THE PHOTON'S WAVEFUNCTION

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We know now that:F is a six-vector and that T = U t . What about A As A must transform in the same way as 0, it is a four-vector. This leaves us, in principle, with eight real quantities to determine in A, as each of its four components is complex. Let us now make a bold and simplifying assumption. Let us assume that A is antiHermitian (i.e., At = -A). This reduces the number of real quantities in A by half: from eight, we have only four real functions to determine now. This bold assumption is consistent with relativity, as the property of a four-vector of being Hermitian or anti-Hermitian does not change from one Lorentz frame to another. But the ultimate test will come in the end, when we will see that such assumption does not lead to any inconsistency. To sum up our discussion so far: We have a double biquatemion wavefunction candidate, consisting of the six-vector:F and the four-vector A. These biquatemions obey Lanczos equations (3.42) and (3.43) and have the following additional properties: (i.e., :F is a pure biquatemion or vector), (i.e., A is anti-Hermitian).

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Another interesting feature of (3.42) and (3.43) is that like the Klein-Gordon equation and the Dirac equation, they connect n to the rest mass through the single parameter JL. In the nonrelativistic SchrOdinger equation, this is not so, as n appears linearly in the time derivative term and quadratically in the spatial derivative terms. Relativity is revealing a deep secret of nature here: If the rest mass vanishes, the quantum-mechanical wave equation will not contain n. This means that for vanishing rest mass, when we take the limit of n, ---+ 0, the same wave equation remains valid in the realm of classical physics. We will see later that when we take JL ---+ 0, (3.42) and (3.43) coincide with the Maxwell equations, which are both the quantum and the classical equations for a photon (if it is, indeed, massless). Now before we can call this a proper quantum wave equation, we must show that it leads to a probability density and that the probability is conserved in time. To do so, we must find a continuity equation relating the probability density to a probability current density. It is this continuity equation that will allow us to identify what is the probability density. Moreover, the probability density must be positive definite and the probability interpretation must not depend on the choice of Lorentz frames. For the latter to hold, the probability density and the probability current density must be, respectively, the time and spatial components of a four-vector. This can be seen by noticing that if the Hermitian quatemion Z = D + L~=l iniCn,

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the transformation because U* zoU t = zoU* U t = zoo Moreover. the pure biquatemion part of Z (i.e.. Z1 h + z2h + z3i3) transforms them into another pure biquatemion. An easy way to see this is to take the square of any transformed quatemion-imaginary unit. The square of the transformed quatemion-imaginary unit i 1. for example. is given by

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(U*i1U t )2 = U*h UtU* i1Ut = U*ili1 2Ut = - U*U t "-v-' "-v-'

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= -1.

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So. as its square is -1. U*i1 ut cannot be 1 but rather. must be one of the quatemion-imaginary units it. h. or h. The same holds for the transformation of the other two quatemion-imaginary units. hand h.

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EXTREME QUANTUM THEORY OF LIGHT WITH A TWIST

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where D, Ct. C 2 , and C 3 are real, is a four-vector, then 1)2 is a six-vector (i.e., the scalar component of 1) 2 is the same in all Lorentz frames). This scalar component is just i {(1/c)8D/8t + \1. C}, where C is the vector formed by ct. c 2 , and C3 . So if S (1)2) = 0, the time and spatial components of the Hermitian four-vector 2 will satisfy the continuity equation (1/c)8D/at + \1 . C = 0 in every Lorentz frame. If we are looking for something bilinear in the wavefunction that forms a fourvector, as ;:: is a six-vector and A is a four-vector, the only combinations that would obviously lead to a four-vector are At;::, n, and their Hermitian conjugates. Unfortunately, none of them leads to a positive-definite time component. 24 So we must look for something else. Consider the biquaternion

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As A and 0 are four-vectors and;:: is a six-vector, U is a four-vector. Moreover, as nO is the quaternionic four-momentum operator, U has dimensions of momentum in units of n. From (3.42) and (3.43), we find that

;::to;:: = (;::to);:: +;::t (0;::)

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