On the behavior of an electron in a homogeneous electrical field according to the relativistic Theory of Dirac

Transcription

1 On the behavior of an electron in a homogeneous electrical field according to the relativistic Theory of Dirac Fritz Sauter in Munich. With 6 Illustrations. April 21st, 1931 Abstract We give the solutions of the Equation of Dirac with the potential V = vx and discuss their properties. In addition to the solutions in the non-relativistic calculation, the Dirac solution contains a region in which the momentum and velocity of the electron have the opposite sign. We calculate the probability for an electron to change from the regime of negative momentum to the area of positive momentum. It turns out that the transition probability takes finite values only if the increase of the potential over distance of the Compton wavelength becomes comparable with the rest energy of the electron. The large values of the transition probability of a potential step of the dimension of the double rest energy calculated by O. Klein may therefore be understood as limiting values for an infinitely sharp potential step. Some time ago, an interesting work was appeared by 0. Klein 1 on the reflection of electrons at a potential step. The calculations followed the relativistic theory of Dirac showed the following result: If the height of the step P is increased starting from 0, the reflection coefficient R rises from 0 up to the value 1, which is reached at P = E E 0. (E is the relativistic energy of the electron, E 0 is its rest energy.) By further increasing P, R remains 1 until the value P = E + E 0 is reached. Upon a further increase of the height of the potential step, the value of the reflection coefficient decreases and approaches E cp E+cp in the limit P =. (p = Impulse of the electron before the potential step). Thus, in the Dirac theory, an electron has a finite probability of going through a very high, classically totally reflecting, potential step. More surprisingly, the electron has, after going through the potential step, a velocity (group velocity) pointing opposite to its momentum. 1 O. Klein, ZS f. Phys. 53, 157,

2 The appearance of a negative momentum is not too surprising, after having learned how to deal with negative energies 2. More remarkable is the large probability found by O. Klein for the transition from a positive to a negative momentum. N. Bohr mentioned that this high value is due to a very high potential step, and that finite transition probabilities are obtained only if the potential grows by mc 2 over a range of the size of the Compton wavelength h/mc 3. It is the aim of this paper to verify this explanation by N. Bohr. For this purpose, the rectangular potential jump AB C D (Fig. 1) of O. Klein will be replaced by a potential ABCD consisting of two areas (I and III) of a constant potential and an area of (II) linearly rising potential, corresponding to a constant electric field. We shall calculate the transition probability for an electron going from area I to area III. For this we first solve the Dirac equation in a homogeneous electrical field. The first three sections deal with finding the solution and their discussion while in the fourth section deals with the problem of the transition probability of an electron from positive to negative momentum. 1 Solution of the Dirac-Gleichung The potential V is given in the form and the Diracs equation reads V = vx (1) with the abbreviations The ansatz 2 Cmp. Theory of Dispersion, like I. Waller (ZS.f.Phys. 61,837,1930 showed, where the states of negative energy have a special meaning. 3 I want to thank Prof. Heisenberg for this remark 2

3 changes (2) into This equation is rearranged by setting where γ 5 anticommutes with γ 1, γ 4 and satisfies γ 2 5 = 1. After multiplication (5) from the left-hand side by γ 1, it becomes We introduce natural units. With we obtain from (5a) One could now integrate this equation by introducing for the γ ν a particular set of 4 4 Dirac matrices and for χ an array of four functions leading to a system of four differential equations of first order. More conveneintly, however, (10) is integrated without specializing the γ ν 4. For this we write χ in the general form where f and g do not contain any γ ν, and Γ is an arbitrary operator constructed out of the γ ν s. If one introduces this ansatz into (10), one can factor out 1+iγ 1 γ 4 and obtains 4 F. Sauter, ZS. f. Phys. 63,803,1930; 64,295,1930 3

4 If one multiplies this equation from the left by 1 ± iγ1 γ4, one recognizes that the expressions in the two square brackets must both vanish; These equations can easily be solved. If F (α, γ, x) denotes the degenerated hypergeometric function, defined by the expansion converging in the entire complex plane one can write the two solutions of (12) in the form: Note that, where the star indicates complex conjugation. In the following, the series expansion (14) will often be replaced by an integral representation. Consider the function Due to (13), it has the expansion 4

