenional Helmholtz equation: 4 T E z = g 2 E z; 4 T B z = g 2 2 B z; (2.8) where g 2 = k 2 h 2 : (2.9) Now we dene two function F from E z and B z with g 2 6=0a F (r;t)= g 2 6=0 [F (r;t)+c:c:]= 4 g 2 6=0 [ ~ F (r) e i!t +c:c:]; (2.20)
where F = E z =g 2 ; F 2 = B z =g 2 2 : (2.2) The function F and their component F atify the ame equation a the z component of the eld: 4F = @ F c 2 ; 4 T F = g 2 @ F (2.22) t From Eq. (2.2) and (2.22) the component F i a olution of the Poion equation 4 T F = E z ; 4 T F 2 = B z (2.23) and then the function 4 T F atify 4 T F = 4 T F 2 = g 2 6=0 (g 2 F +c:c:)= g 2 2 6=0 (g 2 2 F 2 +c:c:)= g 2 6=0 (E z +c:c:); g 2 2 6=0 (B z +c:c:): (2.24) On the other hand, if there i a component E z or B z with g 2 = 0, Eq. (2.8) reduce to the two dimenional Laplace equation: 4 T E z =0or4 T B z =0. Now dene E 0z and B 0z a E 0z = (E z +c:c:); B 0z = (B z +c:c:); (2.25) g 2 =0 g2 2 =0 which atify 4 T E 0z =0; 4 T B 0z =0: (2.26) Then we have a ueful expreion for E z and B z ; for any E z and B z there exit function F given by Eq. (2.20) and function E 0z and B 0z are given by Eq. (2.25) uch that E z = 4 T F + E 0z ; B z = 4 T F 2 + B 0z : (2.27) It i worth emphaizing again that the function F are contructed with the component E z and B z with g 2 6= 0, while E 0z and B 0z conit of the component with g 2 =0. Uing Eq. (2.27) and dening F by we can rewrite Eq. (2.2) a F = e z F ; (2.28) 4 T E rrf +r@ t F 2 = rre0z +r@ t B 0z ; (2.29) 4 T B c 2 r@ tf rrf 2 = c 2r@ te 0z rrb 0z ; (2.30) where E 0z = e z E 0z and B 0z = e z B 0z. Let u dene E 0 and B 0 a the quantitie in the parenthee at the left hand ide in Eq. (2.29) and (2.30), repectively. The reult of thi ection are then ummarized in the following theorem. 5
Theorem : The eld can be decompoed into three component a follow: E = rrf r@ t F 2 +E 0 ; B= c 2r@ tf +rrf 2 +B 0 ; (2.3) where E 0 and B 0 atify 4 T E 0 = rre 0z +r@ t B 0z ; 4 T B 0 = c 2r@ te 0z rrb 0z : (2.32) Theorem play a central role in performing eld quantization in thi paper. Take the z component of Eq. (2.3), we nd that the z component of E 0 and B 0 are given by E 0z and B 0z in Eq. (2.27), repectively. Furthermore, Eq. (2.32) i conitent with Eq. (2.2). Let u conider the phyical meaning of the above decompoition formula. Firt it i worth emphaizing that each term in Eq. (2.3) i a olution to Maxwell' equation. That i, each term containing F (F term) atie Eq. (2.) - (2.4), which i eaily hown by uing the fact that F atie the wave equation (2.22). A a reult, the third term E 0 ; B 0 i alo a olution to Maxwell' equation, o that it atie Eq. (2.2), a mentioned above. Next take a particular example of the boundary condition atifying E 0z ;B 0z = 0. Then the eld E 0 ; B 0 decribe o-called the TEM (tranvere electromagnetic) mode [5]. Since the F term ha no z component of the magnetic eld, it become TM (tranvere magnetic), while the F 2 term become TE (tranvere electric), becaue it doe not contain the z component of the electric eld. In the above particular cae where E 0z ;B 0z = 0, Maxwell' equation for the eld E 0 ; B 0 reduce to r T E 0 =0; r T E 0 =0; r T B 0 =0; r T B 0 =0; r z E 0 +@ t B 0 =0; r z B 0 c 2@ te 0 =0: (2.33) 3 Determination of Function F We are conidering here the cavity encloed by a circular cylindrical wall with radiu a and height L: r < a,0 <z<l. We aume that the cavity ha perfectly conducting wall at z =0;L and at r = a. The tangential component of the electric eld Ej tan and the normal component of the magnetic eld Bj norm mut accordingly vanih at the boundary of the cavity. The above boundary condition reduce to that for the z component E z =0; @ r B z =0; (r=a); (3.) B z =0; @ z E z =0; (z=0; L): (3.2) 6
