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I have a question involving the representation of operators in momentum representation and position representation. The question is a little long, so I'll do my best to explain it.

We are given an operator $g$ from $\mathcal{l}^2(\mathbb{Z})$ to $\mathcal{l}^2(\mathbb{Z})$, i.e., the space of functions that are square summable over $\mathbb{Z}$ such that $$(g\phi)=\phi(n+1)+\phi(n-1)-2\phi(n).$$ We are then asked to find a value $x$ such that $((g)e^{ink})=xe^{ink}$. A quick calculation yields $x=(e^{ik}+e^{-ik}-2)$. It is obvious that $e^{ink}$ is not an element of $\mathcal{l}^2(\mathbb{Z})$, however.

Using the Fourier Transform $\mathcal{F}:L^2([0,\pi)]\rightarrow L^2([0,2\pi)]$, where $L^2([0,\pi)]$ is the space of functions that are square integrable over $(0,2\pi)$, we can then put $g$ in momentum representation using $\mathcal{F}^{-1}g\mathcal{F}$, and we get that $$(\mathcal{F}^{-1}g\mathcal{F}\phi)(n)=(e^{ik}+e^{-ik}-2)\sum<e^{ink},\phi>e^{ink}$$ (where the sum is over the integers).

We are then asked to relate $e^{ink}$ from the first part of the the question to the operator in momentum representation; obviously, the sequence appears in the sum, and we take $$(\mathcal{F}^{-1}g\mathcal{F}e^{ink})=(e^{ik}+e^{-ik}-2)e^{ink},$$ thus we get the same value for $x$ in both parts.

We are then asked if $g\geq0$, which is where I'm having trouble. My intuition tells me it's not, as for example, in the momentum representation $\mathcal{F}^{-1}g\mathcal{F}e^{2\pi i}=-4$ and for $e^{ink}$ in both spaces they have the same value for $x$; but I have no idea if this is correct, or if I'm getting the gist of the question. Any help is appreciated.

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    $\begingroup$ Can you please clean up your formatting a little? This is a bit hard to read with your math and text jumbled together. $\endgroup$ – Cameron Williams Oct 9 '14 at 23:10
  • $\begingroup$ Cleaned it up a little bit, hope it helped. $\endgroup$ – TinaBelcher Oct 9 '14 at 23:30

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