Particle in a 1-Dimensional box: Difference between revisions
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We will now begin to solve for the constants '''A''' and '''B'''. | We will now begin to solve for the constants '''A''' and '''B'''. | ||
Recall that we have imposed the boundary conditions of the probability of finding the particle to be 0 anywhere outside of <math>x = 0</math> to <math>x = L</math>. This means that the wave function has to be continuous at the boundary. | Recall that we have imposed the boundary conditions of the probability of finding the particle to be 0 anywhere outside of <math>x = 0</math> to <math>x = L</math>. This means that the wave function has to be continuous at the boundary. | ||
First, consider the point <math>x = 0</math>. Here, <math>\psi(x) = 0</math>. For this to be true, we need to let '''B = 0'''. | |||
First, consider the point <math>x = 0</math>. Here, <math>\psi(x) = 0</math>. For this to be true, we need to let '''B = 0'''. This is because <math>sin(0) = 0</math>, but <math>cos(0) = 1</math>. If the <math>Bcos(kx)</math> term exists in our <math>\psi(x)</math>, then our wave function would not be cointinuous at the point <math>x = 0</math>. | |||
===A Computational Model=== | ===A Computational Model=== | ||
Revision as of 00:11, 23 April 2022
claimed by Eathan 4/14/2022 The time-independent Schrodinger Equation is a partial differential equation whose solutions describe the wave function of a quantum system. It is given in the following form. [math]\displaystyle{ -\frac{\hbar^2}{2m}\frac{\partial^2\psi(x)}{\partial x^2} + V(x)\psi(x)= E \psi(x) }[/math]
where
[math]\displaystyle{ \hbar }[/math] is the reduced Planck constant
[math]\displaystyle{ m }[/math] is the mass of the particle
[math]\displaystyle{ \psi(x) }[/math] is the wave function
[math]\displaystyle{ V(x) }[/math] is the potential of the system
[math]\displaystyle{ E }[/math] is the energy of the system
The particle in a 1-Dimensional box is a quantum system in which a particle is bounded in a well with infinite energy at the barrier. Since the barriers of the well are infinite, there should be a 0% chance of finding the particle outside the well. We will apply this boundary condition later as we derive the solution to the particle in a 1-Dimensional box
The Main Idea
Imagine the quantum system shown above, where the particle is bounded by infinite potential energy at [math]\displaystyle{ x = 0 }[/math] and [math]\displaystyle{ x = L }[/math]. The system also has 0 potential energy within the boundaries. We will use these as our boundary conditions to solve for the wave function [math]\displaystyle{ \psi(x) }[/math] below.
A Mathematical Model
We will begin to solve the Schrodinger equation with the boundary conditions as shown above. From the points [math]\displaystyle{ x = 0 }[/math] to [math]\displaystyle{ x = L }[/math], there is 0 potential energy. Using the time-independent Schrodinger Equation with [math]\displaystyle{ V(x) = 0 }[/math], we arrive at the following Schrodinger Equation: [math]\displaystyle{ -\frac{\hbar^2}{2m}\frac{\partial^2\psi(x)}{\partial x^2}= E \psi(x) }[/math]
Isolating the second partial derivative term gives us the following: [math]\displaystyle{ \frac{\partial^2\psi(x)}{\partial x^2}= -\frac{2mE}{\hbar^2}\psi(x) }[/math]
To simplify this, let us define the constant k as the following: [math]\displaystyle{ k = \frac{\sqrt{2mE}}{\hbar} }[/math]
Rewriting the Schrodinger Equation with the constant k gives us [math]\displaystyle{ \frac{\partial^2\psi(x)}{\partial x^2} + k^2\psi(x) = 0 }[/math]
The general solution to this partial differential equation is very well known and is given here: [math]\displaystyle{ \psi(x) = Asin(kx) + Bcos(kx) }[/math]
We will now begin to solve for the constants A and B. Recall that we have imposed the boundary conditions of the probability of finding the particle to be 0 anywhere outside of [math]\displaystyle{ x = 0 }[/math] to [math]\displaystyle{ x = L }[/math]. This means that the wave function has to be continuous at the boundary.
First, consider the point [math]\displaystyle{ x = 0 }[/math]. Here, [math]\displaystyle{ \psi(x) = 0 }[/math]. For this to be true, we need to let B = 0. This is because [math]\displaystyle{ sin(0) = 0 }[/math], but [math]\displaystyle{ cos(0) = 1 }[/math]. If the [math]\displaystyle{ Bcos(kx) }[/math] term exists in our [math]\displaystyle{ \psi(x) }[/math], then our wave function would not be cointinuous at the point [math]\displaystyle{ x = 0 }[/math].
A Computational Model
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