Least Action

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Archive for February 2012

Why V(x) = -1/x^2 has no bound state

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Here we present a scaling-based argument to show that the attractive potential

V(x) = -\frac{\lambda}{x^2}

(\lambda > 0), has no bound states (i.e. states with energy E < 0). Consider the Time Independent Schrodinger equation for this potential, which is the eigenvalue equation for the corresponding Hamiltonian,

-\frac{\hbar^2}{2m}\frac{d^2\psi}{dx^2} - \frac{\lambda}{x^2}V(x)\psi = E\psi

This can be rearranged as

\frac{d^2\psi}{dx^2} + \frac{2m\lambda}{\hbar^2}\frac{1}{x^2}\psi = -\frac{2mE}{\hbar^2}\psi

Now, it is easy to see that the quantity

\frac{2m\lambda}{\hbar^2}

is dimensionless. So, this problem has no independent scale, even though naively one might think that \lambda specifies a scale for this problem. We claim that for such a system, there can be no bound state. This is proved below.

Suppose we perform the scale transformation x \rightarrow \alpha x where \alpha is some nonzero real number, we see from the equation above that if E is an eigenvalue, then so is \alpha^2 E.

Suppose now that a bound ground state exists, with energy E_{G}. By definition E_{G} < 0. Then scale invariance implies that \alpha^2 E_{G} must also be the energy of some bound state. But

\alpha^2 E_{G} < E_{G}

as multiplying the negative ground state energy by a positive number only makes it more negative. This contradicts the fact that the ground state has energy E_{G}. In fact, we can make a stronger statement, viz. the ground state energy can be made as small as we want. Therefore, there is no finite energy ground state for this system, and consequently there can be no bound states.

Note that there is no such problem with scattering states, i.e. states with positive energy. One can take an arbitrarily small positive energy scattering state and from it obtain valid energies of the continuum of higher energy scattering states by performing a scale transformation.

Incidentally, this is why potentials like -\lambda[\delta(x)]^2 and -\lambda\delta(x)/x also have no bound states.

Written by Vivek

February 8, 2012 at 08:48

Posted in Mathematics, Physics, Quantum Mechanics