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Thanks in advance.

- Thread starter Sferics
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- #1

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Thanks in advance.

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[tex]\langle x | y\rangle =\delta(x-y)[/tex]

For discrete quantities, the Kronecker delta is either one or zero. The Dirac delta function used here has to be something more like an infinite weight, in the handwaving way we tend to talk about these things. As it's neither a function, nor a number, there's no sense in which we can just divide by [itex]\delta(0)[/itex] to make the "norm" of such a state one.

Hope that helps.

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Fredrik

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$$\|g\|^2=\int(Ae^{ipx})^*(Ae^{ipx})dx=|A|^2\int dx.$$ There's clearly no choice of A that makes the right-hand side =1, since ##\int dx=\infty##.

Note that the space of wavefunctions is the semi-inner product space of square-integrable ##\mathbb C##-valued functions on ##\mathbb R##. What I did above shows that a momentum "eigenfunction" isn't square-integrable. This means that it's not actually a member of that semi-inner product space. So it's actually more appropriate to say that the momentum operator doesn't

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- #5

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If this isn't clear, think about some totally arbitrary continuum of values f, and think about how the identity operator acts on one specific eigenstate with eigenvalue [itex]f_0[/itex]:

[tex]|f_0\rangle=\int df \langle f|f_0\rangle |f\rangle [/tex]

- #6

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Hello!arbitraryoperator with a continuous spectra, it's eigenfunctions are not normalizable?

In this tread https://www.physicsforums.com/showthread.php?p=4036684#post4036684 tom.stoer linked an article that you could find interesting: http://http://arxiv.org/abs/quant-ph/9907069v2.

Ilm

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