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When dealing with potentials, quantum wells, etc. I've usually used the following conditions for assuring the continuity of a wave function: $$1. \ \psi_I(x)|_{x=0} = \psi_{II}(x)|_{x=0}$$ $$2. \ \frac{d \psi_I(x)}{dx}|_{x=0} = \frac{d\psi_{II}(x)}{dx}|_{x=0}$$

Now I'm working on a problem in which an electron is in a system of two layers. In one layer its effective mass is equal to $m^*=m_1>0$, whereas in the other one $m^*=-m_2<0$.

In the solution the second condition is written as: $$\frac{1}{m_1}\frac{d \psi_I(x)}{dx}|_{x=0} = -\frac{1}{m_2} \frac{d\psi_{II}(x)}{dx}|_{x=0}$$

I've never seen such a condition, for sure it takes into account that the mass is different but could somebody explain why it takes such form?

Qmechanic
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mcas
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1 Answers1

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The continuity of $$x~\mapsto~\pi(x)~:=~\frac{1}{m^{\ast}(x)}\frac{d\psi(x)}{dx}\tag{1}$$ and $$x~\mapsto~\psi(x)\tag{2}$$ follows from a mathematical bootstrap argument similar to my Phys.SE answer here.

Proof of (1). Rewrite the TISE $$ - \frac{d}{dx} \frac{\hbar^2}{2m^{\ast}(x)}\frac{d\psi(x)}{dx} +V(x)\psi(x) ~=~E \psi(x) \tag{3}$$ as a differential-integral equation $$ \frac{\hbar^2}{2}\pi(x) ~\equiv~\frac{\hbar^2}{2m^{\ast}(x)}\frac{d\psi(x)}{dx} ~=~\int^x\!\mathrm{d}y ~(V(y)-E)\psi(y). \tag{4} $$ If we assume that $V,\psi \in {\cal L}^2_{\rm loc}(\mathbb{R})$ are locally square integrable functions, then the product $(V-E)\psi\in {\cal L}^1_{\rm loc}(\mathbb{R})$ due to Cauchy–Schwarz inequality. Then the integral $x\mapsto \int^{x}\mathrm{d}y\ (V(y)-E)\psi(y)$ is continuous. Hence the LHS of eq. (4) is continuous as well. $\Box$

Proof of (2). Rewrite eq. (1) as an integral equation $$ \psi(x)~=~ \int^x\!\mathrm{d}y ~m^{\ast}(y)\pi(y).\tag{5} $$ If we assume that $m^{\ast} \in {\cal L}^2_{\rm loc}(\mathbb{R})$, then we can repeat the previous proof technique to conclude that the LHS of eq. (5) is continuous as well. $\Box$

Qmechanic
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