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  1. All the linearly independent eigenfunctions of the parity operator $\mathcal{P}$ form an infinite set and all the linearly independent eigenfunctions of the unit operator $\bf 1$ also form an infinite set. Is the cardinality of these two sets same or different?

  2. Every function is an eigenfunction of the unit operator $\bf 1$. Right? Can any arbitrary function be written as the linear combination of the eigenfunctions of $\mathcal{P}$? If not, doesn't this imply that the set linearly independent eigenfunctions of unit operator $\bf 1$ forms a much bigger set than that of $\mathcal{P}$?

EDIT: I just want to know since $[\mathcal{P},\bf{1}]=0$, whether the total number of linearly independent eigenvectors of $\mathcal{P}$ and $\bf{1}$, forms a basis i.e., do they have same number of linearly independent eigenvectors and and whether both spans the space?

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  • $\begingroup$ The set of all possible combinations of a countable (Aleph-null) set is Aleph-one, but the mere fact that a countable set can be mapped to a subset (the selected linear combinations) doesn't make the two sets' sizes different. $\endgroup$ – Carl Witthoft Mar 18 '14 at 19:02
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To answer your second question: Yes, you can write any arbitrary function $f(x)$ as a linear combination of eigenfunctions of $\mathcal{P}$:

$$f(x)=\underset{=f_+(x)}{\underbrace{\frac{f(x)+f(-x)}{2}}}+\underset{=f_-(x)}{\underbrace{\frac{f(x)-f(-x)}{2}}}$$

with $\mathcal{P}f_+(x)=f_+(x)$ and $\mathcal{P}f_-(x)=-f_-(x)$.

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