I'm reading 'Teach Yourself Quantum Mechanics' by Alexandre Zagoskin.

In chapter 2 he introduces Hilbert spaces by starting with the fact that a function may be defined by its Fourier coefficients. There is at this point a diagram implying that a point in Hilbert space may be defined using the coordinates on the axes, where these axes (the orthonormal bases of the Hilbert space?) are labelled sin(2.pi.x/L),sin(4.pi.x/L) etc.

What's not clear to me is whether in practice the Fourier coefficients are the usual way of defining the function in a Hilbert space, or whether this is one way of doing it, or whether it's one example of a manner in which to construct a Hilbert space.

Sorry if my terminology is imprecise.

  • $\begingroup$ Short overview: Hilbert spaces can be understood as infinite-dimensional vector spaces where each element is a function $f(x)$ that can be represented in different bases. The expansion in Fourier series is only one of them. Another example are Hermite polynomials. Which basis to use depends on what functions you want to approximate. For periodic functions, you would rather use a Fourier series. For functions that vanish quickly for $x \to \pm \infty$, you might use Hermite Polynomials. Do you have a more specific question? $\endgroup$
    – Cream
    Commented Apr 3, 2023 at 13:12
  • 1
    $\begingroup$ Many thanks Cream. Possibly my more specific question was that when I see a Dirac ket, and this appears from context to be equivalent to a wave function, is it ok to view the ket as a set of Fourier coefficients, which enables me to see straight away why they're equivalent. From your comment and the answer from @AwkwardWhale I think I understand that Fourier coefficients are as good as any basis. $\endgroup$
    – D R Ball
    Commented Apr 3, 2023 at 15:04

1 Answer 1


It's one particular orthonormal basis for the Hilbert space of $L^2$ functions on an interval. If you recall from linear algebra, if we have a given vector $\vec{v}$ in our space, and a fixed basis $\hat{e}_1,\hat{e}_2 \ldots$, then the vector has a unique representation in terms of its coefficients, $$ \vec{v} = \sum_i v_i \hat{e}_i. $$ However, our choice of basis is arbitrary, and we can obtain a new basis $\hat{f}_1,\hat{f}_2\ldots$ by a linear transformation $U$, $$ \hat{f}_i = \sum_j U_{ij} \hat{e}_j $$ The vector $\vec{v}$ will then have new coefficients for its representation in the new basis $$ \vec{v} = \sum_i w_i \hat{f}_i, $$ and the new coefficients can be obtained using the inverse of the transformation matrix, $$ w_i = \sum_j U_{ij}^{-1} v_j. $$ The key takeaway is the same holds analogously for a Hilbert space, but now our basis is a set of orthonormal functions. A given Hilbert space still has an arbitrary basis though.

Let's consider for example for $L^2$ functions on the interval $[-1,1]$. One choice of basis would be monic polynomials, $1,x,x^2\ldots$. A function can then be expressed in the form $$ f(x) = \sum_{n=0}^{\infty} f_n x^n. $$ However, we could also use Fourier series, giving us a different basis. In terms of Fourier series, the same function has a different representation $$ f(x) = \sum_{n=-\infty}^{\infty} a_n e^{i n \pi x}. $$

An illustrating example might be to take an exponential function $e^x$ for example. In terms of the monic polynomials, we can use Taylor expansion to find the expansion coefficients, which are just $$ f_n = \frac{1}{n!}. $$ For the Fourier series, you get instead $$ a_n = \frac{i\sinh(1 - in\pi)}{i + n\pi}. $$ The key thing to take away though is that both expansions give the same function $e^x$, and the Hilbert space generated by both choices of basis functions is the same, its just a linear change of basis.

  • $\begingroup$ To be clear then, the ket is a list of numbers, and they could be complex, and there could be an infinite number of them? $\endgroup$
    – D R Ball
    Commented Apr 6, 2023 at 13:12
  • $\begingroup$ No, taking once again functions on the interval, the ket is the function e^x itself. When you consider a particular basis of kets, you can find the coefficient of e^x by projecting the ket representing e^x onto the bra representing a given basis function. This braket is then evaluated as an inner product, which ends up being your usual orthogonality integral (such as the ones you use to find Fourier coefficients) $\endgroup$ Commented Apr 7, 2023 at 14:56
  • $\begingroup$ Using these projections, you can then expand the ket representing e^x into a sum of basis kets times the appropriate expansion coefficient. $\endgroup$ Commented Apr 7, 2023 at 14:57

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