5 Here we insert for 1/Γ(γ + 1) the known integral representation where integration contour is runs around t = 0 in the direction of the positive real axis to infinity. The sum over ν can be carried out and yields One gets a symmetric form by the transformation this yields 5 The integration path can be seen from Fig. 2. The arguments of s ± i/2 go from 0 to 2π. By introducing ϕ into (14), we obtain the integral representations for the solutions 5 The integral representation of W. Gordon (Ann.d. Phys. (5) 2, 1031, 1929) is inconvenient for the following calculations. 5

6 One verifies easily, that these expressions satisfy the equations (12). The absolute sign of ξ is necessary, to have the symmetry of the functions f and g with respect to ξ = 0, implied by (14), and to ensure the continuity of the wave function at this point. 6 2 Expansion of f and g for large k For the physical application the knowledge of the function f and g is required. The sum expansion of (14) using (13) converges only for very small values of ξ and k, where we may consider only the first two terms in the sum. This development is unsuitable for larger values of ξ and k. Let us estimate the order of magnitude of k, which we can write as [see (6) and (9)] If the momentum components in the y- and z- direction are small compared to E 0 /c, then k depends only on the value of the potential rise. By using the numerical values and expressing v/e in Volt/cm one recognizes that even for the highest experimentally possible electrostatic fields of some millions of Volt/cm k is some orders of magnitude greater than 1. Only for practically never reachable extreme high fields of Volts/cm will k be comparably to 1. In the following we shall deal with the case k 1. To have a fast convergence, we expand f and g in powers of inverse powers of k which is possible using the Debye Saddle Point Method. The functions in (17) have the generic form 6 Note that ξ = 0 is a singular point in the integral representation (16a). 6

7 where G(s) changes slowly changing compared to the first two factors of the integrand. With the abbreviation F can be written as This integral is evaluated by seeking a saddle-point at which the integrand has the sharpest possible maximum, then changing the integration contour to run through this point, and expanding the integrand around it. The result is a quickly converging sum which one can truncate after the first term. The position of the saddle points are obtained by solving With these two points These lie, on the real or imaginary axis, depending on whether ξ is smaller or larger than k The points ξ = ±k play mathematically a special role. One verifies easily that they are also physically special. Due to (6), (8) and (9), they are positions satisfying the condition i.e. at these points disappears (classically speaking) the momentum component p x in the field direction. The points represent the turning points of the classic path in the regime ξ > k. The regime ξ < k is classically forbidden. For the calculation of (18a), we still must specify the contour of integration near the saddle points. As it is well known, the contour is placed with advantage 7

8 so that the real part of the exponent R(h) decreases as quickly as possible, meaning that the imaginary part of h(s) remains constant. By writing s = σ+iτ, this gives the condition and thus the equation We treat the two cases ξ < k and ξ > k separately: 1. ξ < k The saddle point is on the real axis. The integration paths are given by Its path is shown in Fig. 3. (The real axis σ = 0 is also a branch of the integration path) The arrows mean decreasment of the real part of h, that is increasing the integrand. A suitable integration path, which transforms steadily into the integration path of Fig. 2, can be recieved by going from + straightly to s 2, from there on the curve drawn in a positive manner over s 1 around the two branch points ±i/2 around and again from s 2 to +. (compare the dotted curve in Fig. 3, which is drawn for clarity besides the true integration paths. The straight parts of the integration path run on different Riemann sheets). The integrand takes the maximum value at the position 7 7 The prime at the root indicates that the positive value is to be used 8

9 the expansion of h(s) in this saddle point is Due to the fixing about the arguments of s ± i/2 made above one gets where Arc sin and Arc cos indicate the main values of the cyclical metric functions (between 0 and π/2). For the calculation of F one must expand besides e h(s) also G(s) for powers of s s1 and integrate by parts. We want to confine ourselves to the first sum member. (The sum expands with increasing powers of 1/(ξ 2 ), it therefore converges very quickly at sufficiently large ξ 2 > 1). G(s) can be factored in front of the integral as constant and it is valid The integration path can in the same approximation being replaced by its tangent in the saddle point. Then the integration is carried out from s1 + i straight to s1 i. Yielding This result is valid until the root in the denominator is greater or is comparable with 1 since in the development (20a) this root appears in the denominators. The case ξ k must be treated separately. Here the two saddle points come closer to the coordinate origin and one expands with advantage starting from ξ = k. Since we don t need this expansion in the following, it s derivation may be suppressed for brevity. The result is 9