It i eay to get the econd condition in Eq. (3.) if we take the ' component of Eq. (2.4). Equation (2.) lead to the econd condition in Eq. (3.2). Before obtaining the component E z and B z,we how that the folowing lemma. Lemma : E 0 =0,B 0 =0. Proof: Firt we how that E 0z ;B 0z =0. Conider the component E 0z, which atie the Laplace equation 4 T E 0z = 0 [ee Eq. (2.26)] with the boundary condition E 0z =0at the boundary (r = a). Then we havee 0z = 0; thi i a well known property of the Laplace equation. Next conider the component B 0z atifying 4 T B 0z = 0 with the boundary condition @ r B 0z =0(r=a). Then B 0z mut be a contant, i.e., independent of the variable r and ': B 0z = f(z; t). Applying the two-dimenional Gau' theorem to rb 0 = r T B 0T +@ z B 0z =0, we get @ z B 0z = 0, becaue B 0r =0(r=a) and, a a reult, B 0z i independent of the variable r, ', and z. On the other hand, B 0z mut be zero at z =0;L, which lead to B 0z =0. A mentioned in the preceding ection, the eld E 0 ; B 0 become TEM atifying Eq. (2.33), becaue E 0z ;B 0z = 0. There exit no TEM in the preent cavity. That i, from Eq. (2.33) we haver T B 0 = 0 and r T B 0 = 0. A a reult, there exit a function ' uch that B 0 = r T ', which give B 0 =0by the boundary condition at r = a. Similarly, we getavector function A uch that E 0 = ra. The boundary condition lead u to E 0 =0. Q.E.D. Since E 0 ; B 0 = 0, Theorem how that the eld i contructed from the function F. Conequently, from Eq. (2.20), the eld only contain the component with g 2 6=0. Moreover, it i eay to prove that g 2 > 0by uing Gau' theorem and the boundary condition. Next we olve the Helmholtz equation (2.5) for the component E z and B z under the boundary condition (3.) and (3.2). Since g 2 > 0, the olution i given by E z (r;t)=c (t)j m ( r=a) e im' co(nz=l); B z (r;t)=c 2 (t)j m ( 2 r=a) e im' in(nz=l); (3.3) where the mode index i =(m; ; n) (m=0;;2;; =;2;3;;n =0;;2;), J m i the Beel function of the rt kind, m i the th zero point of J m, and 2 m2 i the th zero point ofjm, 0 the derivative ofj m. It follow from Eq. (2.4) that C (t) / exp( i! t). From the olution (3.3) we have which give 4 T E z = (k 2 (n=l) 2 )E z = ( =a) 2 E z ; 4 T B z = (k 2 2 (n=l) 2 )B z = ( 2 =a) 2 B z ; (3.4) g 2 =( =a) 2 ; k 2 =( =a) 2 +(n=l) 2 : (3.5) Let u next obtain the function F, which are dened in Eq. (2.20) and (2.2); they are given by F (r;t)= a2 E z (r;t) 2 h! i a (t) (r); 2" 0 7
where we haveintroduced a and : F 2 (r;t)= a2 B z (r;t) h!2 2 i a 2 (t) 2 (r); (3.6) 2 2" 0 a (t) =a (0)e i!t ; (3.7) (r;t)=c J m ( r=a) e im' co(nz=l); (3.8) 2(r;t)=c 2 J m ( 2 r=a) e im' in(nz=l); (3.9) where c are normalization contant. The function have the orthonormality property, which i ued in quantization in the next ection. Lemma 2: where R c dr = R cavity Z c dr (r) 0 (r) = 2 jc j 2 V 0; (3.0) rdrd'dz, V i the cavity volume, and m = J 2 m+ ( m); 2 m2 = J 2 m( m2 ) J 2 m+( m2 ): (3.) Here the quantity co(nz=l) in Eq. (3.8) mut be changed to = p 2 when n =0. Proof: The orthonormality property (3.0) i eaily derived from the following equality for the Beel function: Z a 0 dr rj m ( m r=a)j m ( m0 r=a) = 2 a2 0: (3.2) To prove Eq. (3.2), let u rt oberve the