Proving Completeness Relation for Box Normalization

[geoduck]

In one dimension the normalized momentum eigenstate for a particle with periodic boundary conditions of length L is: [tex]\psi_k(x)=\frac{1}{\sqrt{L}}e^{ikx} [/tex].

Is the completeness relation obvious:

[tex]\Sigma \psi_k(x)\psi_{k}(0)=\frac{1}{L}\Sigma e^{ikx}e^{-ik0}=\frac{1}{L}\Sigma e^{ikx}=\delta(x) [/tex]

where the sum is over discrete eigenstates k?

How would you go about proving that sum?

 

[Simon Bridge]

See pp3-4
http://hitoshi.berkeley.edu/221a/delta.pdf

 

[Bill_K]

Is the completeness relation obvious:
[tex]...\frac{1}{L}\Sigma e^{ikx}=\delta(x) [/tex]
where the sum is over discrete eigenstates k?

Hm, well the first thing to do is recognize that what you've written is false. The LHS is a periodic function with period L, while the RHS is not.

 

[Ravi Mohan]

The schrodinger equation is a Sturm–Liouville equation and thus its solutions must form a complete basis.
So what is wrong with his equation Bill?

 

[Bill_K]

The schrodinger equation is a Sturm–Liouville equation and thus its solutions must form a complete basis.
So what is wrong with his equation Bill?

The equation he's trying to prove is purely a mathematical identity involving nothing but exponentials and delta functions. Whether it is true or not cannot depend on how we were led to it, e.g. via Sturm-Liouville theory or Schrodinger quantum mechanics.

When you say the solutions must form a complete basis, the question is, basis in what space. And the answer is only the space of periodic functions with period L. The completeness relation you write down must adhere to this.

 

[geoduck]

When you say the solutions must form a complete basis, the question is, basis in what space. And the answer is only the space of periodic functions with period L. The completeness relation you write down must adhere to this.

I can see that if you have a periodic function, and integrate it with respect to the delta function representation sum that I have, you'll get the value of the function at x=0 (you just write the periodic function as a sum of the allowed plane waves and use orthonormality).

I just don't find it obvious that the infinite sum is zero everywhere outside of x=0. I guess you are adding a lot of phases and they have the potential to cancel. O well.

 

[Bill_K]

I just don't find it obvious that the infinite sum is zero everywhere outside of x=0. I guess you are adding a lot of phases and they have the potential to cancel. O well.

It isn't. It is nonzero at all points x = 0, ±L, ±2L, etc. What you should have written on the RHS is an infinite sum of delta functions, ∑ δ(x + nL). I know you are thinking that x is bounded in the range 0 to L, but a delta function δ(x) does not know that!

 

[Ravi Mohan]

When you say the solutions must form a complete basis, the question is, basis in what space. And the answer is only the space of periodic functions with period L. The completeness relation you write down must adhere to this.

You are right. I should have been more careful.

 

[Avodyne]

To "prove" the original identity: given a function f(x) on an interval -L/2 < x <L/2, we can compute the coefficients of a Fourier series that represents this function on this interval, and is periodic with period L outside of the interval. If we do this computation for δ(x), we get the OP's completeness relation (with appropriate factors of 2π).

I put "prove" in quotes because we have to worry about convergence of the series and other mathematical details.

 

[Bill_K]

To "prove" the original identity: given a function f(x) on an interval -L/2 < x <L/2, we can compute the coefficients of a Fourier series that represents this function on this interval, and is periodic with period L outside of the interval. If we do this computation for δ(x), we get the OP's completeness relation (with appropriate factors of 2π).

Nonsense, Avodyne. The function that is "δ(x) on the interval -L/2 < x <L/2 and periodic with period L outside" is not what anyone else would recognize as an ordinary delta function.

 

[Bill_K]

Here's a brief note on the properties of this sum of exponentials, pointing out as I said that it has an infinite number of equally spaced peaks.

 

[andrien]

Is the completeness relation obvious:

[tex]\Sigma \psi_k(x)\psi_{k}(0)=\frac{1}{L}\Sigma e^{ikx}e^{-ik0}=\frac{1}{L}\Sigma e^{ikx}=\delta(x) [/tex]

where the sum is over discrete eigenstates k?

How would you go about proving that sum?

If I remember it correctly,then it can be proved by going to L→∞,In that way the sum may be represented by Fourier integral using Ʃ→∫Ldk/2∏ and hence
1/L∫Ldk/2∏ e^(ikx) which is δ(x).But this integral conversion can only be justified on convergence and periodicity condition.

 

[Bill_K]

If I remember it correctly,then it can be proved by going to L→∞,In that way the sum may be represented by Fourier integral using Ʃ→∫Ldk/2∏ and hence
1/L∫Ldk/2∏ e^(ikx) which is δ(x).But this integral conversion can only be justified on convergence and periodicity condition.

But of course L=infinity is only a special case, and does not help to address the more general question when L is finite.

 

[andrien]

But of course L=infinity is only a special case, and does not help to address the more general question when L is finite.

yes,you are right.I will have to see some reference for the finite case.

 

[Avodyne]

Nonsense, Avodyne. The function that is "δ(x) on the interval -L/2 < x <L/2 and periodic with period L outside" is not what anyone else would recognize as an ordinary delta function.

I meant the original identity restricted to -L/2 < x <L/2.

 

[andrien]

Here is how we will prove it.We take an arbitrary well behaved function which can be expanded in terms of the eigenfunctions one got.We will take the normalized eigenfunction so it will be (1/√L)e^ikx,which we will denote by ψ_n(x) for different k.So
f(x)=Ʃ_nc_nψ_n(x),we multiply by ψ_m*(x) on both sides and integrate and use the orthonormality condition on right side and we get,
c_n=∫ψ_n*(x')f(x')dx',putting into f(x) we have
f(x)=Ʃ_n∫ψ_n*(x')ψ_n(x)f(x')dx',if we are allowed to change the order of summation and integration we have,
f(x)=∫f(x')dx'Ʃ_nψ_n*(x')ψ_n(x),and so it is evident that
Ʃ_nψ_n*(x')ψ_n(x)=δ(x-x'),putting the ψ_n(x) as we have written and putting x'=0,we get
(1/L)Ʃ_ne^ikx=δ(x)