10 valid for ξ 2 k ξ > k. The saddle points are on q the imaginary axis. The integration 2 path is yielded by σ1,2 = 0, τ1,2 = ±1/2 1 kξ2 (Fig. 4, the arrows have the same meaning as in Fig. 3) A useable integration path can by recieved in the following way: Coming from +, going around the branchpoint +i/2 in a positive manner, over the saddle point s1 and again to + back; then over the second saddle point s2 around the branchpoint i/2 again in a positive manner around to + (see the dotted curve in Fig. 4). This integration path can be transformed into this of Fig. 2 smoothly if the path s1 and s2 are on the same Riemann sheet. To get F one has to add the contributions of the integral in the surroundings of the two saddle points. (The upper sign refers to s1, the lower to s2 ). For h(s) one receives the expansion 10

11 The interation paths cross themselfs at the saddle points the imaginary axis by 45, we put in the surroundings of the first and second saddle point So in the considered approximation the integrations can be carried from t = to t = +. The result is This expansion is in powers of k12, therefore converges for arbitrary values of ξ sufficiently quickly, so that one can use the first approximation. It loses like in the first case its validity at the approach of ξ to k where we have to use the expansion (20b) in the limiting case. By putting (20a) and (20c) into (17), one gets 11

12 here w is the abbreviation 1 k2 ξ. The corresponding expansion for the second 2 solution system we receive from this due to the relations (15): f 2 = g 1, g 2 = f 1 3 Discussions of the solutions The functions f and g may be examined. Taken from (14), they are symmetric or antisymmetric with respect to the point ξ = 0. One of them is even the other one is odd. The functions are retrieved by the expansions (21) and (22). They are like a oscillation function with variable frequency and amplitude for ξ < k and ξ > +k, indicated schematically in Fig. 5. (The functions are fundamentally complex, Fig. 5 therefore only serves for a raw orientation). In the middle area its absolute amount decreases from the point ξ = k on approximated by a power of e until a value of 1 or 0 [compare (14)] and rises then exponentially again. It must be noticed that this exponential decreasement until the point ξ = 0 only is caused by the special choice of our solution systems. If we use a integration path where the second saddle point in Fig. 3 provides the essential contribution to the integrand, we would receive an exponential increasment in the functions to zero. Such an integration path is given e.g. by a loop coming like in Fig. 2 from positive real infinity going around only one of the two branchpoints. A suitable linear combination of our two solution systems must therefore give the mentioned increasement. The dotted curve in Fig. 5 shall indicate this possibility schematically. Comparing these relationships with those of the not relativistic wave mechanics: The Schroedinger equation for the one-dimensional problem is 8 : 8 In the following we set p y = p z = 0. Then due to (6) and (9) k = p x/ve 0 = q 2π hcv mc2 12

13 By the ansatz (E=non-relativistic energy) one gets, by introducing for x the new variable for χ the differential equation Its solution is if Z p (x) is an arbitrary solution of the Bessel differential equation The function χ is presented in Fig. 6 schematically: For negative values of ξ the argument of Z 1/3 is real, therefore the functions are periodical. The argument for positive ξ is complex, χ appears to be overlapping of one exponentially rising and a branch falling exponentially (for splitting of Z 1/3 into the two Hankel functions H (1) 1/3 and H(2) 1/3 ). To carry out the transition from the relativistic to the non-relativistic case we take into account, that E = Ē + E 0 ξ = ξ k The coordinate system of Fig. 5 is therefore moved to the left by k compared to Fig. 6, the point ξ = 0 of Fig. 6 corresponds to the point A(ξ = k) in Fig

