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Previous relativity treatments

The relativity Hamiltonian according to the classical theory for a point electron moving in an arbitrary electro-magnetic field with scalar potential $A_{0}$ and vector potential $\boldsymbol{A}$ is $$F\equiv\left(\frac{W}{c}+\frac{e}{c}A_{0}\right)^{2}+\left(\boldsymbol{p}+\frac{e}{c}\boldsymbol{A}\right)^{2}+m^{2}c^{2}\text{,}$$ where $\boldsymbol{p}$ is the momentum vector. It has been suggested by Gordon that the operator of the wave equation of the quantum theory should be obtained from this $F$ bu the same procedure as in non-relativity theory, namely, by putting $$\begin{aligned} & W=ih\frac{\partial}{\partial t},\\ & p_{r}=-i\hbar\frac{\partial}{\partial x_{r}}, & & r=1,2,3, \end{aligned}$$ in it. This gives the wave equation $$F\psi=\left[\left(ih\frac{\partial}{c\partial t}+\frac{e}{c}A_{0}\right)^{2}+\sum_{r}\left(-ih\frac{\partial}{\partial x_{r}}+\frac{e}{c}A_{r}\right)^{2}+m^{2}c^{2}\right]\psi=0,$$ the wave function \psi being a function of $x_{1},x_{2},x_{3},t$. This gives rise to two difficulties.

The first is in connection with the physical interpretation of $\psi$. Gordon, and also independently Klein, from considerations of the conservation theorems, make the assumption that if $\psi_{m}$, $\psi_{n}$ are two solutions $$\rho_{mn}=-\frac{e}{2mc^{2}}\left\{ ih\left(\psi_{m}\frac{\partial\psi_{n}}{\partial t}-\bar{\psi}_{n}\frac{\partial\psi_{m}}{\partial t}\right)+2eA_{0}\psi_{m}\bar{\psi}_{n}\right\} $$ and $$\boldsymbol{I}_{mn}=-\frac{e}{2m}\left\{ -ih\left(\psi_{m}\GRADIENT\bar{\psi}_{n}-\bar{\psi}_{n}\GRADIENT\psi_{m}\right)+2\frac{e}{c}\boldsymbol{A}_{m}\psi_{m}\bar{\psi}_{n}\right\} $$ are to be interpreted as the charge and current associated with the transition $m\to n$. This appears to be satisfactory so far as the emission and absorption of radiation are concerned, but is not so general as the interpretation of the non-relativity quantum mechanics, which has been developed sufficiently to enable one to answer the question: What is the probability of any dynamical variable at any specified time having a value lying between any specified limits, when the system is represented by a given wave function $\psi_{n}$? The Gordon-Klein interpretation can answer such questions if they refer to the position of the electron (by the use of $\rho_{nm}$), but not if they refer to its momentum, or angular momentum or any other dynamical variable. We should expect the interpretation of the relativity theory to be just as general as that of the non-relativity theory.

The general interpretation of non-relativity quantum mechanics is based on the transformation theory, and is made possible by the wave equation being of the form $$\left(H-W\right)\psi=0\text{,}$$ i.e., being linear in $W$ or $\partial/\partial t$, so that the wave function at any time determines the wave function at any later time. The wave equation of the relativity theory must also be linear in $W$ if the linear interpretation is to be possible.

The second difficulty in Gordon's interpretation arises from the fact that if one takes the conjugate imaginary of equation , one gets $$\left[\left(-\frac{W}{c}+\frac{e}{c}A_{0}\right)^{2}+\left(-\boldsymbol{p}+\frac{e}{c}\boldsymbol{A}\right)^{2}+m^{2}c^{2}\right]\psi=0,$$ which is the same as one would get if on put $-e$ for $e$. The wave equation thus refers equally well to an electron with charge $e$ as to one with charge $-e$. If one considers for definiteness the limiting case of large quantum numbers one would find that some of the solutions of the wave equation are wave packets moving in the way a particle of charge $-e$ would move on the classical theory, while others are wave packets moving in the way a particle of charge $e$ would move classically. For this second class of solutions $W$ has a negative value. One gets over the difficulty on the classical theory by arbitrarily excluding those solutions that have a negative $W$. One cannot do this on the quantum theory, since in general a perturbation will cause transitions from states with $W$ positive to states with $W$ negative. Such a translation would appear experimentally as the electron suddenly changing its charge sign from $-e$ to $e$, a phenomenom which has not been observed. The true relativity wave equation should thus be such that its solutions split up into two non-combining sets, referring respectively to the charge $-e$ and the charge $e$.

