7
$\begingroup$

The translation operator in one dimension is defined as $$ \hat T \psi(x) = \psi(x-\alpha) . $$ This can be written as an integral transformation, $$ \begin{align*} \hat T\psi(x) = \langle x|\hat T|\psi\rangle = \int\mathrm{d} x' \langle x|\hat T|x'\rangle\langle x'|\psi\rangle = \int\mathrm{d} x' T(x,x')\psi(x') = \int\mathrm{d} x' \delta(x-\alpha-x')\psi(x') = \psi(x-\alpha) \end{align*} $$ $$ \implies T(x,x') = \langle x | \hat T | x' \rangle = \delta(x' - (x-\alpha)). $$

On the other hand, the same operation can be represented by a differential operator using the series expansion of $\psi(x-\alpha)$, $$ \psi(x-\alpha) \approx \psi(x) - \alpha\frac{\partial\psi}{\partial x} + \frac{(-\alpha)^2}{2!}\frac{\partial^2\psi}{\partial^2x} + ... = \sum\limits_{k=0}^{\infty}\frac{(-\alpha)^k}{k!}\frac{\partial^k\psi}{\partial^k x}\\ = \left[ \sum\limits_{k=0}^{\infty} \frac{1}{k!} \left(-\alpha\frac{\partial}{\partial x}\right)^k\right]\psi(x) = e^{-\alpha\frac{\partial}{\partial x}}\psi(x). $$

So, we have two different but equivalent representations of $\hat T$, $$ \hat T[\psi] = \int\mathrm{d}x'T(x,x')\psi(x') = \int\mathrm{d}x'\delta(x' - (x-\alpha))\psi(x') \\ \hat T[\psi] = e^{-\alpha\frac{\partial}{\partial x}}\psi(x). $$ Questions:

  • How can I determine if an integral transformation with kernel $K(x,x')$ has a corresponding differential operator?

  • If there exists a differential operator for $K(x,x')$, how can I calculate it?

  • How to calculate $K(x,x')$ if the differential operator is known (e.g. $\hat K[\psi] = a\partial_x \psi(x) + b e^{-\partial^2_x}\psi(x) + U(x)\psi(x)$) ?
$\endgroup$
5
$\begingroup$

"Differential" may be a bit of a red herring, as the generic $\hat T$ operator may have strings of x s and $\partial_x$s. There is no good reason to hyperfocus on derivatives, which, in the coordinate representation you are using, correspond to the momentum operator $\hat p$, $$ \hat p = \int dx |x\rangle \frac {\hbar} {i} \partial _x \langle x|, $$ so $$ \langle x|\hat p|x'\rangle = \frac{\hbar}{i} \partial_x \langle x| x'\rangle = \frac{\hbar}{i} \partial_x \delta (x-x'). $$ (Do you now see that contrasting $\partial_x$ to convolution with an integral kernel $\partial_x \delta(x-x')$ is a false dichotomy?)

Consequently, $$ \langle x|e^{-ia \hat {p} /\hbar }|x'\rangle = e^{-a \partial_x } \langle x| x'\rangle= e^{-a \partial_x } \delta (x-x') = \delta (x-a-x'), $$ a Lagrange shift.

But also, acting on this by $\int dx' \langle x'| \psi\rangle $, $$ \int dx' \delta (x-a-x') \psi (x')= \psi (x-a). $$

In general, performing the formal manipulations to evaluate $T(x,x')= \langle x | \hat T | x\rangle$ as an elegant function may well depend on your prowess and dexterity with Fourier transforms, etc. Likewise, inverting it: $\int dx dx' ~ |x\rangle ~T(x,x')~\langle x'| = \hat T$.

The δ-function you started with has a Taylor series expansion in a with an infinity of differential operators, the above Lagrange shift. Your kernel "function" amounts to a pseudo-differential operator acting on a function. They dub this "microlocal analysis", $$ \int dx' \delta(x-a-x') \psi (x') = \int dx' e^{-a\partial_x }\delta(x-x') \psi (x')= e^{-a\partial_x } \psi(x). $$

Conversely, you evaluate the matrix element, if you can, as above, to obtain the integral kernel, which is basically the Green's function of the inverse of the pseudodifferential operator you start with.

The conceit is that all operators in this space are expressible in $\hat x, \hat p$.

To test-drive this with your given $$ \hat K = ia\hat p /\hbar + b e^{\hat p ^2/\hbar^2 } + U(\hat x) , $$ note $$ K(x,x') = a\partial _x \delta (x-x') +b e^{-\partial_x^2} \delta (x-x') + U(x) \delta (x-x'). $$

The middle term with coefficient b is readily recognizable as the formal inverse of the Weierstrass transform, $$ e^{-\partial_x^2} ~~\frac {1}{\sqrt{4\pi }} \int dy f(y) e^{-(x-y)^2/4} = f(x) . $$

So, tweaking your example to a low pass filter operator, $\hat L= e^{-\hat p^2/\hbar^2 }$, you have $$ e^{\partial_x^2} \psi(x) = \int dx' \frac{e^{-(x-x')^2/4}}{\sqrt{4\pi}} \psi (x'), $$ essentially linked to the Green's function for diffusion.

$\endgroup$
2
$\begingroup$

I assume that in transformation you restrict yourself to unitary transformations, such that probability is maintained $|| \hat{T} |\psi\rangle || = 1$. I will further assume that you limit the question to forms of transformation that are position-based somehow, so the notion of "Kernel" $\hat{K}(x,x')$ makes sense (so for example rotations in spin-space are not part of the discussion here). Then the existence of what you call "a differential" representation will depend (I think), on the following:

1) The transformation has to be continuous, that is it has to depend on some parameters $a_i$ that take a continuous range of values. So rotations about a point are ok (with the angles being the parameters) but reflection is not.

2) There is a set of parameters $a^0_i$ for which the transformation is the identity. Usually these will be $a^0_i = 0$. In your notation, that restrict to changes of the wave-function, we can will say that $\hat{T}_{a^0_1, a^0_2,\ldots,a^0_N}(\vec{r},\vec{r}') = \delta(\vec{r}-\vec{r}')$. So for example if we'll think of a bizarre transformation such as $$\hat{T}_a\psi(x) = \sqrt{\frac{a}{2\pi}}\int\! dx' e^{iaxx'}\psi(x')$$ it will not fulfill it even though it depends on a continuous variable $a$.

3) We can expand about the identity in a continuous manner. So $\hat{T}_{a^0_1+\delta a_1, \ldots, a^0_N+\delta a_N}\psi(\vec{r})$ will describe a small change, on the scale of $\delta a$. This means that the Kernel $T_{a_i}(\vec{r},\vec{r}')$ should be differentiable. Note that for our discussion, the delta function is differentiable (as we can always define its derivatives under the integral using integration by parts).

$\endgroup$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.