Schrodinger equation in momentum representation with position-dependent effective mass I'm trying to convert Schrodinger equation with position-dependent effective mass (PDEM) to momentum representation, and I'm not sure how to apply the kinetic energy operator.
In position representation the operator has the following form:
$$\hat{T} \Psi(\vec{r})=- \frac{\hbar^2}{2} \nabla \left(  \frac{1}{m^*(\vec{r})}   \nabla \Psi(\vec{r}) \right) \tag{1}$$
Which could be also written as:
$$\hat{T} \Psi(\vec{r})=\frac{1}{2}\hat{p} \left(  \frac{1}{m^*(\vec{r})}   \hat{p} \Psi(\vec{r}) \right) \tag{2}$$
Now I would like to obtain this operator in momentum representation. For constant mass, it would be:
$$\hat{T} \Psi(\vec{p})=\frac{1}{2 m^*} p^2 \Psi(\vec{p}) \tag{3}$$
How do I Fourier transform the operator in case of PDEM? I suspect there's convolution theorem involved, but not sure how to apply the rule correctly when derivatives are involved.
It is enough to consider one-dimensional case first, so I need to transform:
$$\hat{T}(x) \Psi(x)=-\frac{\hbar^2}{2}\frac{\mathrm d}{\mathrm d x} \left(\frac{1}{m^*(x)} \frac{\mathrm d \Psi(x)}{\mathrm d x} \right) \tag{4}$$
to:
$$\hat{T}(p) \Psi(p)$$
For potential we have by convolution theorem:
$$U(x) \Psi(x) \rightarrow \int_{-\infty}^\infty U(p-p') \Psi(p') dp' \tag{5}$$
 A: As my2cts hints, this is basically the same equation you get for light with a position-dependent dielectric constant, and you might gain some insight by looking into that literature. If the dielectric constant (or effective mass) is piecewise continuous, then you can solve for $\psi$ in each piece and stitch the solutions together with proper boundary conditions.
That said, you can take the Fourier transform of that equation if you want; you're right that it will contain convolutions (since the Fourier transform of a product of functions is a convolution of their Fourier transforms), which may make the resulting equation of limited use. I'm going to use the relationships in this table, and I make no guarantee that the following is error free.
First, for simplicity, define $b(x) = 1/m(x)$. Then your equation 4 becomes (ignoring the $-\hbar^2/2$)
$$\frac{db}{dx}\frac{d\psi}{dx} + b \frac{d^2 \psi}{dx^2}$$
Using relation 109 from the table, the Fourier transform of the above expression is
$$\frac{1}{2\pi}\left( \hat{\frac{db}{dx}} * \hat{\frac{d\psi}{dx}} + \hat{b} * \hat{\frac{d^2 \psi}{dx^2}} \right),$$
where $*$ indicates a convolution and $\hat{}$ indicates Fourier-transformed quantities. Then using relation 106 (and a little simplification), that becomes
$$-\frac{1}{2\pi}\left[ \left(\omega\hat{b}\right) * \left(\omega\hat{\psi}\right) + \hat{b} * \left(\omega^2\hat{\psi}\right) \right],$$
which I think is the answer to your question. There may be a way to simplify the expression further (you can write the convolutions as integrals and combine some terms), but that's up to someone else.
A: If the m(r) would be general then the kinetic energy is no longer diagonal in k space. As you probably want to match materials with different effective mass, I would treat this like fresnel did for light: find solutions in each medium and match. If this is your choice and you publish your result, please reference me. 