In the present paper we shall be concerned only with the removal of the first of these two difficulties. The resulting theory is therefore still only an approximation, but it appears to be good enough to account for all the duplexity phenomena without arbitrary assumptions.

The hamiltonian for no field

Our problem is to obtain a wave equation of the form which shall be invariant under a Lorentz transformation and shall be equivalent to in the limit of large quantum numbers. We shall consider first the case of no field, when reduces to $$\left(-p_{0}^{2}+\boldsymbol{p}^{2}+m^{2}c^{2}\right)\psi=0$$ if one puts $$p_{0}=\frac{W}{c}=ih\frac{\partial}{c\partial t}\text{.}$$

The symmetry between $p_{0}$ and $p_{1},p_{2},p_{3}$ required by relativity shows that, since the Hamiltonian we want is linear in $p_{0}$, it must also be linear in $p_{1},p_{2},p_{3}$. Our wave equation is therefore of the form $$\left(p_{0}+\alpha_{1}p_{1}+\alpha_{2}p_{2}+\alpha_{3}p_{3}+\beta\right)\psi=0,$$ where for the present all that is known about the dynamical variables or operators $\alpha_{1},\alpha_{2},\alpha_{3},\beta$ is that they are independent of $p_{0},p_{1},p_{2},p_{3}$, i.e., that they commute with $t,x_{1},x_{2},x_{3}$. Since we are considering the case of a particle moving in empty space, so that all points in space are equivalent, we should expect the Hamiltonian not to involve $t,x_{1},x_{2},x_{3}$. This means that $\alpha_{1},\alpha_{2},\alpha_{3},\beta$ are independent of $t,x_{1},x_{2},x_{3}$, i.e., that they commute with $p_{0},p_{1},p_{2},p_{3}$. We are therefore obliged to have other dynamical variables besides the co-ordinates and momenta of the electron, in order that $\alpha_{1},\alpha_{2},\alpha_{3},\beta$ may be functions of them. The wave function \psi must then involve more variables than merely $t,x_{1},x_{2},x_{3}$.

leads to $$\begin{aligned}0 & =\left(-p_{0}+\alpha_{1}p_{1}+\alpha_{2}p_{2}+\alpha_{3}p_{3}+\beta\right)\left(p_{0}+\alpha_{1}p_{1}+\alpha_{2}p_{2}+\alpha_{3}p_{3}+\beta\right)\psi\\ & =\left[-p_{0}^{2}+\sum\alpha_{1}^{2}p_{1}^{2}+\sum\left(\alpha_{1}\alpha_{2}+\alpha_{2}\alpha_{1}\right)p_{1}p_{2}+\beta^{2}+\sum\left(\alpha_{1}\beta+\beta\alpha_{1}\right)p_{1}\right]\psi \end{aligned} $$ where the $\sum$ refers to cyclic permutation of the suffices 1,2,3. This agrees with if $$\left.\begin{aligned} & \alpha_{r}^{2}=1, & & \alpha_{r}\alpha_{s}+\alpha_{s}\alpha_{r}=0\;\left(r\ne s\right)\\ & \beta^{2}=m^{2}c^{2}, & & \alpha_{r}\beta+\beta\alpha_{r}=0 \end{aligned} \right\} \ \ r,s=1,2,3\text{.}$$ If we put $\beta=\alpha_{4}mc$, these conditions become $$\alpha_{\mu}^{2}=1\ \ \ \ \alpha_{\mu}\alpha_{\nu}+\alpha_{\nu}\alpha_{\mu}=0\ (\mu\ne\nu)\ \ \ \ \mu,\nu=1,2,3,4\text{.}$$

