Operator relation involving the logarithm of an operator?

Question

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 $\ln$?

Answer 1

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.

Answer 2

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:

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

Update:

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

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