14 For the transition to the non-relativistic case meaning the limit k, that is E 0 and k. By this transition the positions O and B of Fig. 5, having the abscisses k and 2k related to the non relativistic system with A as an origin, to positiv infinity. The right half of Fig. 5 therefore vanishs and the left half is streched from to + from which the function path of Fig. 5 apparently arises. While the area ξ < 0 is therefore corresponding to the scope of the Schroedinger Wave Functions the area ξ > 0 has no non-relativistic analogon. This area is indicated, as proved soon, by a wave mechanical vector of the momentum is contrary directional to the velocity. In the Theory of Dirac the three-dimensional velocity u v corresponds to the operator icγ ν (ν = 1, 2, 3). One gets the wave mechanical expectation value of this operator in the form Here Ψ one means the adjoint wave function, which suffices equation To get a relation between the size given by (23) [u x ] and the expectation value [p x ] of the momentum in the field direction, we multiply the equation (2) from the left with Ψγ 4 γ 1, equation (2a) from the right with γ 1 γ 4 Ψ and adding them; we get At infinitely sharply certain momentum in the y and z direction the elements with y and z disappear [the dependency of y and z given by (4) vanishes in the quadratic expression Ψγ 4 γ 1 γ 2,3 Ψ]. Since we look at a stationary state of the energy E, one can carry out the time differentiations and one gets 14

15 The second term can be written due to (23) For the interpretation of the first expression we take into account, that the momentum operator in the Theory of Dirac is given by h 2πi γ 4 xν ; the first term of (24) represents therefore the symmetrisized expectation value of the momentum [p x ], multiplied by the factor 2πi/h. (24) therefore can be written This equation therefore means, that for E > vx (or because of (8) for ξ < 0) the momentum and velocity have the same sign, while they are judged oppositely for E < vx (or ξ > 0) as it had to be proved. One takes into account by the way that (25) represents the wave mechanical translation of the relativistic connection between the momentum and the velocity because p x = mu x 1 u2 c 2 For the application in the next chapter the value of the velocity [u x ] in dependence of f and g will be calculated. Therefore we introduce for Ψ resulting from (4) and (11) the expression: it is seen easily, that the corresponding expression for Ψ is given by 15

16 By setting up the density Ψγ 4 Ψ one gets is a constant operator. In analogy one gets [u x ] is also spatially constantly, as one can prove due to (12). This already results from the divergence condition for the current S ν = e[u ν ], because we look at a stationary condition, where the current components are constant in the y and z direction (flat wave). One most simply calculates the value of (28) due to this constancy for the position ξ = 0. By combining the general functions Ψ as a linear combination from the partial functions Ψ 1 and Ψ 2, say one simply proves due to (14) since f 1 (0) = g 2 (0) = 1, f 2 (0) = g 1 (0) = 0. The currents resulting from the two partial solutions Ψ 1 and Ψ 2 combine without interference to the complete current in the x direction. 4 Transition of an electron through a contra directional field We turn to the problem specified in the introduction of calculating the probability for an electron going from the area of positive momentum to the area of negative momentum 9. For this purpose we look at the potential path represented into Fig. 1 by the line ABCD: 9 By positive and negative momentum it is understood, that the momentum and velocity are parallel resp. anti-parallel. In the first case the kinetic energy is positiv, in the latter case negativ 16

17 We must now solve the Equation of Dirac in the three cases separatly and then attach the three wave functions to each other in the points x 1 and x 2 steadily. The integration can be carried out in all three cases according to the same method, then the connection is made easier. We make the ansatz where γ 5 is given by (6) and (7). By introducing a new variable as above one obtains for f and g, exactly as in the first section, the equations k is defined by (9), ξ 1 and ξ 2 result from (8) by the introduction of the values x 1 and x 2 instead of x. For the solutions for the second equation pair (33) we write using of the two constants α and β in the form (29a), in which f 1, g 1 and f 2, g 2 are given by (14). For the first and third area one gets of course flat waves of the form e iq1,2ξ in which q 1 and q 2 are given by the equations 10 One calculates (33). For the coefficient ratio the solutions arise with the arbitrary constants a ν and b ν v = 1 apply to the area I, v = 2 for area III. We receive a steady connection of the wave functions in x 1 and x 2, resp. ξ 1 and ξ 2 if we attach the functions f and g to each other one by one. This leads to the equations 10 q ν depends with the usual momentum p ν through the relation q ν = p x/vcp ν. q ν is real, due to ξ ν > k and we set q ν > 0. 17