following two integral [6] Z dx xj m ( i x)j m ( j x) = h i i J m+ ( i )J m ( j ) j J m+ ( j )J m ( i ) 0 Z 0 = dx xj 2 m( i x) = 2 2 i 2 i 2 j 2 j h i J 0 m ( i)j m ( j )+ j J m ( i )J 0 m ( j) i (3.3) hj 2 m( i ) J m ( i )J m+ ( i ) i h J 2 m( i ) J 2 m+( i ) 2J 0 m( i )J m+ ( i ) i ; (3.4) = 2 where i (i =;2;) i a poitive real number and i 6= j. Conidering that m i the th zero point ofj m (x) and that J m ( m )+J m+ ( m )= 0, from Eq. (3.3) and (3.4), we have Z 0 dx xj m ( m x)j m ( m0 x) = 2 m 0; (3.5) where m i given by Eq. (3.). If we change the variable x to r=a, Eq. (3.5) give Eq. (3.2) for =. Similarly, Eq. (3.3) and (3.4) lead to Z 0 dx xj m ( m2 x)j m ( m0 2x) = 2 m2 0; (3.6) where m2 i given by Eq. (3.). Equation (3.6) give u Eq. (3.2) for = 2. Q.E.D. 8
4 Vector Mode Function and Field Quantization After dening the vector mode function, we quantize the eld and obtain the quantized Hamiltonian. The mode function are contructed according to the decompoition formula given in Eq. (2.3) In the cavity with the perfectly conducting wall, the decompoition (2.3) in Theorem reduce to E = rrf r@ t F 2 ; B= c 2r@ tf +rrf 2 ; (4.) becaue of E 0 ; B 0 = 0 [ee Lemma ]. Subtituting the function F in Eq. (3.6) into E in Eq. (4.), we nd E(r;t)=i h! where the vector mode function u are given by 2" 0 h a (t)u (r) a (t)u (r) i ; (4.2) u = rre z ; u 2 =i! 2 re z 2 : (4.3) Similarly, ubtituting F into B and conidering the equality we obtain ru =k 2 re z ; (4.4) h h i a (t)ru (r)+a (t)ru (r) : (4.5) 2" 0! B(r;t)= Before obtaining the quantized eld and Hamiltonian, let u preent ome propertie of the mode function. Lemma 3: ru =0; (4.6) (4 + k 2 )u =0; (4.7) u j tan =0; ru j norm =0; (on wall); (4.8) Z c dru (r) u 0 0(r) = 0 0; (4.9) Proof: We give a proof of the orthonormality property (4.9), becaue it i obviou to prove the other equation. From Eq. (4.3), the mode function u i rewritten a u = r(re z ) 4e z =k 2 e z +r(@ z ); (4.0) which give u u 0 = h k 2 k2 0 k 2 h2 0 i 0 h +k 2 0 zh @ (@z ) i + r 0 (@z )r@ z 0 i : (4.) 9
Taking into account Gau' theorem and the boundary condition, we nd that the lat two term in Eq. (4.) have no eect on the volume integration of u u 0. Equation (4.) give the orthonormality relation for u : Z c dru (r)u 0 (r)= 0; (4.2) where we have ued Eq. (3.0) and et 2c c = 2 a 2 V 2 : (4.3)!2 In the cae of u 2,we make ue of the equality (re z 2)(re z 0 2) = e z 2 (rre z 0 2)+r he i z 2 (re z 0 2) = 2 0 2 a 2 2 2+r( 0 2 r T 0 2): (4.4) The orthonormality property ofu 2 i then obtained from Eq. (3.0) if we et the normalization contant: 2a c 2 = 2 V 2 2 : (4.5) 2! 2 2 Finally we how that the mode function u and u 0 2 are orthogonal to each other. Since u 2 ha no z component, we ee that u u 0 2 = i! 0 2 k 2 e z + r@ z (re z 0 2) = i! 0 2r(@ z ) (re z 0 2) = i! 0 2r which ha no eect on the volume integration of u u 0 2. Q.E.D. h (@z )(re z 0 2) i ; (4.6) We are now ready to carry out the eld quantization. To get the quantized eld the function a (t) are regarded a annihilation operator, which annihilate photon with indice and atify the commutation relation: [a (t);a y 0 0(t)] = 0 0; (4.7) where a y 0 0(t) are creation operator. Then the eld given by Eq. (4.2) and (4.5) become operator. The quantized electromagnetic Hamiltonian H R i obtained by uing the equality Z c dr (ru (r)) (ru 0 0(r)) = k2 0 0 Thee are ummarized in the following theorem. Theorem 2: The quantized eld and the Hamiltonian are given by E(r;t)=i B(r;t)= H R = 2 h! h! Z c dru (r) u 0 0(r): (4.8) 2" 0 h a (t)u (r) a y (t)u (r)i ; (4.9) h h a (t)ru (r)+a y (r)i 2" 0! (t)ru : (4.20) a y a + a a y = : (4.2) 0 h! a y a + 2
