Proofs with current density and wavefunctions

In summary, the conversation involves a problem that is similar to a homework problem, but the person is unsure about certain aspects of it. They are wondering why \phi=\phi* in the problem and how e^{-i\phi}e^{i\phi}+e^{i\phi}e^{-i\phi} cancels out along with 1/2i. They have attached the proof to the forum and are looking for help in understanding the math behind it. Another individual points out that there may be a missing factor of i when plugging back in and explains that e^{i\phi}e^{-i\phi} = e^{0} = 1 using properties of exponentials and Euler's formula. They
  • #1
Mary
11
0

Homework Statement



So I was able to find a problem that was kind of similar to a homework problem that I am working on. Unfortunately, I'm not quite sure what is going on partially within the problem.

In the problem they state that [itex]\phi[/itex]=[itex]\phi[/itex]*, but it does not state why. I was wondering if someone could explain this to me?

Also, later on in the problem e[itex]^{-i\phi}[/itex]e[itex]^{i\phi}[/itex]+e[itex]^{i\phi}[/itex]e[itex]^{-i\phi}[/itex] cancels out along with the 1/2i. I'm not quite sure how this happened either. I have tried to review the euler's equation to understand where this comes from because I'm supposing it is being canceled out based on its properties.

I have attached the proof to this forum.


I would greatly appreciate some help because the problem I am working on involves proving that the integral of the current density over a surface is equal to the momentum operator/m. I understand conceptually what this means but I need a push in the write direction with the math behind it.
 

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  • #2
You seem to be missing a factor of [itex]i[/itex] when you plug back in. The [itex]i[/itex] factors come from the [itex]\frac{\partial}{\partial x}[/itex].

Also,

[itex]e^{i\phi}e^{-i\phi} = e^{0} = 1[/itex].

This comes from the properties of exponentials, but you can show this using Euler's formula as well. Note [itex]\left(a + ib\right)\left(a-ib\right) = a^2 + b^2[/itex], and [itex]\sin\left(-\phi\right) = -\sin\phi[/itex].
 

Related to Proofs with current density and wavefunctions

1. What is the significance of current density in proofs involving wavefunctions?

Current density is a measure of the flow of electric charge within a material. In proofs involving wavefunctions, current density is used to calculate the probability of an electron being in a particular location. This is because the magnitude of the wavefunction squared represents the probability density of finding an electron at a given point in space.

2. How is current density related to the wavefunction?

Current density is directly proportional to the gradient of the wavefunction. This means that as the wavefunction changes in space, the current density also changes, indicating the flow of charge. In other words, the current density is a measure of how quickly the wavefunction changes.

3. How do we incorporate current density into Schrödinger's equation?

Schrödinger's equation is a fundamental equation in quantum mechanics that describes how the wavefunction of a quantum system evolves over time. In this equation, the term for current density is included by multiplying the gradient of the wavefunction by the charge of the particle and dividing by its mass.

4. Can we use current density to predict the behavior of an electron?

Yes, current density can be used to predict the behavior of an electron by calculating the probability of finding it at a given location. This can help us understand how electrons move and interact with other particles in a material, allowing us to make predictions about its properties and behavior.

5. How is current density related to electrical conductivity?

Electrical conductivity is a measure of how well a material can conduct electricity. Current density is directly related to this, as it represents the flow of charge within a material. Materials with higher current density have higher electrical conductivity, meaning they are better conductors of electricity.

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