Projection of wavefunction onto basis function

Question

I am given to believe that one way that one would could represent a wavefunction is by the expansion

$$\Psi(x) = \Sigma_n \Psi_n(x) = \Sigma_n f_n\phi_n(x) \tag{1}$$

where $\{\phi_n (x) \}$ is an orthonormal base of functions.

$$f_n = proj_{\phi_n} \Psi_n(x) \tag{2}$$

Firstly, is this correct? Or is the projection that is described supposed to be the following:

$$ f_n = proj_{\phi_n} \Psi(x) \tag{3} $$

Secondly, I am also given to believe that one way one can represent the projection described is

$$ f_n = \langle \phi_n(x),\Psi_n(x) \rangle \tag{4} $$

in Dirac notation. So, then when I am projecting functions onto basis functions and the basis functions are not normalized while the wavefunctions are, what is the appropriate expression to describe the projection? Is it like the equation for vectors, namely:

$$ proj_u v = \frac{u \cdot v}{u \cdot u} u \tag{5}$$

Like how is it expanded to functions?

Answer 1

Think of your functions just like vectors. You can add two (square integrable) functions and get another (square integrable) function. You can multiply a (square integrable) function by a scalar (pointwise) and get a (square integrable) function. So they add like vectors, they can be scaled like vectors. You even have a scalar product. So they are just like vectors. They might not live in a finite dimensional space, but they are vectors. The big difference is that if you have a maximal orthogonal set of vectors, an arbitrary vector might not be written as a finite linear combination, but instead only as an infinite linear combination.

So let's look at what you have:

$$\Psi(x) = \Sigma_n \Psi_n(x) = \Sigma_n f_n\phi_n(x) \tag{1}$$

where $\{\phi_n (x) \}$ is an orthonormal base of functions.

The above is fine, note that $\Psi,$ $\Psi_n,$ and $\phi_n$ are all functions (i.e. vectors) and $f_n$ is a scalar. Further note that $\Psi_n(x)=f_n\phi_n(x)$, and actually we'll see that $\Psi_n(x)=\operatorname{proj}_{\phi_n} \Psi(x)$.

Is $f_n = \operatorname{proj}_{\phi_n} \Psi_n(x)$ correct?

Not quite. Projection gives you a vector, whereas $f_n$ is a scalar. A correct equation is

$$\Psi_n = \operatorname{proj}_{\phi_n} \Psi(x).$$

Another correct equation is

$$ f_n = \langle \phi_n(x),\Psi_n(x) \rangle, \tag{4}$$

But what you probably meant is

$$ f_n = \langle \phi_n(x),\Psi(x) \rangle . $$

And that corresponds to $u \cdot v$, where $u=\phi_n$ is the thing to project onto, and $v=\Psi$ is the thing to be projected. So to finish, take:

$$ \operatorname{proj}_u v = \frac{u \cdot v}{u \cdot u} u \tag{5},$$

and get

$$ \operatorname{proj}_{\phi_n} \Psi = \frac{\langle\phi_n, \Psi\rangle}{\langle\phi_n, \phi_n\rangle} \phi_n . $$

Answer 2

In the position basis. $\Psi_n(x) = \langle \phi _n, \Psi \rangle \phi_n(x)$, and if the set of $\phi_n$ is a complete orthonormal set of functions, then:

$$\Psi(x) =\sum_n \langle \phi _n, \Psi \rangle \phi_n(x)$$

where $\langle \cdot , \cdot \rangle$ is an inner product, and is not really Dirac notation. $\langle \cdot | \cdot \rangle$ is the same thing in Dirac notation, but there's a bit more that's very very interesting to Dirac notation, which is when you write $\langle \cdot |$ alone.

You do not usually have an unnormalized complete basis. If you do, your first step is to normalize them, but yes, in the position basis you do wind up with the usual projection:

$$\Psi(x) =\sum_n \frac{\langle \phi _n, \Psi \rangle}{\langle \phi _n, \phi_n\rangle} \phi_n(x)$$ $$\Psi_n(x) =\frac{\langle \phi _n, \Psi \rangle}{\langle \phi _n, \phi_n\rangle} \phi_n(x)$$

in the position basis.

In Dirac notation, not in the position basis, if $|\phi _n\rangle$ is a complete set of orthonormal basis vectors.

$$\mathbf{1}=\sum_n |\phi _n\rangle \langle \phi _n|$$

Applying the identity to the wavefunction $|\Psi\rangle$. $$|\Psi\rangle=\mathbf{1}|\Psi\rangle=\sum_n |\phi _n\rangle \langle \phi _n|\Psi\rangle$$

Now moving to the position basis:

$$\Psi(x)=\langle x|\Psi\rangle=\sum_n \langle x|\phi _n\rangle \langle \phi _n|\Psi\rangle=\sum_n \phi _n(x)\langle \phi _n|\Psi\rangle$$

which is the same equation as before.

Answer 3

How is it expanded to functions?

I'm watching the Superbowl, so I'll give you a partial answer. For functions $f,g$ I would write $$\operatorname{proj}_gf=\frac{1}{||g||^2}\left(\int dx\,\bar{g}f\right)g$$ Also, TEX tip: use \operatorname{proj}.

Answer 4

Equation (3) is correct, (2) is not. But the $f_n$ is just the amplitude with which the basis function appears in the development of $\Psi$. The projection of $\Psi$ on $\phi_n$ is $f_n \phi_n $.

Eq. (4) is wrong, the correct amplitude of the projection is $f_n = \langle \phi_n(x),\Psi_n(x) \rangle $ .

Eq. (5) is correct.

About your last question it is not so clear. I just guess that you mean how will look like an expansion in an orthogonal, but non-normalized base. Well, it will look as follows

$$ v = \sum _n \frac{u_n \cdot v}{ u_n \cdot u_n} u_n \tag{5}$$