How to derive the Schrodinger equation for a system with position dependent effective mass? For example, I encountered this equation when I first studied semiconductor hetero-structures. All the books that I referred take the equation as the starting point without at least giving an insight into the form of the Schrodinger equation which reads as

$$\big[-\frac{\hbar^2}{2}\nabla \frac{1}{m^*}\nabla + U \big]\Psi ~=~ E \Psi. $$

I feel that it has something to do with conserving the probability current and should use the continuity equation, but I am not sure.

  • 1
    $\begingroup$ Hi ballkikhaal - I edited out the part of your question asking about a book, because we limit the number of book recommendation questions on the site. See if anything in this question helps you. $\endgroup$
    – David Z
    Commented Oct 10, 2012 at 16:23

5 Answers 5


A Hamiltonian must be self-adjoint. The equation must also reduce to the familiar equation in the case of a constant mass. Now the form of the operator is already determined as the only simple self-adjoint generalization of the position-independent Schroedinger equation to the position dependent case.

If you specialize to 1 dimension, you get the Sturm-Liouville equation. At
you can find a discussion of its self-adjointness. Everything genaralizes to the PDE case.


For derivation of the PDM Schrodinger equation see K. Young, Phys. Rev. B 39, 13434–13441 (1989) "Position-dependent effective mass for inhomogeneous semiconductors". Abstract.:A systematic approach is adopted to extract an effective low-energy Hamiltonian for crystals with a slowly varying inhomogeneity, resolving several controversies. It is shown that the effective mass $m_R$ is, in general, position dependent and enters the kinetic energy operator as $ -\nabla({m_R-1})\nabla/2$. The advantage of using a basis set that exactly diagonalizes the Hamiltonian in the homogeneous limit is emphasized.


The derivation is straightforward if you consider the source of the effective mass is a slowly varying hopping parameter on a tight-binding (lattice particle) model. Here you have a particle on a square lattice with a probability amplitude to go left, right, up and down, forward, backwards. The main physical requirement is Hermiticity, which in 1d can be used (with a phase choice on the wavefunction) to turn the phase everywhere real.

Once you do this, there is a real amplitude at site n to hop one square to the right r(n) and an amplitude to hop one square to the left, which by hermiticity and reality, must be r(n-1)--- it must be the complex conjugate of the amplitude to hop right from position n-1. So the amplitude equation is

$$ i{dC\over dt} = r(n-1) C_{n-1} - (r(n-1)+r(n))C_n + r(n) C_{n+1} $$

This is, when r is slowly varying, equivalent to the continuum equation found by Taylor expanding and keeping only the most relevant terms:

$$ i {d\psi \over dt} = {1\over 2} {\partial \over \partial x} (r(x) {\partial\over \partial x} \psi(x)) $$

As Feynman noted but never published (Dyson published this comment posthumously, in a paper in American Journal of Physics titles something like "Feynman's derivation of the Maxwell equations from Schrodinger equation"), Dirac's phase trick doesn't work in higher dimensions, because you can't fix all the phases. Then the commutators have a magnetic field addition, and to make it consistent, the magnetic field has to end up obeying Maxwell's equation, since the phase rotation gives a U(1) symmetry. This is not a true derivation of Maxwell's physics from quantum mechanics, it is just a way of showing you need the extra assumption of CP invariance to make the hopping hamiltonian real (which is true).

Then with the extra assumption, you just get

$$ i {d\psi\over dt} = {1\over 2} \nabla \cdot (t(x) \nabla \psi) + V(x) \psi $$

Where I have added back the potential. This is the continuum limit of a tight binding model with spatially slowly varying hopping, or inverse effective mass.