18 Here we set due to (14) and use (15). The system of equations (36) for the regulation of the six values a ν, b ν, α, β is still undetermined; to the mathematical continuity terms a physical condition has to be added which fixes the behavior of the wave function in the infinity. The probability that an electron breaks from area I with a positive momentum to area III with a negative momentum is to be calculated. So we must consider an incident particle current from the left, which partly goes through the area II and enters into area III (from left to right), partly reflected at the separation areas at x 1 and x 2 and representing a back-moving current in the area I. The solution must be chosen in a manner, that only a current runs from left to right in III in the direction and no-one in the reversed direction. The current component in the x-direction for a wave function represented by the expression (32) was found (equation (28)) in the third section. Applied to the represented by (35) flat wave yields the particle density is given (to 27). For ν = 1 we get because of ξ 1 < q 1 an incoming current of the strength and a reflected current for ν = 2 due to ξ 2 > q 2 18

19 the desired outgoing current, during the incoming current must vanish in this area. One obtains therefor for the physical condition a2 = 0. The transmission coefficient is to be calculated to The refelction coefficient is given resp. to where R + D = 1. By setting a2 = 0 into (36b), the dissolving of these two equations for α and β. F2 F2 G2 G 2 = 1 See (37) and (13) By going with these two expression into (36a) and solving for a1 and b1, one gets For simplicity we assume symmetrical relations to the point ξ = 0: This results in 19

20 Now we use the in section 2 given expansion (22) for F0 and G0 ; 11 Valid is The result is Due to (38) and (39) one gets the transmission and reflection coefficient 11 By the use of the expansions (22), which only applies to case that ξ is not too near to k, for ξ k one has to applicate sums of the form (20b), the validity of the result is restricted to the cases of high velocity of the electron running in and out. For small velocitys the problem is less interesting, since Klein s paradoxon disappears due to (45) or (46). 20

21 This result is correct except for higher order members in and was derived under the prerequisite k 2 >> 1. So it was shown, that for all electrical fields with k 2 >> 1, practically for all productable fields, the transmission coefficient is insignificantly small so that practically all electrons are reflected; Transitions into the area of a negative momentum are therefor rare. 12 The value of the transmission coefficient D depends due to (41) in the case of high electron velocity and symmetrical potential in first approximation only on the field strength, therefore only on the rising of the potential. For D one gets only then finite values if k 2 becomes of the order of magnitude 1. In this case the field strengths correspond to Volts/cm. The position k 2 1 has its special physical meaning; in this case l k 2 This is with the assumption of N. Bohr indicated in the introduction in agreement that only then there are finite probabilities for the transition of an electron to the area with a negative momentum if the potential increasment v h mc on a distance of a Compton Wavelength h mc is of the range of the rest energy. Of course it is impossible to produce fields of this strength experimentally. Perhaps one could remember, however, that such fields can inside an atom appear under certain circumstances; this one would get approximated in a pure Coulomb Field by a critical position given (by 43) distance from the strength center; also after the Gamow Model of the nucleus like such high fields should exist in the atom. It would be inappropriately, to draw any conclusions on the behavior of an electron too close to the nucleus, because the Theory of Dirac looses its validity in the surrounding of the nucleus. We want to complete the picture by looking at the case treated by 0. Klein of a potential step, that is the limiting case at infinitely high potential rise. The calculation can then be carried out, that the wave functions valid for the areas I and III are be connected directly to each other. To come to Klein s names, one must set 12 This result is independent of the above acceptance of symmetrical conditions. In general the final formulae become very confusedly. 21

22 where P gives the height of the potential step, E the energy before the transition through the opposite field and p and p are the momentums before and after the transition. One then gets after simple calculation, which shall not be carried out of brevity: One gets the formulae given in the introduction from this for the limiting case of a very high potential jump (P ) To get the analogon to the case of symmetrical potential course treated above one must set ξ 1 = ξ 2 and therefore P = 2E and gets with that It is important to me to thank Professor A. Sommerfeld for his interest in the execution of these examinations very well. Furthermore my gratitude is due to the Austrian German science help for the granting of a Stipendium. 22