We can suppose the $\alpha_{\mu}$'s to be expressed as matrices in some matrix scheme, the matrix elements of $\alpha_{\mu}$ being, say, $\alpha_{\mu}\left(\zeta^{\prime}\ \zeta^{\prime\prime}\right)$. The wave function $\psi$ must now be a function of $\zeta$ as well as $x_{1},x_{2},x_{3},t$. The result of $\alpha_{\mu}$ multiplied into $\psi$ will be a function $\left(\alpha_{\mu}\psi\right)$ of $x_{1},x_{2},x_{3},t,\zeta$ defined by $$\left(\alpha_{\mu}\psi\right)\left(x,t,\zeta\right)=\sum_{\zeta^{\prime}\ \alpha_{\mu}}\alpha_{\mu}\left(\zeta\ \zeta^{\prime}\right)\psi\left(x,t,\zeta^{\prime}\right)\text{.} $$

We must now find four matrices $\alpha_{\mu}$ to satisfy the conditions . We make use of the matrices $$\sigma_{1}=\left(\begin{matrix}0 & 1\\ 1 & 0 \end{matrix}\right)\ \ \ \ \sigma_{2}=\left(\begin{matrix}0 & -i\\ i & 0 \end{matrix}\right)\ \ \ \ \sigma_{3}=\left(\begin{matrix}1 & 0\\ 0 & -1 \end{matrix}\right) $$ which Pauli introduced to describethe three components of spin angular momentum. These matrices have just the properties $$\sigma_{r}^{2}=1\ \ \ \ \sigma_{r}\sigma_{s}+\sigma_{s}\sigma_{r}=0,\ \ \ \ \left(r\ne s\right),$$ that we require for our $\alpha$'s. We cannot, however, just take the $\sigma$'s so be three of our $\alpha$'s, because then it would not be possible to find the fourth. We must extend the $\sigma$'s in a diagonal manner to bring in two more rows and columns, so that we can introduce three more matrices $\rho_{1},\rho_{2},\rho_{3}$ of the same form as $\sigma_{1},\sigma_{2},\sigma_{3}$ but referring to different rows and columns, thus $$\begin{aligned} & \sigma_{1}=\left\{ \begin{matrix}0 & 1 & 0 & 0\\ 1 & 0 & 0 & 0\\ 0 & 0 & 0 & 1\\ 0 & 0 & 1 & 0 \end{matrix}\right\} & & \sigma_{2}=\left\{ \begin{matrix}0 & -i & 0 & 0\\ i & 0 & 0 & 0\\ 0 & 0 & 0 & -i\\ 0 & 0 & i & 0 \end{matrix}\right\} & & \sigma_{3}=\left\{ \begin{matrix}1 & 0 & 0 & 0\\ 0 & -1 & 0 & 0\\ 0 & 0 & 1 & 0\\ 0 & 0 & 0 & -1 \end{matrix}\right\} \\ & \rho_{1}=\left\{ \begin{matrix}0 & 0 & 1 & 0\\ 0 & 0 & 0 & 1\\ 1 & 0 & 0 & 0\\ 0 & 1 & 0 & 0 \end{matrix}\right\} & & \rho_{2}=\left\{ \begin{matrix}0 & 0 & -i & 0\\ 0 & 0 & 0 & -i\\ i & 0 & 0 & 0\\ 0 & i & 0 & 0 \end{matrix}\right\} & & \rho_{3}=\left\{ \begin{matrix}1 & 0 & 0 & 0\\ 0 & 1 & 0 & 0\\ 0 & 0 & -1 & 0\\ 0 & 0 & 0 & -1 \end{matrix}\right\} \text{.} \end{aligned} $$ The $\rho$'s are obtained from the $\sigma$'s by interchanging the second and third rows, and the second and third columns. We now have, in addition to $$\left.\begin{aligned} & \rho_{r}^{2}=1 & & \rho_{r}\rho_{s}+\rho_{s}\rho_{r}=0 & & \left(r\ne s\right)\\ & & & \rho_{r}\sigma_{r}=\sigma_{t}\rho_{r}\text{.} \end{aligned} \right\} $$

