Foldy–Wouthuysen transformation

Clearly, the FW transformation is a continuous transformation, that is, one may employ any value for θ which one chooses. Now comes the distinct question of choosing a particular value for θ, which amounts to choosing a particular transformed representation.

One particularly important representation, is that in which the transformed Hamiltonian operator Ĥ′₀ is diagonalized. Clearly, a completely diagonalized representation can be obtained by choosing θ such that the α · p term in 5 is made to vanish. Such a representation is specified by defining:

${\displaystyle \tan 2\theta \equiv {\frac {|\mathbf {p} |}{m}}}$

(6)

so that 5 is reduced to the diagonalized (this presupposes that β is taken in the Dirac–Pauli representation (after Paul Dirac and Wolfgang Pauli) in which it is a diagonal matrix):

${\displaystyle {\hat {H}}'_{0}=\beta (m\cos 2\theta +|\mathbf {p} |\sin 2\theta )}$

(7)

By elementary trigonometry, 6 also implies that:

${\displaystyle \sin 2\theta ={\frac {|\mathbf {p} |}{\sqrt {m^{2}+|\mathbf {p} |^{2}}}}\quad {\text{and}}\quad \cos 2\theta ={\frac {m}{\sqrt {m^{2}+|\mathbf {p} |^{2}}}}}$

(8)

so that using 8 in 7 now leads following reduction to:

${\displaystyle {\hat {H}}'_{0}=\beta {\sqrt {m^{2}+|\mathbf {p} |^{2}}}}$

(9)

Prior to Foldy and Wouthuysen publishing their transformation, it was already known that 9 is the Hamiltonian in the Newton–Wigner (NW) representation (named after Theodore Duddell Newton and Eugene Wigner) of the Dirac equation. What 9 therefore tells us, is that by applying a FW transformation to the Dirac–Pauli representation of Dirac's equation, and then selecting the continuous transformation parameter θ so as to diagonalize the Hamiltonian, one arrives at the NW representation of Dirac's equation, because NW itself already contains the Hamiltonian specified in (9). See this link.

If one considers an on-shell mass—fermion or otherwise—given by m² = p^σp_σ, and employs a Minkowski metric tensor for which diag(η) = (+1, −1, −1, −1), it should be apparent that the expression

${\displaystyle p^{0}={\sqrt {m^{2}+|\mathbf {p} |^{2}}}}$

is equivalent to the E ≡ p⁰ component of the energy-momentum vector p^μ, so that 9 is alternatively specified rather simply by Ĥ′₀ = βE.

Now consider a fermion at rest, which we may define in this context as a fermion for which |p| = 0. From 6 or 8, this means that cos 2θ = 1, so that θ = 0, ±π, ±2π and, from 2, that the unitary operator U = ±I. Therefore, any operator O in the Dirac–Pauli representation upon which we perform a biunitary transformation, will be given, for an at-rest fermion, by:

${\displaystyle O\to O'\equiv UOU^{-1}=(\pm I)(O)(\pm I)=O}$

(10)

Contrasting the original Dirac–Pauli Hamiltonian operator

${\displaystyle {\hat {H}}_{0}\equiv {\boldsymbol {\alpha }}\cdot \mathbf {p} +\beta m}$

with the NW Hamiltonian 9, we do indeed find the |p| = 0 "at rest" correspondence:

${\displaystyle {\hat {H}}_{0}={\hat {H}}'_{0}=\beta m}$

(11)

Now, consider the velocity operator. To obtain this operator, we must commute the Hamiltonian operator Ĥ′₀ with the canonical position operators x_i, i.e., we must calculate

${\displaystyle {\hat {v_{i}}}\equiv i\left[{\hat {H}}_{0},x_{i}\right]}$

One good way to approach this calculation, is to start by writing the scalar rest mass m as

${\displaystyle m=\gamma ^{0}{\hat {H}}_{0}+\gamma ^{j}p_{j}}$

and then to mandate that the scalar rest mass commute with the x_i. Thus, we may write:

${\displaystyle 0=[m,x_{i}]=\left[\left(\gamma ^{0}{\hat {H}}_{0}+\gamma ^{j}p_{j}\right),x_{i}\right]=\left[\gamma ^{0}{\hat {H}}_{0},x_{i}\right]+i\gamma _{i}}$

(12)

where we have made use of the Heisenberg canonical commutation relationship [x_i,p_j] = −iη_ij to reduce terms. Then, multiplying from the left by γ⁰ and rearranging terms, we arrive at:

${\displaystyle {\frac {d{\hat {x}}_{i}}{dt}}={\hat {v_{i}}}\equiv i\left[{\hat {H}}_{0},x_{i}\right]=\alpha _{i}}$

(13)

Because the canonical relationship

${\displaystyle i\left[{\hat {H}}_{0},{\hat {v}}_{i}\right]\neq 0}$

the above provides the basis for computing an inherent, non-zero acceleration operator, which specifies the oscillatory motion known as zitterbewegung.

In the Newton–Wigner representation, we now wish to calculate

${\displaystyle {\hat {v}}_{i}'\equiv i\left[{\hat {H}}'_{0},x_{i}\right]}$

If we use the result at the very end of section 2 above, Ĥ′₀ = βp₀, then this can be written instead as:

${\displaystyle {\hat {v}}_{i}'\equiv i\left[{\hat {H}}'_{0},x_{i}\right]=i\beta \left[p_{0},x_{i}\right]}$

(14)

Using the above, we need simply to calculate [p₀,x_i], then multiply by iβ.

The canonical calculation proceeds similarly to the calculation in section 4 above, but because of the square root expression in p⁰ = √m² + |p|², one additional step is required.

First, to accommodate the square root, we will wish to require that the scalar square mass m² commute with the canonical coordinates x_i, which we write as:

${\displaystyle 0\equiv \left[m^{2},x_{i}\right]=\left[\left(p^{0}p_{0}+p^{j}p_{j}\right),x_{i}\right]=\left[p^{0}p_{0},x_{i}\right]+2ip_{i}}$

(15)

where we again use the Heisenberg canonical relationship [x_i,p_j] = −iη_ij. Then, we need an expression for [p₀,x_i] which will satisfy 15. It is straightforward to verify that:

${\displaystyle i\left[p_{0},x_{i}\right]={\frac {p_{i}}{p^{0}}}=v_{i}}$

(16)

will satisfy 15 when again employing [x_i,p_j] = −iη_ij. Now, we simply return the iβ factor via 14, to arrive at:

${\displaystyle {\frac {d{\hat {x}}_{i}'}{dt}}={\hat {v}}_{i}'\equiv i\left[{\hat {H}}'_{0},x_{i}\right]=\beta {\frac {p_{i}}{p^{0}}}=\beta v_{i}}$

(17)

This is understood to be the velocity operator in the Newton–Wigner representation. Because:

${\displaystyle i\left[{\hat {H}}'_{0},{\hat {v}}_{i}'\right]=i\left[\beta p_{0},\beta v_{i}\right]=0}$

(18)

it is commonly thought that the zitterbewegung motion arising out of 12 vanishes when a fermion is transformed into the Newton–Wigner representation.