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Why do we have to multiply a proportionality constant in the delta potential $V(x)=-\alpha \delta(x)$ where $\alpha$ is some positive constant?

Isn't that $V(x)=\pm \delta(x)$ already enough to represent a very "high" potential?

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  • $\begingroup$ You given part of the answer when you ask about $\pm \delta (x)$, clearly there is something different from $+ \delta (x)$ and $- \delta (x)$. The presence of $\alpha$ is taken to be a measure of the strength of the potential. $\endgroup$ – jim Jun 18 '17 at 15:24
  • $\begingroup$ Leading question: is $\delta(x)$ a very high potential? Or is it a very narrow potential? $\endgroup$ – rob Jun 18 '17 at 16:27
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If $x$ is the argument of the delta function $\delta(x)$ with dimension $[x]$, then the dimensions of the delta function are,

$$[\delta(x)] = \frac{1}{[x]}.$$

As such, we must have that $V(x) = \alpha\delta(x)$ in order for dimensions to be consistent. To see this is the case, notice that,

$$\int dx \, \delta(\alpha x) f(x) = \frac{1}{|\alpha|}\int dx \, \delta(x)f(x)$$

and this is only dimensionally consistent if $\delta$ has dimensions being the inverse of its argument. The value of $\alpha$ itself matters; if we perturb a Hamiltonian by $V(x)$, the first correction to the energy is,

$$\langle \psi | V | \psi \rangle = \alpha \int |\psi(x)|^2 \delta(x)\, dx = \alpha|\psi(0)|^2$$

and thus while conceptually we may think of $\alpha \delta(x)$ as always being an infinitely large spike regardless of the value of $\alpha$, the actual value itself does affect results.

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  • $\begingroup$ Additionally to the dimensional reasoning, the absolute value of the prefactor plays a role if you have more than one delta peaks in your potential. $\endgroup$ – Photon Jun 18 '17 at 16:23
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Isn't that V(x)=±δ(x) already enough to represent a very "high" potential?

Consider instead, the canonical finite potential well problem.

enter image description here

with the additional constraint that width $L = 2a$ and depth $V_0$ are related rather than being independent parameters:

$$L V_0 = \alpha \Rightarrow V(x) = -\alpha\,\frac{\Theta(x + \frac{L}{2}) - \Theta(x - \frac{L}{2})}{L}$$

where $\alpha$ has units $\mathrm{J\cdot m}$ and $\Theta(x)$ is the Heaviside step function.

Note that as $L \rightarrow 0$, $V_0 \rightarrow \infty$ and, in the limit, we have

$$V(x) = -\alpha\; \delta(x) $$

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The discontinuity in the wave function across a $\delta$-potential $V(x)=-\alpha \delta(x)$ is proportional to $\alpha$, and the energy $E=-\frac{\alpha^2 m}{2\hbar^2}$ for this potential also depends on $\alpha$. Thus, $\alpha$ functions as an tunable parameter for your potential.

Likewise there is an $\alpha$-dependence on other $\delta$-type potentials, such as the double delta $-\alpha (\delta (x-a)+\delta(x+a)$ so one can adjust the energies in the problem to the particular situation at hand.

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