If we now take $$\alpha_{1}=\rho_{1}\sigma_{1},\ \ \ \ \alpha_{2}=\rho_{1}\sigma_{2},\ \ \ \ \alpha=\rho_{1}\sigma_{3},\ \ \ \ \alpha_{4}=\rho_{3,}$$ all the conditions are satisfied, e.g., $$\begin{aligned}\alpha_{1}^{2}= & \rho_{1}\sigma_{1}\rho_{1}\sigma_{1}=\rho_{1}^{2}\sigma_{1}^{2}=1\\ \alpha_{1}\alpha_{2}= & \rho_{1}\sigma_{1}\rho_{1}\sigma_{2}=\rho_{1}^{2}\sigma_{1}\sigma_{1}=-\rho_{1}^{2}\sigma_{2}\sigma_{1}=-\alpha_{2}\alpha_{1}\text{.} \end{aligned}$$ The following equations are to be noted for later reference $$\left.\begin{aligned}\rho_{1}\rho_{2}=i\rho_{3}=-\rho_{2}\rho_{1}\\ \sigma_{1}\sigma_{2}=i\sigma_{3}=--\sigma_{2}\sigma_{1} \end{aligned} \right\} \text{,}$$ together with the equations obtained by cyclic permutation of the suffixes.

The wave equation now takes the form $$\left[p_{0}+\rho_{1}\left(\boldsymbol{\sigma},\boldsymbol{p}\right)+\rho_{3}mc\right]\psi=0,$$ where $\boldsymbol{\sigma}$ denotes the vector $\left(\sigma_{1},\sigma_{2},\sigma_{3}\right)$.

Proof of invariance under a Lorentz transformation

Multiply by $\rho_{3}$ on the left-hand side. It becomes, with the help of , \[ \left[\rho_{3}p_{0}+i\rho_{2}\left(\sigma_{1}p_{1}+\sigma_{2}p_{2}+\sigma_{3}p_{3}\right)+mc\right]\psi=0\text{.} \] Puting \[ p_{0}=ip_{4}, \] \[ \rho_{3}=\gamma_{4},\ \ \ \ \rho_{2}\sigma_{r}=\gamma_{r},\ \ \ \ r=1,2,3, \] we have \[ \left[i\sum\gamma_{\mu}p_{\mu}+mc\right]\psi=0,\ \ \ \ \mu=1,2,3,4. \] The $p_{\mu}$ transform under a Lorentz transformation according to the law \[ p_{\mu}^{\prime}=\sum_{\nu}a_{\mu\nu}p_{\nu}, \] where the coefficients $a_{\mu\nu}$ are c-numbers satisfying \[ \sum_{\mu}a_{\mu\nu}a_{\mu\tau}=\delta_{\nu\tau},\ \ \ \ \sum_{\tau}a_{\mu\tau}a_{\nu\tau}=\delta_{\mu\nu}\text{.} \] The wave equation therefore transforms into \[ \left[i\sum\gamma_{\mu}^{\prime}p_{\mu}^{\prime}+mc\right]\psi=0, \] where \[ \gamma_{\mu}^{\prime}=\sum_{\nu}a_{\mu\nu}\gamma_{\nu}\text{.} \]

Now the $\gamma_{\mu}$, like the $\alpha_{\mu}$, satisfy $$\gamma_{\mu}^{2}=1,\ \ \ \ \gamma_{\mu}\gamma_{\nu}+\gamma_{\nu}\gamma_{\mu}=0,\ \ \ \ \left(\mu\ne\nu\right)\text{.}$$ These relations can be summed up in the single equation $$\gamma_{\mu}\gamma_{\nu}+\gamma_{\mu}\gamma_{\nu}=2\delta_{\mu\nu}\text{.}$$ We have $$\begin{aligned}\gamma_{\mu}^{\prime}\gamma_{\nu}^{\prime}+\gamma_{\nu}^{\prime}\gamma_{\mu}^{\prime} & =\sum_{\tau\lambda}a_{\mu\tau}a_{\nu\lambda}\left(\gamma_{\tau}\gamma_{\lambda}+\gamma_{\lambda}\gamma_{\tau}\right)\\ & =2\sum_{\tau\lambda}a_{\mu\tau}a_{\nu\lambda}\delta_{\tau\lambda}\\ & =2\sum_{\tau}a_{\mu\tau}a_{\nu\tau}=2\delta_{\mu\nu}\text{.} \end{aligned}$$ Thus the $ \gamma_{\mu}^{\prime}$ satisfy the same relations as the $\gamma_{\mu}$. Thus we can put, analogously to $$\begin{aligned}\gamma_{4}^{\prime}=\rho_{3}^{\prime} & & \gamma_{r}^{\prime}=\rho_{2}^{\prime}\sigma_{r}^{\prime}\end{aligned}$$ where the $\rho^{\prime}$'s and $\sigma^{\prime}$'s are easily verified to satisfy the relations corresponding to , and , if $\rho_{2}^{\prime}$ and $\rho_{1}^{\prime}$ are defined by $\rho_{2}^{\prime}=-i\gamma_{1}^{\prime}\gamma_{2}^{\prime}\gamma_{3}^{\prime}$, $\rho_{1}^{\prime}=-i\rho_{2}^{\prime}\rho_{3}^{\prime}$.