Theorem 2 will play a fundamental role when weinvetigate the electromagnetic property of a phyical ytem in the circular cylindrical cavity. There are alo important iue in the preent cavity including, for example, pontaneou (timulated) emiion, aborption, the atomic energy hift, and the Caimir eect; they will be invetigated with the help of Theorem 2. It i neceary to expand the eld in term of the vector mode function a in Theorem 2. According to Fourier analyi, the eld can be expanded in term of, for intance, the periodic function a in ordinary quantum electrodynamic. However, in thi cae, the operator a and a y cannot decribe the phyical photon. Conider the ituation where the eld conit of only one mode correponding to the photon with a u, where u atie the periodic boundary condition. Thi mut be impoible, becaue thi eld doe not atify the boundary condition; the photon i then ctitiou. 5 Concluion The quantization procedure of the eld in the cavity without E 0 ; B 0 i a follow: (a) obtain the decompoition formula (2.3) in the circular cylindrical coordinate; (b) olve the Helmholtz equation (2.5) for the component E z and B z under the boundary condition (3.) and (3.2); (c) determine the function F, which are the olution of the Poion equation (2.23), from E z ;B z ; (d) ubtitute F into the decompoition formula (2.3) and obtain the vector mode function atifying the orthonormality property (4.9); (e) then we arrive at the quantized eld and Hamiltonian in Theorem 2. A ha been hown explicitly in Theorem 2 in Sec. 4, the energy h! of the photon depend on the polarization index a well a the wave number index. Thi i one of the characteritic of the circular cylindrical cavity. The photon energy i independent ofthe polarization index in a rectangular cavity and in free pace. In the entire proce of quantization, the decompoition formula in Theorem play the eential role. We mut nd uch a formula in an appropriate coordinate ytem if we tudy other type of cavitie. Alo, if a cavity allow the third term E 0 ; B 0 in (2.3), we haveto contruct a methodology to deal with the problem. We hope to dicu the quantization of electromagnetic eld inide a pherical cavity in a future publication. Acknowledgment We would like to thank Prof. Y. Ezawa, Dr. R. Ray, Dr. G. Gat, Dr. S. Kobayahi, Dr. Kurennoy, and Mr. K. Ohhiro for numerou valuable dicuion. We are alo grateful to Prof. M. S. Kim for bringing ome of the reference to our attention. Reference [] E. M. Purcell, Phy. Rev. 69, 68 (946); D. Kleppner, Phy. Rev. Lett. 47, 233 (98).
[2] G. Barton and N. S. J. Fawcett, Phy. Report 70, (988); P. Meytre, ibid 29, 243 (992); H. Walther, ibid 29, 263 (992); Cavity Quantum Electrodynamic, edited by P. R. Berman (Academic Pre, New York, 994); P. W. Milonni, The Quantum Vacuum: An Introduction to Quantum Electrodynamic (Academic Pre, New York, 994). [3] C. K. Carniglia and L. Mandel, Phy. Rev. D 3, 280 (97); P. W. Milonni, Phy. Rev. A 25, 35 (982); E. A. Power and T. Thirunamachandran, Phy. Rev. A 25, 2473 (982). [4] K. Kakazu and Y. S. Kim, Phy. Rev. A 50, 830 (994). [5] J. D. Jackon, Claical Electrodynamic (John Wiley and Son, New York, 975). [6] G. N. Waton, A Treatice on the Theory of Beel Function (Cambridge Univerity Pre, Cambridge, 962). 2