# System where the energy is discreet

• GSXR750
In summary, the conversation discusses the process of finding the values of a function in a discrete orthonormal base and a continuous base. The conversation also addresses a potential error in integrating over the wrong variable. The correct method is to integrate over dk and use the orthonormality of the base to find the values of a(k).

#### GSXR750

I know when you have a system where the energy is discreet (i.e. a bound state), there is an discreet orthonormal base, and you can developpe(?) an arbitrary function in this base:

f(x)=Sum{a[n]*f[n](x)}. And you can find the a[n] by multipling with f[n'] and integrate over x. Then you get a[n']=Int{dx*f(x)*f[n'](x)}.

Now I have a continu(?) base and I can do:

f(x)=Int{dk*a(k)*f[k](x)}.

From now on I begin to doubt:

If I multiply with f[k'](x), the RHS becomes:

Int{dk*a(k)*delta(k-k'))=a(k') and thus:

a(k')=f(x)*f[k'](x)

where did I screw up??

Last edited by a moderator:

Originally posted by GSXR750
If I multiply with f[k'](x), the RHS becomes:

Int{dk*a(k)*delta(k-k'))=a(k')

No, it doesn't. Instead, you get

Int(dk*a(k)*f[k](x)*f[k'](x))

Note the integration is over dk, not dx. So you can't simplify, since f[k'](x) does not depend on k.

I think you made the error of integrating over dx instead of dk.

Then how can I find tha a(k) 's??

a(k') = Int{dx * f[k'](x) * f(x)}

Proof:
As you said,
f(x)=Int{dk*a(k)*f[k](x)}.
Multiply by f[k'](x):
f[k'](x) * f(x) = Int{dk*a(k)*f[k](x)* f[k'](x)}.
Integrate over dx:
Int{dx * f[k'](x) * f(x)}=Int{dx *Int{dk*a(k)*f[k](x)* f[k'](x)}}
(Here, some theorem about exchanging integrations is applied)
= Int{dk * a(k) *Int {dx * f[k](x) * f[k'](x)}}
(Now, execute the integration over dx, using orthonormality)
= Int{dk * a(k) * delta(k-k')}
(Now, execute the integration over dk)
= a(k')

Now, drop the ' and there you are. OK?

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OK Thanks

## 1. What is a system where the energy is discreet?

A system where the energy is discreet refers to a system where energy is quantized or exists in discrete packets. This means that energy is not continuous and can only exist in specific amounts.

## 2. How is a system where the energy is discreet different from a continuous system?

In a continuous system, energy can exist in any amount and can be divided into smaller and smaller units. In a system where the energy is discreet, energy exists in specific discrete amounts and cannot be divided into smaller units.

## 3. What are some examples of systems where the energy is discreet?

Some examples of systems where the energy is discreet include quantum systems, such as atoms and subatomic particles, as well as electromagnetic radiation, like photons.

## 4. Why is it important to understand systems where the energy is discreet?

Understanding systems where the energy is discreet is important in many areas of science, including quantum mechanics, atomic physics, and chemistry. It also has practical applications in technologies such as lasers, solar cells, and computer memory.

## 5. How does the concept of a system where the energy is discreet relate to the laws of thermodynamics?

The laws of thermodynamics govern the behavior of energy in all systems, including systems where the energy is discreet. The first law states that energy cannot be created or destroyed, only transferred or converted. The second law states that in any energy transfer or conversion, some energy will be lost as heat. These laws apply to both continuous and discreet systems.