In addition to Claudius' and Ron Maimon's answers, I would like to make three comments:

  1. Classically, the Hamiltonian function for the effective mass approximation reads $$\tag{1} H({\bf r}, {\bf p})~:=~\frac{{\bf p}^2}{2m^*({\bf r})}+V({\bf r}).$$

  2. Quantum mechanically, when one quantizes the classical model (1), one should pick a self-consistent choice for the Hamiltonian operator $\hat{H}$. It is natural to replace the classical variable ${\bf r}$ and ${\bf p}$ in the Hamiltonian (1) with the operators $$\tag{2}\hat{\bf r}~=~{\bf r} \qquad\text{and}\qquad \hat{\bf p}~=~\frac{\hbar}{i}\nabla $$ (in the Schrodinger representation). But which operator ordering prescription should one choose? One natural choice, which (under appropriate boundary conditions) makes the Hamiltonian Hermitian, is $$\tag{3} \hat{H}~:=~\hat{\bf p}\cdot \frac{1}{2m^*(\hat{\bf r})}\hat{\bf p}+V(\hat{\bf r})~=~-\frac{\hbar^2}{2}\nabla\cdot \frac{1}{m^*({\bf r})}\nabla+V({\bf r}).$$

  3. Finally, let us mention a somewhat related/generalized Hermitian Hamiltonian operator $$\tag{4} \hat{H}~=~-\frac{\hbar^2}{2}\Delta_g +V({\bf r}), $$ which may give another useful (anisotropic) effective mass model. Here $\Delta_g$ is the Laplace-Beltrami operator for a Riemannian $3\times 3$ metric $g_{ij}=g_{ij}({\bf r})$, which, roughly speaking, may be viewed as an (anisotropic) effective mass tensor.


I would be very surprised if you managed to find a strict mathematical derivation of the Schrödinger equation anywhere – at least I have not encountered one until now. However, it might be worth pointing out that the ‘general’ time-dependent Schrödinger equation, which is often taken as an axiom of quantum mechanics, is usually

$$i \hbar \partial_t \Psi = \hat H \Psi \quad .$$

In the case of a stationary Hamiltonian (usually $U(x,t) \equiv U(x)$), this equation separates and you get the stationary Schrödinger equation, namely

$$ \hat H \Psi = E \Psi \quad ,$$

that is, an eigenvalue equation for the Hamiltonian.

Given this equation, it is then relatively simple to work out the form of the Hamiltonian (in your case, $-\frac{\hbar^2}{2} \nabla \frac{1}{m^\star} \nabla + U$) and plug it into the equation. The exact form of the Hamiltonian is usually guesswork based on observation and analogies to classical mechanics. In general, we have

$$ \hat H = \hat T + \hat U $$

where $\hat T$ and $\hat U$ denote the operators for kinetic and potential energy, correspondingly.

It is worth noting that you can derive the continuity equation (which is identical to probability conservation in this case) from the Schrödinger equation by adding the complex conjugate of the Schrödinger equation to itself.

  • $\begingroup$ thank you for the answer but what i am asking is how to derive it within the regime of effective mass approximation and that too when the mass has a spatial profile... for example in the case of Al/GaAs high electron mobility transistor we have a position dependent mass. $\endgroup$ Commented Oct 10, 2012 at 16:55
  • $\begingroup$ Are you looking for a derivation of the Hamiltonian $\hat H$ or the Schrödinger equation? I am positive that neither probability conservation nor the continuity equation have anything to do with the earlier. $\endgroup$
    – Claudius
    Commented Oct 10, 2012 at 16:57
  • $\begingroup$ I have come across this equation in Hamaguchi on page 347 <books.google.co.in/…> $\endgroup$ Commented Oct 10, 2012 at 18:13
  • $\begingroup$ Where do you think Schrodinger equations come from, if not by some sort of a derivation? $\endgroup$
    – Ron Maimon
    Commented Oct 10, 2012 at 19:31
  • $\begingroup$ The answer is not by guesswork, it is from the tight-binding approximation with CP invariance to guarantee that the hopping parameter is real, and then Hermiticity guarantees the hopping is symmetric and equal to the given Hamiltonian. If the hopping is slowly locally varying, then you get the Hamiltonian they say. The Schrodinger equation which is axiomatic is not as specific as the Schrodinger equation in space, which has a specific ansatz for the kinetic term which can be justified from tight binding, as Feynman does in his lectures. $\endgroup$
    – Ron Maimon
    Commented Oct 10, 2012 at 19:47

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