Significance of ((H,x),x) double commutator?

[Peeter]

I've just done a textbook exersize to calculate (H,x) and ((H,x),x) for H= p^2/2m +V.

Having done the manipulation, my next question is what is the significance of this calculation. Where would one use these commutator and double commutator relations?

 

[xepma]

One place where the double (and higher) commutator comes into play is when you perform a transformation, such as a rotation or a translation.

Transformations such as these are performed through a unitary transformation [tex]U[/tex]. In turn, such a unitary transformation is said to be generated by the Hermitian operator [tex]A[/tex] if we can write:

[tex]U = e^{iA}[/tex]

Here the exponential of an operator is defined through the Taylor series of the exponential. Now, the effect of this transformation is that it transforms vectors in your Hilbert space as:

[tex]|v\rangle \longrightarrow |v'\rangle = e^{iA}|v\rangle[/tex]

Operators which act on this Hilbert space are also transformed, namely:

[tex]O \longrightarrow O' = U^\dagOU = e^{-iA}Oe^{iA}[/tex]

With this definition the expecation value of the operator [tex]O[/tex] is invariant under the transformation, i.e.

[tex]\langle v|O|v\rangle \langle v'|O'|v'\rangle[/tex]

Anyways, if you expand the transformation of the operator, i.e. if you just use the Taylor series, you will find that it looks like:

[tex]e^{iA}Oe^{-iA} = O + i[O,A] + \frac{i^2}{2!} [[O,A],A] + \ldots[/tex]

where the dots denote the triple and higher commutators.

Long story short, the higher order commutators of two operators determine the transformation properties of one operator with respect to the transformation generated by the other operator.

 

[arkajad]

Adding to the above:

[tex]mx[/tex] is the generator of the Galilean boost (see e.g. "http://www.physics.princeton.edu/~mcdonald/examples/QM/brown_ajp_67_204_99.pdf" "). So if you want to calculate how H changes when you change your reference frame to a moving one with velocity v, you will have to calculate the double commutator. (The higher commutators will be easy in this particular case!)