QM Vector Space-Linear Functions

[MrRobot]

Homework Statement

Hi, am having difficulty with the Linear algebra in QM. We have been given a problem set and one of the questions am struggling with is as follows:

Consider the space of all linear functions $f(x) = ax + b$ (x real) defined over the range $-1 < x < 1$, with the scalar product $\int_{-1}^{1} g(x)f(x) dx$.

Show that any arbitrary function $f(x) = ax + b$ can be written as $f(x) = c_1\psi_1(x) + c_2\psi_2(x)$, finding $c_1$ and $c_2$ using the scalar product. (i.e. show that $\psi_1(x)$ and $\psi_2(x)$ form a basis). $\psi_1(x) = 1/ \sqrt{2}$, $\psi_2(x) = \sqrt{3/2}x$

So i know that functions are complete in Hilbert Space and so can be written as $f(x) = \sum_{n = 1}^{\infty} c_n f_{n}(x)$, but now am not sure how to use this to answer the question or if am suppose to use it, please help. THANK YOU IN ADVANCE.

Homework Equations

The Attempt at a Solution

 

[andrewkirk]

You don't need to use that infinite sum, because the given space is only a tiny subspace of the function space. Intuitively, one can easily 'feel' that it's two-dimensional, since every function in it can be uniquely characterised by the two parameters a and b. By contrast, the unconstrained function space is infinite-dimensional.

To show that the two psi vectors form a basis you need to

(1) show that any function ax + b can be written as a linear sum of the two psis; and
(2) show that the two psis are linearly independent.

Start by showing that $\langle\psi_1|\psi_2\rangle=0$, which proves (2). Then look at the equation they gave you:
$$
f(x) = c_1\psi_1(x) + c_2\psi_2(x)
$$

If you take the scalar product of both sides with some function (what function?) you can, with some very easy integration, get the value of $c_1$. Then doing something similar with a different function can give you $c_2$. Then show that with those values of $c_1, c_2$, which will both be functions of $a$ and $b$, the equation holds, ie the function $f$ is a linear sum of the psis.

The method they have required you to use, involving scalar products, is much longer than the alternative of just equating coefficients, but I guess the idea is to get you used to operating with scalar products.

 

[MrRobot]

You don't need to use that infinite sum, because the given space is only a tiny subspace of the function space. Intuitively, one can easily 'feel' that it's two-dimensional, since every function in it can be uniquely characterised by the two parameters a and b. By contrast, the unconstrained function space is infinite-dimensional.

To show that the two psi vectors form a basis you need to

(1) show that any function ax + b can be written as a linear sum of the two psis; and
(2) show that the two psis are linearly independent.

Start by showing that $\langle\psi_1|\psi_2\rangle=0$, which proves (2). Then look at the equation they gave you:
$$
f(x) = c_1\psi_1(x) + c_2\psi_2(x)
$$

If you take the scalar product of both sides with some function (what function?) you can, with some very easy integration, get the value of $c_1$. Then doing something similar with a different function can give you $c_2$. Then show that with those values of $c_1, c_2$, which will both be functions of $a$ and $b$, the equation holds, ie the function $f$ is a linear sum of the psis.

The method they have required you to use, involving scalar products, is much longer than the alternative of just equating coefficients, but I guess the idea is to get you used to operating with scalar products.

Thank you very much andrewkirk this helped, It would have been very straight forward indeed to just match up the coefficients because I then did the matching after the long inner product calculation and my answers matched.

I used each of the psis to take inner product with the function $f$ as follows
$$
\langle\psi_1|f(x)\rangle = \langle\psi_1|c_1\psi_1 + c_2\psi_2\rangle
$$
$$
\int_{-1}^{1} \left(1/\sqrt{2}\right) \left(ax + b\right) dx = \langle\psi_1|c_1\psi_1\rangle + \langle\psi_1|c_2\psi_2\rangle
$$
and after some computation i got $c_1 = \sqrt{2}b$, did a similar thing for $c_2$, this time using $\psi_2$ and got $c_2 = \sqrt{2/3}a$ and substituting back really gave me $f(x) = ax + b$