We shall now show that, by a canonical transformation, the $\rho^{\prime}$'s and $\sigma^{\prime}$'s may be brought into the form of the $\rho$'s and $\sigma$'s. From the equation $\rho_{3}^{\prime}=1$, it follows that the only possible characteristic values for $\rho_{3}^{\prime}$ are $\pm1$. If one applies to $\rho_{3}^{\prime}$ a canonical transofrmation with the transformation function $\rho_{1}^{\prime}$, the result is $$\rho_{1}^{\prime}\rho_{3}^{\prime}\left(\rho_{1}^{\prime}\right)^{-1}=-\rho_{3}^{\prime}\rho_{1}^{\prime}\left(\rho_{1}^{\prime}\right)^{-1}=-\rho_{3}^{\prime}\text{.}$$ Since characteristic values are not changed by a canonical transformation, $\rho_{3}^{\prime}$ must have the same characteristic values as $-\rho_{3}^{\prime}$. Hence the characteristic values of $\rho_{3}^{\prime}$ are $+1$ twice and $-1$ twice. The same argument applies to each of the other $\rho^{\prime}$'s, and to each $\sigma^{\prime}$'s.

Since $\rho_{3}^{\prime}$ and $\sigma_{3}^{\prime}$ commute, they can be brought simultaneously to the diagonal form by a canonical transformation. They will then have for their diagonal elements each $+1$ twice and $-1$ twice. Thus, by suitably rearranging the rows and columns, they can be brought into the form $\rho_{3}$ and $\sigma_{3}$ respectively. (The possibility $\rho_{3}^{\prime}=\pm\sigma_{3}^{\prime}$ is excluded by the existence of matrices that commute with one but not with the other.)

Any matrix containing four rows and columns can be expressed as $$c+\sum_{r}c_{r}\sigma_{r}+\sum_{r}c_{r}^{\prime}\rho_{r}+\sum_{rs}c_{rs}\rho_{r}\sigma_{s}$$ where the sixteen coefficients $c$, $c_{r}$, $c_{r}^{\prime}$, $c_{rs}$ are c-numbers. By expressing $\sigma_{1}^{\prime}$ in this way, we see, from the fact that it commutes with $\rho_{3}^{\prime}=\rho_{3}$ and anticommutesWe say that $a$ anticommutes with $b$ when $ab=-ba$. with $\sigma_{3}^{\prime}=\sigma_{3}$, that it must be of the form $$\sigma_{1}^{\prime}=c_{1}\sigma_{1}+c_{2}\sigma_{2}+c_{31}\rho_{3}\sigma_{1}+c_{32}\rho_{3}\sigma_{2},$$ i.e., of the form $$\sigma_{1}^{\prime}=\left\{ \begin{matrix}0 & a_{12} & 0 & 0\\ a_{21} & 0 & 0 & 0\\ 0 & 0 & 0 & a_{34}\\ 0 & 0 & a_{43} & 0 \end{matrix}\right\} $$ The condition $\sigma_{1}^{\prime2}=1$ shows that $a_{12}a_{21}=1$, $a_{34}a_{43}=1$. If we now apply the canonical transformation: first row to be multiplied by $\left(a_{21}/a_{12}\right)^{\frac{1}{2}}$ and third row to be multiplied by $\left(a_{43}/a_{34}\right)^{\frac{1}{2}}$, and first and third columns to be divided by the same expressions, $\sigma_{1}^{\prime}$ will be brought into the form of $\sigma_{1}$, and the diagonal matrices $\sigma_{3}^{\prime}$ and $\rho_{3}^{\prime}$ will not be changed.

