How Is k Normalized in Schrodinger's Equation?

In summary: I don't understand how they got to that.In summary, the conversation discusses solving Schrodinger's equations for a potential distribution divided into three regions. The solution for one of the regions is of the form Ae^{kx}. When substituted into Schrodinger's equation, the value of k is found to be k = \frac{\sqrt{2mE}}{h}. However, the concept of normalizing k is introduced, which involves defining [E] as \frac{\pi^{2}h^{2}}{2ML^{2}} and finding an expression for k_hat. The conversation ends with a question about how [E] was derived and what the purpose of k_hat is.
  • #1
bumclouds
25
0
Hey guys,

Homework Statement


I have an assignment, which is to Solve Schrodinger's equations, for a certain potential distribution, which can be divided up into three regions.

A solution for one of the regions is of the form: Ae[tex]^{kx}[/tex]



If you substitute this into Schrodinger's equation (time independant, one dimension) and solve for k, you get this:

Schrodingers:
[tex]\frac{-h}{2m} \frac{d^{2}}{dx^{2}} \Psi (x) + Vo \Psi (x) = E \Psi (x)[/tex]

Solve for k:
k = [tex]\frac{\sqrt{2mE}}{h}[/tex]

I know this part is right because I've seen it written on the board a couple of times, and it's also what I get on paper.

But then there's the next bit, which I don't get. Apparently it's 'normalising k' which I just don't get..

k = [tex]\frac{\sqrt{2mE}}{h}[/tex]

[tex]k\overline{^}\overline{}[/tex] = [tex]\frac{\sqrt{2mE}}{h}.\frac{L}{2}[/tex]

[E] = [tex]\frac{\pi^{2}h^{2}}{2ML^{2}}[/tex]

k_hat = [tex]\frac{L}{2} . \frac{\sqrt{2m}}{h} . \sqrt{E}[/tex]

k_hat = [tex]\frac{L}{2} . \frac{\sqrt{2m}}{h} . \sqrt{\frac{E[E]}{[E]}}[/tex]

k_hat = [tex]\frac{L}{2} . \frac{\sqrt{2m}}{h} . \sqrt{[E]} \sqrt{Ehat} [/tex]

k_hat = [tex]\frac{\pi}{2} \sqrt{Ehat}[/tex]

Homework Equations



None.


The Attempt at a Solution



If I rearrange k = [tex]\frac{\sqrt{2mE}}{h}[/tex] and make E the subject,

I get ..

E = [tex]\frac{h^{2}khat^{2}}{2m}[/tex]

and maybe this is where I go wrong.. because I assume E = [E] ?

Subtituting [E] into the second last step in section 2 above yields:

[tex]khat = \frac{L}{2} khat \sqrt{Ehat}[/tex]

and that doesn't equal the last step >_<

EDIT: ahh.. if I use their definition of [E], [E] = [tex]\frac{\pi^{2}h^{2}}{2ML^{2}}[/tex]
I arrive at the right answer..


so how did they come up with [E] = [tex]\frac{\pi^{2}h^{2}}{2ML^{2}}[/tex]??
 
Last edited:
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  • #2
What is k_hat and what are you trying to achieve, an expression for E?
 
  • #3
Sorry! I've edited my above post.. Now it makes more sense :)

What I'm really after is how did they (my teacher) come up with [E] and what is it?


k_hat is apparently a normalised k
 

Related to How Is k Normalized in Schrodinger's Equation?

1. What is a 'normalisation' step in science?

A 'normalisation' step in science refers to the process of standardizing data or samples to a common scale or reference point.

2. Why is 'normalisation' important in scientific research?

'Normalisation' is important in scientific research because it allows for accurate comparisons between different data sets or samples. It also helps to remove any biases or variations that may exist between the data or samples.

3. How is 'normalisation' carried out in scientific experiments?

'Normalisation' can be carried out in various ways, depending on the type of data or samples being studied. Some common methods include using internal controls, statistical methods, or mathematical transformations to standardize the data.

4. What are the potential challenges or limitations of 'normalisation'?

One potential challenge of 'normalisation' is determining the appropriate reference point or scale to use. Additionally, some data sets or samples may be difficult to normalise due to inherent variations or biases that cannot be easily removed.

5. Can 'normalisation' be applied to all types of data or samples?

Yes, 'normalisation' can be applied to various types of data or samples, including numerical, categorical, and biological data. However, the specific method of normalisation may differ depending on the data type and research context.

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