Quantum mechanics for wave equation solution

In summary, the conversation discusses finding the values of 'a' for which the equation e^(a*x) is a valid solution to the differential equation d²ψ(x)/dx²=k²ψ(x). The participants also clarify the concept of a solution and how to test its validity.
  • #1
bfed
9
0
1. Homework Statement

consider the differential d²ψ(x)/dx²=k²ψ(x); for which values of a is the equation e^(a*x) is a solution to the above equation.

2. Homework Equations



3. The Attempt at a Solution
I have been working on this problem but I do not know how relate the 2 equations, or if I should use the Schrodinger equation. Any help is greatly appreciated.
 
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  • #2
Welcome to PF!

Hi bfed ! Welcome to PF! :smile:

(try using the X2 tag just above the Reply box :wink:)
bfed said:
consider the differential d²ψ(x)/dx²=k²ψ(x); for which values of a is the equation e^(a*x) is a solution to the above equation..

Just put ψ(x) = eax into d²ψ(x)/dx²=k²ψ(x), and solve for a. :wink:
 
  • #3
thanks tiny-tim,
so i should take the second derivative of ψ(x) = eax before I substitute it into d²ψ(x)/dx²=k²ψ(x) and solve for a?
 
  • #4
bfed said:
thanks tiny-tim,
so i should take the second derivative of ψ(x) = eax before I substitute it into d²ψ(x)/dx²=k²ψ(x) and solve for a?

uhhh? :confused:

just put it in! :smile:
 
  • #5
I think your confusion may come from from the idea of solutions to differential equations.
A solution is anything that we can put into this equation where after taking the derivatives and everything, the two sides equal each other. i.e. we are testing to see if our proposed solution of [tex]\Psi(x)[/tex] is valid. A good example to show would be one where the proposed solution was not valid. let's consider the proposed solution:

[tex]\Psi(x) = x^2[/tex]

Now if we plug this into our differential equation we obtain:


[tex]2 = k^2 x^2[/tex]

If k is a constant this is obviously not always true(example x=0). Therefore our proposed solution is not valid. In reality we want to find solutions that are valid. Let's propose a solution:


[tex]\Psi(x) = e^{ax}[/tex]

Now you go as I did in the invalid case, and plug [tex]\Psi(x)[/tex] into our differential equation. You will find that 'a' must equal something for the solution to be valid.
 
  • #6
thanks all, got'er done with your help!
-bfed
 

1. What is the Schrodinger equation and how does it relate to quantum mechanics?

The Schrodinger equation is a mathematical equation that describes how quantum states of a physical system evolve over time. It is a fundamental equation in quantum mechanics and allows us to calculate the probability of finding a particle in a particular state. It relates to quantum mechanics by providing a way to calculate the behavior of quantum systems, such as particles and atoms.

2. What is the difference between a wave function and a wave equation in quantum mechanics?

A wave function is a mathematical function that describes the quantum state of a physical system. It is used to calculate the probability of finding a particle at a particular location or with a particular energy. On the other hand, a wave equation is a differential equation that describes the behavior of a wave function over time. It is derived from the Schrodinger equation and is used to solve for the wave function in different systems.

3. How is the wave equation solved in quantum mechanics?

The wave equation is typically solved using mathematical techniques such as separation of variables, series expansion, and eigenvalue methods. These methods allow us to break down the complex equation into simpler equations that can be solved for the wave function. The solutions to the wave equation are then used to calculate the behavior of quantum systems.

4. Can the wave equation predict the exact behavior of quantum systems?

No, the wave equation can only predict the probability of finding a particle in a particular state. This is due to the probabilistic nature of quantum mechanics, where the exact behavior of a particle cannot be determined. The wave equation gives us information about the possible outcomes of a measurement, but it cannot predict the exact behavior of a particle.

5. How does the solution to the wave equation change in different environments?

The solution to the wave equation can change depending on the environment in which the quantum system is located. Factors such as the potential energy of the system, external forces, and interactions with other particles can all affect the behavior of the wave function. This is why the wave equation must be solved for each specific system in order to accurately predict the behavior of the quantum system.

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