If we now express $\rho_{1}^{\prime}$ in the form and use the conditions that it commutes with $\sigma_{1}^{\prime}=\sigma_{1}$ and $\sigma_{3}^{\prime}=\sigma_{3}$ and anticommutes with $\rho_{3}^{\prime}=\rho_{3}$, we see that it must be of the form $$\rho_{1}^{\prime}=c_{1}^{\prime}\rho_{1}+c_{2}^{\prime}\rho_{2}\text{.}$$ The condition $\rho_{1}^{\prime2}=1$ shows that $c_{1}^{\prime2}+c_{2}^{\prime2}=1$, or $c_{1}^{\prime}=\cos\theta$, $c_{2}^{\prime}=\sin\theta$. Hence $\rho_{1}^{\prime}$ is of the form $$\rho_{1}^{\prime}=\left\{ \begin{matrix}0 & 0 & e^{-i\theta} & 0\\ 0 & 0 & 0 & e^{-i\theta}\\ e^{i\theta} & 0 & 0 & 0\\ 0 & e^{i\theta} & 0 & 0 \end{matrix}\right\} $$ If we now apply the canonical transformation: first and second rows to be multiplied by $e^{i\theta}$ and first and second columns to be divided by the same expression, $\rho_{1}^{\prime}$ will be brought into the form $\rho_{1}$, and $\sigma_{1},\sigma_{3},\rho_{3}$ will not be altered. $\rho_{2}^{\prime}$ and $\sigma_{2}^{\prime}$ must now be of the form $\rho_{2}$ and $\sigma_{2}$, on account of the relations $i\rho_{2}^{\prime}=\rho_{3}^{\prime}\rho_{1}^{\prime},i\sigma_{2}^{\prime}=\sigma_{3}^{\prime}\sigma_{1}^{\prime}$.

Thus by a succession of canonical transformations, which can be combined to form a single canonical transformation, the $\rho^{\prime}$'s and $\sigma^{\prime}$'s can be brought into the form of the $\rho^{\prime}$'s and $\sigma^{\prime}$'s. The new wave equation can in this way be brought back into the form of the original wave equation or , so that the results that follow from this original wave equation must be independent of the frame of reference used.

The Hamiltonian for an arbitrary field

To obtain the Hamiltonian for an electron in an electromagnetic field with scalar potential $A_{0}$ and vector potential $\boldsymbol{A}_{0}$, we adopt the usual procedure of substituting $p_{0}+e/cA_{0}$ for $p_{0}$ and $\boldsymbol{p}+e/c\boldsymbol{A}$ for $\boldsymbol{p}$ in the Hamiltonian for no field. From we thus obtain $$\left[p_{0}+\frac{e}{c}A_{0}+\rho_{1}\left(\boldsymbol{\sigma},\boldsymbol{p}+\frac{e}{c}\boldsymbol{A}\right)+\rho_{3}mc\right]\psi=0\text{.}$$ This wave equation appears to be sufficient to account for all the duplexity phenomena. On account of the matrices $\rho$ and $\sigma$ containing four rows and columns, it will have four times as many solutions as the non-relativity wave equation, and twice as many as the previous relativity wave equation . Since half the solutions must be rejected as referring to the charge $+e$ on the electron, the correct number will be left to account for duplexity phenomena. The proof given in the preceding section of invariance under a Lorentz transformation applices equally well to the more general wave equation .

