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I have read in my textbook that dirac delta function is regarded as wave function. But isn't it violating the condition required for a wave function ie .

It becomes infinite $ δ(x-x_0)$ at $x=x_0$ but wave function should be finite everywhere How can this be explained??enter image description here

Book- Quantum physics by Hc verma, page 27, reprint 2018 edition.

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  • $\begingroup$ A delta function is sometimes considered as a potential well or potential barrier $V(x)$ in the Schrodinger equation. Could that be what you read, rather than $\psi(x)$ being a delta function? $\endgroup$ – G. Smith Aug 23 '19 at 18:40
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    $\begingroup$ In the book they clearly state that the wavefunction is not realistic; however, it may be useful to conceptually understand how a fully 'localised' particle wavefunction would look like if you collapsed it to a definite position. $\endgroup$ – Akerai Aug 23 '19 at 19:00
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    $\begingroup$ Who says the wavefunction has to be finite everywhere? $\endgroup$ – Javier Aug 23 '19 at 19:01
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    $\begingroup$ You are used to delocalized plane waves in coordinate space. Consider such in momentum space and inspect their Fourier transform, representing them in coordinate space. $\endgroup$ – Cosmas Zachos Aug 23 '19 at 19:20
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    $\begingroup$ @dmckee I understand that. My question was more of a rhetorical one; the point I would have made (if OP replies) is that a finiteness requirement is a bit naive: the value of a PDF at a point doesn't matter, what you want is square integrability. $\endgroup$ – Javier Aug 23 '19 at 19:49
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The delta function here can be thought of a position eigenstate, but as it is not square-integrable, it cannot be an actual wave function. Hence, "unrealistic". An actual, square-integrable wave function can be thought as being made of a superposition of dirac deltas, however: $$ \psi(x) = \int \psi(x_0) \delta(x - x_0) dx_0. $$

You can think of this as a superposition of dirac deltas at every possible $x_0$ along the real axis, with weight $\psi(x_0)dx_0$

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