# Why is the energy shift due to a 'sagging' potential negative and independent of box size?

Consider a box of width $L$ and the composed of the following potential $$V(x)=\frac{V_0x(x-L)}{L^2}, x\in[0,L]$$ and $V(x)=\infty$ elsewhere. Using perturbation theory - with a square box as the similar potential - one finds that the perturbation potential for $x\in[0,L]$ is simply the potential describing the sagging box (which makes sense because $V(x)=0$ for normal box).

When calculating the nth state energy shift to first order one also finds that $$\Delta E_n^{(1)}=-V_0\Big({1\over6}+{1\over2n^2\pi^2}\Big).$$

1. It is negative,
2. It is independent of the width of the box.

The first is easier to come up with an explanation for. My attempt is that the potential of the sagging box is such that the box appears to sag into the negative territory of the potential axis, thus making the change in energy negative.

Edit: Another potential explanation (pardon the pun) that I just thought of is similar: Because the box sags down this creates more space for lower energy eigenstates within the box, thus creating a net energy shift of negative value.

The second I am struggling to come up with an explanation for. Perhaps a sagging potential is one of ratios rather than sizes. One thought is that, visually, if you were to stretch the box then the 'sag' would move upwards (i.e. be less sagged), such as one would expect when pulling either side of an elastic band or something. Thus counteracting any dependancy on the width of the box.

Any suggestions?

Edit

Here is the calculation of $\Delta E_n^{(1)}$:

We consider the perturbation potential $\Delta V=V(x)-V_0(x)$, where $V_0(x)$ is some potential that is visually similar to the potential given. A very similar potential (you can check this by plotting both of them as boxes) is the simple situation of a 'box' (i.e. $V(x)=0$ for $x\in [0,L]$). Thus $$\Delta V(x)=V(x)-V_0(x)=\frac{V_0x(x-L)}{L^2}-0=\frac{V_0x(x-L)}{L^2}.$$ The energy shift is then given by $$\Delta E=\left< \Delta V\right>=\int_0^Ldx\Phi^*\Delta v\Phi.$$ Using the stationary state solutions to the Schrodinger equation, $\Phi_n(x)=\sqrt{L/2}\sin{\frac{n\pi x}{L}}$, we get $$\Delta E=\int_0^Ldx\frac{L}{2}\sin^2\Big(\frac{n\pi x}{L}\Big)\frac{V_0x(x-L)}{L^2}=...=-V_0\Big(\frac{1}{6}+\frac{1}{2n^2\pi^2}\Big).$$

• Given the form of $V(x)$, something sounds odd about your assertion that $\Delta E_n^{(1)}$ is independent of $L$. Can you outline your calculation? – udrv May 4 '16 at 11:52
• I have edited the question. – ODP May 4 '16 at 20:26

Your conclusion is correct, but the idea is not restricted to the sagging well. This is a neat exercise in scaling, and it all comes from the choice of parametrization of the potential for given length $L$.
To see this better in the present case, just rewrite the hamiltonian in terms of the scaled coordinate $\xi = \frac{x}{L}$, using $\frac{d}{dx} = \frac{1}{L}\frac{d}{d\xi}$, $${\hat H} = -\frac{\hbar^2}{2m}\frac{d^2}{dx^2} + V_0\frac{x}{L}\left( \frac{x}{L} - 1\right) = -\frac{\hbar^2}{2mL^2}\frac{d^2}{d\xi^2} + V_0\xi \left( \xi - 1\right) = \frac{1}{L^2} {\hat h}_0 + V_0{\hat u}$$ where both ${\hat h}_0$ and ${\hat u}$ are now scale-independent. In this form it is clear that for given $V_0$ the entire dependence on $L$ comes from the original infinite box, and that this holds whenever the potential is of the form $V(x) = V_0 \;u(x/L)$.
There is also another interesting choice of scaling. If we take further $$V_0 = \frac{{\bar V}_0}{L^2}$$ the eigenvalue problem ${\hat H}\psi_n = E_n\psi_n$ can be rewritten in the scale-independent form $$\left[{\hat h}_0 + {\bar V}_0{\hat u}\right]\psi_n = \epsilon_n\psi$$ Here $\epsilon_n = L^2E_n$ is evidently scale-independent as well, $\psi_n = \psi_n(\xi)\equiv \psi(x/L)$, and for given ${\bar V}_0$ $$E_n = \frac{\epsilon_n}{L^2}$$ preserves the same dependence on $L$ as the eigenvalues of the infinite box, $E^0_n = \frac{n^2\pi^2\hbar^2}{2mL^2}$