# Foldy Wouthuysen Frustration

1. Nov 6, 2012

### kbarger

I'm looking at the original Foldy Wouthuysen article found here http://www.physics.drexel.edu/~bob/Quantum_Papers/Foldy-Wouthuysen.pdf,
and have some question regarding eq. 31 in this paper. Would anyone be able to explain how this is derived? In particular the part with the time derivative of the exp(-iS) term. The wiki-article on FW transformation is rank with typos and also glosses over this point.

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2. Nov 6, 2012

### dextercioby

It's all in the formulas 3 and 4 and in any textbook on QM which deals with unitary transformations (such as shifting between pictures: Schrödinger vs Heisenberg vs interaction (Dirac, Tomonaga, Schwinger)).

3. Nov 6, 2012

### kbarger

Why does the time derivative of the S operator in the FW original paper have a + sign while everywhere else it is derived there is a - sign? With my naive commutator skills I get what you see in the attachment. I'm looking for something a little more than "Go look in a physics book", please.

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4. Nov 6, 2012

### kbarger

Here's my naive calculation. What's wrong with this?

\begin{array}{l}
- ie^{iS} \frac{{\partial e^{ - iS} }}{{\partial t}} = - ie^{iS} \left( { - i\frac{{\partial S}}{{\partial t}}} \right)e^{ - iS} = \\
- i\left( {1 + iS - \frac{1}{2}S^2 - \frac{i}{6}S^3 + \frac{1}{{24}}S^4 + - - \cdots } \right)\left( { - i\frac{{\partial S}}{{\partial t}}} \right)\left( {1 - iS - \frac{1}{2}S^2 + \frac{i}{6}S^3 + \frac{1}{{24}}S^4 + - - \cdots } \right) = \\
- \left( {1 + iS - \frac{1}{2}S^2 - \frac{i}{6}S^3 + \frac{1}{{24}}S^4 + - - \cdots } \right)\frac{{\partial S}}{{\partial t}}\left( {1 - iS - \frac{1}{2}S^2 + \frac{i}{6}S^3 + \frac{1}{{24}}S^4 + - - \cdots } \right) = \\
- \left( {1 + iS - \frac{1}{2}S^2 - \frac{i}{6}S^3 + \frac{1}{{24}}S^4 + - - \cdots } \right)\left( {\dot S - i\dot SS - \frac{1}{2}\dot SS^2 + \frac{i}{6}\dot SS^3 + \frac{1}{{24}}\dot SS^4 + - - \cdots } \right) = \\
\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\, - \left( {\dot S - i\dot SS - \frac{1}{2}\dot SS^2 + \frac{i}{6}\dot SS^3 + \frac{1}{{24}}\dot SS^4 + - - \cdots } \right) \\
\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\, - \left( {\,\,\,\,\,\,\,iS\dot S\,\, + S\dot SS - \frac{i}{2}S\dot SS^2 - \frac{1}{6}S\dot SS^3 + \frac{i}{{24}}S\dot SS^4 + - - \cdots } \right) \\
\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\, - \left( {\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\, - \frac{1}{2}S^2 \dot S\,\, + \frac{i}{2}S^2 \dot SS + \frac{1}{4}S^2 \dot SS^2 - \frac{i}{{12}}S^2 \dot SS^3 - \frac{1}{{48}}S^2 \dot SS^4 + + - - \cdots } \right) \\
\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\, - \left( {\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\, - \frac{i}{6}S^3 \dot S\, - \frac{1}{6}S^3 \dot SS + \frac{i}{{12}}S^3 \dot SS^2 + \frac{1}{{36}}S^3 \dot SS^3 - - + + \cdots } \right) - - - \cdots = \\
\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\, - \dot S - i\left[ {S,\dot S} \right] + \frac{1}{2}\left[ {S,\left[ {S,\dot S} \right]} \right] + \frac{i}{6}\left[ {S,\left[ {S,\left[ {S,\dot S} \right]} \right]} \right] - - + + \cdots \\
\end{array}

5. Nov 6, 2012

### Bill_K

The first equality is wrong. The assumption is that S and ∂S/∂t do not commute. So when you calculate (∂/∂t)(e-iS), you need to expand the exponential first:
(∂/∂t)(1 - iS -½SS + ...) = -i(∂S/∂t) - ½(∂S/∂t)S - ½S(∂S/∂t) + ...

6. Nov 7, 2012

### kbarger

Thank you Bill_K. That clears that up to my great satisfaction. I now get the desired result. I'm a little curious how the original paper has the wrong sign for the S time derivative. Thanks again.

Kevin