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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?

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  • $\begingroup$ Why don't you put numbers to your equations? To your first question, the 2nd answer is correct. $\endgroup$ – Sofia Feb 1 '15 at 23:08
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    $\begingroup$ Next, no, $f_n = \langle \phi _n, \Psi \rangle $. $\endgroup$ – Sofia Feb 1 '15 at 23:11
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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

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

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  • $\begingroup$ Thank you! Makes a lot more sense now that you have made the distinction between vectors and scalars here. $\endgroup$ – NewDogOldTricks Feb 2 '15 at 6:56
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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.

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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}.

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  • $\begingroup$ So it goes when you try to do two things at once. (They say that Napoleon was able to do so.) What you complicate the fellow? Don't you see the his base is discrete? I wish you a lot of enjoyment with the Superbowl and tomorrow tell us if you were pleased with the result! $\endgroup$ – Sofia Feb 1 '15 at 23:58
  • $\begingroup$ @Sofia: Saw "functions" and thought he wanted integrals. Did I misread that? F*ck the Patriots. $\endgroup$ – Ryan Unger Feb 2 '15 at 0:02
  • $\begingroup$ @Sofia The answer is fine, the integral is from the inner product. The integral gives a number, which is exactly what you want for a discrete basis. $\endgroup$ – Timaeus Feb 2 '15 at 0:12
  • $\begingroup$ @Timaeus : I am not sure that 0celo7's answer is fine. I will tell him too. When I have time, I am going to post such a question. Putting it simply, the functions associated with the continuous spectrum of Hamiltonians are problematic. I have a severe problem in my own research, with such functions. And the spectral theorem doesn't help much when the spectrum is continuous. I am afraid that for such functions one cannot rely on their orthogonality without being cautious and proving it for the particular base with which one works. $\endgroup$ – Sofia Feb 2 '15 at 13:24
  • $\begingroup$ @0celo7 : what happens when $||g||$ is $\infty$ ? I looked again at the spectral theorem, and it is not of big help. I am going to post a question. $\endgroup$ – Sofia Feb 2 '15 at 13:48
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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

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

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  • $\begingroup$ Equation 3 is wrong, $f_n$ is a scalar, but the projection should give a vector. And equation 4 isn't wrong in the sense that it is technically correct, but it is just not helpful (except for showing that the projection is idempotent). $\endgroup$ – Timaeus Feb 2 '15 at 0:08
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    $\begingroup$ The right hand sides of equation 2 and equation 3 are equal as are the left hand sides, so you can't say one is right and one is wrong (and they are both wrong). And equation 4 is correct, and it's also what you wrote to replace it with, so again you can't say that it is both wrong and correct (they are both correct but unhelpful) $\endgroup$ – Timaeus Feb 2 '15 at 0:34
  • $\begingroup$ @This is why you gave me a minus? Can't you see that I explained him what you said in your 1st comment ? And your 2nd comment I'll read later. Anyway I have objections to both, but not now. $\endgroup$ – Sofia Feb 2 '15 at 0:44
  • $\begingroup$ I think you are helpful to stackexchange, and thus I'm more than happy to change my vote if you correct your answer. I also thought your equation 5 was helpful, but I thought your comments about which equations are right or wrong was unhelpful. Specifically, (2) and (3) are equally true/false (and both are false) and (4) and your replacement of 4 are equally true/false (and both are true, but unhelpful). The community wiki answer made the same errors, and I edited the answer to be correct, so if you truly think I'm wrong you should consider editing that answer instead of editing your answer. $\endgroup$ – Timaeus Feb 2 '15 at 0:57
  • $\begingroup$ @Timaeus thank you both for the good will and for the appreciation. It is known to me that the projection has to comprise the vector too. But it's matter of slang. Please see an example. Given the vector $\vec r = x \vec i + y \vec j$ what do we say that it is projection on the axis $x$ ? Many times we say it's $x$, and on the axis $y$ we say it's $y$. This is why I didn't know what to do with this, and I explained him in other words, as you saw. Of course I would be glad to change, but I am afraid that I may confuse him. What you say? I'd sincerely appreciate advice. $\endgroup$ – Sofia Feb 2 '15 at 1:08

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