I have studied that a wave function should vanish at the location of an infinite potential. Consider a direct Delta delta potential at $x=0$. Why does does function not become zero here at $x=0$?
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I have studied that a wave function should vanish at the location of an infinite potential.
This isn't really true. Whenever you encounter an infinite potential in QM, it's a good idea to think of it as a limiting case of a finite potential.
For example, consider a finite potential well such that $V(x)=0$ inside the well and $V_0>0 $ outside. If you solve for the (bound) energy eigenstates, you find that they decay exponentially outside the well. If you take the limit as $V_0\rightarrow \infty$, then those states go to zero outside the well while remaining sinusoidal inside it.
Similarly, you could consider a rectangular potential barrier of width $a$ and height $\lambda/a$ centered at $x=0$. If you solve for the energy eigenstates (which would be non-normalizable in this case), you can take the limit as $a\rightarrow 0$. In this limit, the rectangular barrier becomes a delta function $\lambda \delta(x)$, and you will find that the eigenstates do not vanish at $x=0$.
Of course, if you want a more mathematical explanation, the answer is that infinite potentials simply don't make any sense in the first place. They certainly don't satisfy the requirements we place on self-adjoint operators, which rules them out as candidates for the Hamiltonian.
Whenever you encounter an infinite potential, you should understand that it's really a physically-motivated shorthand for a particular Hilbert space with particular boundary conditions on the energy eigenstates. For example, the infinite potential well is really the free particle on an interval. The delta function potential is really the free particle on a disconnected line.
Whether you choose to think of infinite potentials as limiting cases of finite ones or as simply as shorthand for somewhat more exotic Hilbert spaces is a matter of taste - the former being more widely adopted by physicists. Its certainly true that these models are very useful, both pedagogically and as crude approximations to real physical scenarios.
J. Murray
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Thanks sir great explanation, can you also tell that the condition that wave function has sharp edges at location of infinite potential is true or not? – Jan 15 '21 at 15:52
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@LalitHooda From the Schrodinger equation, we have that $\psi'' \propto (V(x)-E) \psi$. So for example, if $V$ has a step discontinuity, then $\psi''$ will have a step discontinuity, which means that $\psi'$ is continuous but has a corner. Note that this is true for energy eigenstates, not for general wavefunctions. – J. Murray Jan 15 '21 at 16:02
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Well, let's review OP's argument for the 1D TISE at some fixed energy $E$. Consider first a finite tall wall $V\gg E$ of some fixed non-zero thickness $>0$. We find that the wavefunction has some characteristic penetration length (quantum tunnelling), which becomes smaller and smaller as the wall grows taller. Hence the wave function has to vanish inside an infinite wall of non-zero thickness $>0$.
But the Dirac delta potential is not of the above form. The support of a Dirac delta distribution is a single point of zero thickness. And in fact one may show that the wavefunction doesn't have to vanish at the support.
See also e.g. my related Phys.SE answer here.
Qmechanic
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in a 1D box, walls are of infinite height. When we solve TISE for it, we remove one constant from the general solution by arguing that w.f. has to be zero as walls are of infinite height.(At least I have been told so in classes), so are walls in 1 D box of finite width or the argument used to remove constant (what I know) is wrong.? – Jan 15 '21 at 17:04
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It is only guaranteed to apply to an infinite wall of non-zero thickness. – Qmechanic Jan 15 '21 at 17:49