# 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 position dependent mass is just a potential. Are you sure you do not also make the momentum position dependent? That is also a potential. – my2cts Jan 28 '19 at 13:27
• @my2cts, I do not understand your comment. Schrodinger equation for position dependent mass in coordinate representation is $$- \frac{\hbar^2}{2} \nabla \left( \frac{1}{m^*(\vec{r})} \nabla \Psi(\vec{r}) \right)+U(\vec{r}) \Psi(\vec{r})=E \Psi(\vec{r})$$ where the PDM and potential do not enter in the same way at all. Position dependent mass is not equivalent to a potential. As for whether PDM makes the momentum operator itself position dependent, I'm not sure. I think it should still have the same position representation $\hat{p}=-i \hbar \nabla$ – Yuriy S Jan 28 '19 at 13:39
• Please number you equations. – my2cts Jan 28 '19 at 13:43
• That is because you only take its inertial effect onto account, not its energy effect. This is correct as you are, likely, discussing effective mass in some kind of condensed matter medium. – my2cts Jan 28 '19 at 13:52
• @my2cts, yes, this is for semiconductor heterostructures application – Yuriy S Jan 28 '19 at 13:55

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.

• Thank you for the answer! Just to be sure, does this mean what I think it means? $$\left(k \hat{b} (k)\right) * \left(k \hat{\psi} (k)\right)=\int (k-k') \hat{b} (k-k') k' \hat{\psi} (k') dk'$$ I understand convolution, just wanted to be sure about the meaning of the brackets in this case. Here $\hat{b}(k)$ means that $\hat{b}$ is a function of $k$. – Yuriy S Jan 29 '19 at 8:55
• Yes, that's correct. – lnmaurer Jan 29 '19 at 14:08

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.