Trying to understand Schrodinger's Equation

In summary: You have to understand that the n-state wave function is the solution of Schrodinger's equation. So, it must be satisfied with the boundary conditions (at x=0 and x=L).In summary, the conversation discusses the time-independent version of Schrodinger's equation and the derivation of a wavefunction for a particle in a confined box with infinite walls. The biggest wavelength is found to be 2L due to the requirements of the wave function being zero outside the box and continuous without any abrupt jumps. The value of n in the energy level is considered to be a state number and is a discrete value that satisfies the boundary conditions. En is the total energy and in this case, it is equal to the kinetic energy since the potential of
  • #1
hover
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Hello,

It's only been recently that I have acquired the math skills to deal with the time independent version of Schrodinger's Equation which is:

[tex] \frac{-\hbar^2}{2m} \frac{d^2}{dx^2}\Psi(x) + U(x)\Psi(x) = E\Psi(x)[/tex]

I tried to derive a wavefunction that deals with a particle in a confined box with infinite walls as shown http://hyperphysics.phy-astr.gsu.edu/hbase/quantum/pbox.html#c1"

I was able to get something similar to:

[tex] \Psi(x) = A*sin(kx)[/tex]

What somewhat baffles me is how they define lambda( the wavelength within k as shown http://hyperphysics.phy-astr.gsu.edu/hbase/quantum/schr.html#c2"). Perhaps I am missing the obvious but why would the biggest wavelength be
[tex] \lambda = 2L [/tex]
Why not have
[tex] \lambda = 4L [/tex]
Or
[tex] \lambda = 7L [/tex]
What makes [tex] \lambda = 2L [/tex] the biggest wavelength value. This would clear up a ton.
The second and last question I have(so far) is what particular value of n does a particle need to take? What does the value of n depend upon?

Thanks!
 
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  • #2
hover said:
[itex]\frac{-\hbar^2}{2\m} \frac{d^2}{dx^2}\Psi(x) + U(x)\Psi(x) = E\Psi(x)[/itex]
fx'd

In case you want to know, I'll show you a full proof of this.

First, let's define a function for this infinite potential well

U(x)={∞; x<0 or x>L}={0; 0≤x≤L}

So the schrodinger's equation for 0≤x≤L become:
(-ħ2/2m)ψ''=Eψ

We have
ψ'' + (2mE/ħ2)ψ = 0

The solution of an equation is
ψ=Acos(kx) + Bsin(kx)
where k=sqrt(2mE/ħ2) and A,B are constant.

from the condition: ψ(0)=0;
ψ(0) = Acos(0) + Bsin(0) = A = 0

Thus the equation become;
ψ=Bsin(kx)

also; ψ(L)=0;
ψ(L) = Bsin(kL) = 0
But for B≠0;
kL=n¶ or
L=n¶/k

but for k=2¶/λ

so
L = nλ/2

for the minimum state n; n=1
λ = 2L #

and that's the reason for this particular number.
guess it's clear :)
 
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  • #3
hover said:
What somewhat baffles me is how they define lambda( the wavelength within k as shown http://hyperphysics.phy-astr.gsu.edu/hbase/quantum/schr.html#c2"). Perhaps I am missing the obvious but why would the biggest wavelength be
[tex] \lambda = 2L [/tex]

It follows from these requirements:

1. The wave function must be zero outside the box, because of the "infinitely-high" walls.

2. The wave function must be continuous with no abrupt "jumps" from one point to the next.

This means that the equation you use for the wave function inside the box must evaluate to zero at the walls. What's the longest wavelength that gives you a node (zero point) at both walls?
 
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  • #4
Thanks for the responses! I like both solutions that are proposed here. On one hand, Black Integra shows how the math just makes the λ=2L fall right out. On the other hand, jtbell shows how the λ=2L falls right out by stating 2 requirements. The longest wavelength must be λ=2L since the wave gives you a zero point starting at a the beginning of wavelength and if you add a half the wavelength you end up at zero again. Very nice solutions.

Just one more question. How does one determine what value of n the wave is in? Since n relates to the amount of energy, does n always take the lowest possible value?

http://hyperphysics.phy-astr.gsu.edu/hbase/quantum/schr.html#c3"

[tex] E_n = \frac{n^2h^2}{8mL^2}[/tex]

Also is En a kinetic energy or potential energy? I assume it is a kinetic energy since that is how it seems to be described in the link.
 
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  • #5
No problems, hover :)

If you still prefer equations for an explanation, here we go.

What is the value of En?
From the relation I've said
L=n¶/k ----> k=n¶/L
and
(kħ)2=2mE

So we have a Total Energy at state n

En=(1/2m)(n¶ħ/L)2=(1/8m)(nh/L)2
It's total energy, due to the variable E in Schrodinger's equation.
Anyway, the potential of system is zero at 0≤x≤L.
So, in this case, total energy is equal to kinatic energy.

What exactly is the value n?
n is considered to be a state number of each system.

For example, the value of n in the energy level of hydrogen atom.
In this system, i.e. infinite square well, there're many states depending on the value n(or total energy).
Each energy will give different wave function(different n).
At n=1, it will give the lowest energy. This energy is called 'zero-point energy'

The value n is a discrete value. So it could not be any value. It just like a resonance of LRC-circuit or spectrum of Hg-atom.
 

FAQ: Trying to understand Schrodinger's Equation

1. What is Schrodinger's Equation and why is it important in science?

Schrodinger's Equation is a mathematical formula that describes how the quantum state of a physical system changes over time. It is important because it is a fundamental tool used in quantum mechanics to understand the behavior of particles and systems at the atomic and subatomic level.

2. How does Schrodinger's Equation relate to the famous "cat in a box" thought experiment?

The "cat in a box" thought experiment, also known as Schrodinger's cat, was proposed by physicist Erwin Schrodinger to illustrate the concept of superposition in quantum mechanics. Schrodinger's Equation is used to mathematically describe the probability of the cat being in a state of being alive or dead while inside the box.

3. Is Schrodinger's Equation difficult to understand for non-scientists?

Yes, Schrodinger's Equation involves complex mathematical concepts that can be difficult to grasp for non-scientists. However, there are simplified explanations and visual representations available that can help in understanding the basic principles behind it.

4. Can Schrodinger's Equation be applied to macroscopic objects?

No, Schrodinger's Equation is only applicable to microscopic particles and systems at the quantum level. It does not accurately describe the behavior of macroscopic objects, which are governed by classical mechanics.

5. How has Schrodinger's Equation been used in practical applications?

Schrodinger's Equation has been used in various practical applications, such as in the development of quantum computers, understanding the properties of materials at the nanoscale, and in the study of chemical reactions. It has also been used in fields such as biology and neuroscience to better understand the behavior of biological systems.

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