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Dirac gives the relation: $\exp(iaq)f(q,p) = f(q, p - a\hbar)\exp(iaq)$ where $\hbar$ is Planck's constant. Can anybody give me the corresponding relation when the $\exp$ function is a $\ln$?

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You say anticommutator but this relation is a simple consequence of the commutation relations $[q, p] = i \hbar$. Let us compute $$[q^n, p] = q [q^{n-1}, p] + [q, p] q^{n-1} =$$ $$ =q(q[q^{n-2}, p] + [q,p] q^{n-2}) + i \hbar q^{n-1} = \cdots = i \hbar n q^{n-1}$$ Next, suppose a function $g(q) = \sum_{n=0}^{\infty} {a_n \over n!} q^n$ is analytic in $q$ around 0 (it also works for function analytic around other points but it would make the derivation a little messier). Then we have $$[g(q), p] = \sum_{n=0}^{\infty} {a_n \over n!} [q^n, p] = i \hbar \sum_{n=1}^{\infty} {a_n \over (n-1)!} q^{n-1} = i \hbar g'(q) $$ For $g(q) = \exp(iaq)$ we get $[\exp(iaq), p] = - \hbar a \exp(iaq)$ (this is already your relation for $f(p,q) \equiv p$).

Next we compute (by successive commutations of the exponential to the right using the relation above) $$\exp(iaq) p^n = (p - \hbar a) \exp(iaq))p^{n-1} = \cdots = (p - \hbar a)^n \exp(iaq)$$ Finally, assuming $f(q,p) = \sum_{n=0}^{\infty} {b_n(q) \over n!} p^n$ is analytic around 0 in $p$ we have $$\exp(iaq) f(q,p) = \sum_{n=0}^{\infty} {b_n(q) \over n!} \exp(iaq) p^n= $$ $$ =\sum_{n=0}^{\infty} {b_n(q) \over n!} (p - \hbar a)^n \exp(iaq) = f(q, p-\hbar a) \exp(iaq) $$ which is your claimed relation.

Now, there is no reason to expect that $\log$ will give any reasonable relation. Exponential was special because it maps $q$ (which is a generator of the Lie algebra of the Heisenberg group in this representation) to an element that performs shift in momentum (which is precisely what your relation says).

To demonstrate this, if we instead chose above $g(q) = \log(iaq)$ we would only get $[\log(iaq), p] = - {a \over q} $ but this obviously can't lead to any similar relation like the one we had above.

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thank you Marek for your answer. i got as far as your last relation and was wondering about how to go further. – Fosco Jul 21 '11 at 5:23

You can get an integral representation of the result using the following integral representation of the natural logarithm.

$\ln(x) = \int_0^{\infty}\frac{e^{-s}-e^{-sx}}{s} ds$

This representation converges for $0 < |x|< \infty$, however, extra care must be taken when used as an operator identity. At least for plane waves in the momentum space, the result is trivially convergent.

Applying the identity to the function we obtain:

$\ln(iaq)f(q, p) = \int_0^{\infty}\frac{e^{-s}f(q, p)-f(q, p-sa\hbar)e^{-iasq}}{s}ds$


This is a technical method to perform the computation of $ \ln(q) \ln(p)$. However, it must be emphasized that the logarithm is a multiple valued function and the validity of the expression must be tested on the class of functions you want to apply this operator identity on.

The natural logarithm of a number can be defined as:

$\ln(x) = \frac{d}{dt} x^t|_{t=0}$

Using this formula we may write:

$\ln(q)\ln(p) = \frac{d}{ds}\frac{d}{dt} q^s p^t|_{s=t=0}$

If s and t were integer we would have the following relation:

$q^s p^t = (i\hbar)^s\frac{t!}{s!} p^{t-s} q^s$

We continue this relation to non-integer exponents as:

$q^s p^t = (i\hbar)^s\frac{\Gamma(1+t)}{\Gamma(1+s)} p^{t-s} q^s$

where $\Gamma$ is the gamma function.

By taking the derivatives with respect to s and t for both sides and the limits to zero, we obtain:

$\ln(q)\ln(p) = (\ln(i\hbar p) - \gamma) (\gamma + \ln(q) - \ln(p))$

where, $\gamma$ is the Euler constant, and the following identities were used:

$\frac{d}{dx} \Gamma(x) = \Gamma(x) \psi(x)$

where $\psi$ is the digamma function, and

$\psi(1) = - \gamma$.

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thank you David for your answer. the problem i am working with is much more special case: just ln(q)ln(p) and i am looking for a relation with ln(p)ln(q). by the way i hope this is not a stupid question. – Fosco Jul 21 '11 at 5:36
I understand that you wat to express $ \ln(q) \ln(p)$, such that all dependence on q is moved to the right. I added an new update in which a close form is technically obtained. – David Bar Moshe Jul 21 '11 at 9:01
thank you David, this is great! – Fosco Jul 21 '11 at 9:27
David, can i ask one more question: what about ln(p)ln(q)? if you have time to answer that would be great. – Fosco Jul 21 '11 at 11:43
Since $[q,p] = i\hbar$ while $[p,q] = -i\hbar$, you just have to interchange $p$ and $q$ and change the sign of $\hbar$. – David Bar Moshe Jul 21 '11 at 11:48

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