We can obtain a rough idea of how differs from the previous relativity wave equation by multiplying it up analogously to . This gives, if we write $e^{\prime}$ for $e/c$ $$\begin{aligned}0 & =\left[-\left(p_{0}+e^{\prime}A_{0}\right)+\rho_{1}\left(\boldsymbol{\sigma},\boldsymbol{p}+e^{\prime}\boldsymbol{A}\right)+\rho_{3}mc\right]\times\left[\left(p_{0}+e^{\prime}A_{0}\right)+\rho_{1}\left(\boldsymbol{\sigma},\boldsymbol{p}+e^{\prime}\boldsymbol{A}\right)+\rho_{3}mc\right]\psi\\ & =\left[-\left(p_{0}+e^{\prime}A_{0}\right)^{2}+\left(\boldsymbol{\sigma},\boldsymbol{p}+e^{\prime}\boldsymbol{A}\right)^{2}+m^{2}c^{2}+\rho_{1}\left\{ \left(\boldsymbol{\sigma},\boldsymbol{p}+e^{\prime}\boldsymbol{A}\right)\left(p_{0}+e^{\prime}A_{0}\right)-\left(p_{0}+e^{\prime}A_{0}\right)\left(\boldsymbol{\sigma},\boldsymbol{p}+e^{\prime}\boldsymbol{A}\right)\right\} \right]\psi\text{.} \end{aligned} $$ We now use the general formula, that if $\boldsymbol{B}$ and $\boldsymbol{C}$ are any two vectors that commute with $\boldsymbol{\sigma}$ $$\begin{aligned}\left(\boldsymbol{\sigma},\boldsymbol{B}\right)\left(\boldsymbol{\sigma},\boldsymbol{C}\right)= & \sum\sigma_{1}^{2}B_{1}C_{1}+\sum(\sigma_{1}\sigma_{2}B_{1}C_{2}+\sigma_{2}\sigma_{1}B_{2}C_{1}\\ = & \left(\boldsymbol{B},\boldsymbol{C}\right)+i\sum\sigma_{3}\left(B_{1}C_{2}-B_{2}C_{1}\right)\\ = & \left(\boldsymbol{B},\boldsymbol{C}\right)+i\left(\boldsymbol{\sigma},\boldsymbol{B}\times\boldsymbol{C}\right)\text{.} \end{aligned} $$ Taking $\boldsymbol{B}=\boldsymbol{C}=\boldsymbol{p}+e^{\prime}\boldsymbol{A}$, we find $$\begin{aligned}\left(\boldsymbol{\sigma},\boldsymbol{p}+e^{\prime}\boldsymbol{A}\right)^{2}= & \left(\boldsymbol{p}+e^{\prime}\boldsymbol{A}\right)^{2}+i\sum\sigma_{3}\left[\left(p_{1}+e^{\prime}A_{1}\right)\left(p_{2}+e^{\prime}A_{2}\right)-\left(p_{2}+e^{\prime}A_{2}\right)\left(p_{1}+e^{\prime}A_{1}\right)\right]\\ = & \left(\boldsymbol{p}+e^{\prime}\boldsymbol{A}\right)^{2}+he^{\prime}\left(\boldsymbol{\sigma},\CURL\boldsymbol{A}\right)\text{.} \end{aligned} $$ Thus becomes $$\begin{aligned}0= & \left[-\left(p_{0}+e^{\prime}A_{0}\right)^{2}+\left(\boldsymbol{p}+e^{\prime}\boldsymbol{A}\right)^{2}+m^{2}c^{2}+e^{\prime}h\left(\boldsymbol{\sigma},\CURL\boldsymbol{A}\right)-ie^{\prime}h\rho_{1}\left(\boldsymbol{\sigma},\GRADIENT A_{0}+\frac{1}{c}\frac{\partial\boldsymbol{A}}{\partial t}\right)\right]\psi\\ = & \left[-\left(p_{0}+e^{\prime}A_{0}\right)^{2}+\left(\boldsymbol{p}+e^{\prime}\boldsymbol{A}\right)^{2}+m^{2}c^{2}+e^{\prime}h\left(\boldsymbol{\sigma},\boldsymbol{H}\right)+ie^{\prime}h\rho_{1}\left(\boldsymbol{\sigma},\boldsymbol{E}\right)\right]\psi\text{, } \end{aligned} $$ where $\boldsymbol{E}$ and $\boldsymbol{H}$ are the electric and magnetic vectors of the field.

This differs from by the two extra terms $$\frac{eh}{c}\left(\boldsymbol{\sigma},\boldsymbol{H}\right)+\frac{ieh}{c}\rho_{1}\left(\boldsymbol{\sigma},\boldsymbol{E}\right)$$ in $F$. These two terms, when divided by the factor $2m$, can be regarded as the additional potential energy of the electron due to its new degree of freedom. The electron will therefore behave as though it has a magnetic moment $\frac{eh}{2mc}\boldsymbol{\sigma}$ and an electric moment $\frac{ieh}{2mc}\rho_{1}\boldsymbol{\sigma}$. This magnetic moment is just that assumed in the spinning electron model. The electric moment, being a pure imaginary, we shold not expect to appear in the model. It is doubtfull wether the electric moment has any physical meaning, since the Hamiltonian in that we started from is real, and the imaginary part only appeared when we multiplied it up in an artificial way in order to make it resemble the Hamiltonian of previous theories.

The angular momentum integrals for motion in a central field

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