Particle in a box derivation question

Question

When deriving the energy equation for a particle in a box, the wavefunction for a particle was given as:

$$\Psi = A\sin(kx)+B\cos(kx)$$ Does anyone know how this function is arrived at? Was it experimental?

Answer 1

Does anyone know how this function is arrived at? Was it experimental?

Note that the wave function is not an observable and so cannot be arrived at experimentally.

The Schrödinger equation for the 1D potential well is given by:

$$-\frac{\hbar}{2m}\frac{\text{d}^2 \psi}{\text{d}x^2}=E\psi$$

Slightly re-written:

$$\psi''+k^2\psi=0\tag{1}$$

where:

$$k^2=\frac{2mE}{\hbar}$$

$(1)$ is a second order, linear, homogeneous differential equation and any decent math textbook will tell you it has the following solution, thus not experimentally obtained:

$$\psi(x)=A\sin kx + B\cos kx$$

where $A$ and $B$ are integration constants.

One of these values is determined by means of the boundary conditions $\psi(0)=\psi(L)=0$ (assuming the wave function acts on the domain$[0,L])$. So-called normalisation then determines the other. $k$ is also determined from the boundary conditions and so the quantisation of energy levels is established.

Answer 2

The eigenvalue equation obeyed by a particle in a box is a second order homogeneous linear differential equation of the form:

$$ -\frac{\hbar^2}{2m}\frac{d^2\psi(x)}{dx^2}-E\psi(x)=0. $$

The answer to your question is purely mathematical. More generally, for a homogeneous linear differential equation of any order:

$$ a_n\frac{d^ny}{dx^n}+a_{n-1}\frac{d^{n-1}y}{dx^{n-1}}+\cdots+a_1\frac{dy}{dx}+a_0y=0, $$

the trial solution $e^{qx}$ gives:

$$ a_nq^ne^{qx}+a_{n-1}q^{n-1}e^{qx}+\cdots+a_1qe^{qx}+a_0e^{qx}=0. $$

As each term now has the factor $e^{qx}$, we can reduce this equation to an $n$-order polynomial equation for $q$:

$$ a_nq^n+a_{n-1}q^{n-1}+\cdots+a_1q+a_0=0. $$

Therefore, all homogeneous linear differential equations of any order are solved in the same way: the exponential trial solution allows you to turn the differential equation into a polynomial equation, and then you only have to solve the latter for the roots of $q$.

Going back to your equation, the exponential trial solution $e^{qx}$ will lead to imaginary roots for $q$, so it is typically written as $e^{ikx}$, where $q=ik$, to get an equation for $k$ rather than $q$ with real roots. Plugging this into the eigenvalue equation for the energy gives:

$$ -\frac{\hbar^2}{2m}(ik)^2-E=0\Longrightarrow k=\pm\frac{\sqrt{2mE}}{\hbar}. $$

As expected from a second order equation, we get two roots, and calling the positive root $+k$ and the negative root $-k$, the general solution to your problem is:

$$ \psi(x)=Ae^{ikx}+Be^{-ikx}. $$

After you have the general solution, you can find your specific solution by applying boundary conditions. In the case of an inifinite square well between $-a/2$ and $a/2$, these would be $\psi(-a/2)=\psi(a/2)=0$. After applying boundary conditions, you can get to your solution in terms of sines and cosines (rather than exponentials) by using the standard relation between sines and cosines and complex exponentials.

I recently went through the maths solving the particle in a box in some detail, where I explicitly show all the steps in detail. You can find it here.

Answer 3

It is the simplest solution of the quantum mechanical differential equation with the potential shown below,

Schrodinger's equation is a wave equation, and the functions you show describe waves.

The experimental data , for example the double slit interference patterns for single electrons, showed that the probability of finding the particles has the interference patterns of waves. This simple solution was assigned to describe the wavefunction $Ψ$ of a particle in a box like potential well , with the interpretation that the $Ψ^*Ψ$ is the probability of finding the particle at an (x,y,z). The function you show is a general solution , before imposing boundary condition, which bounds will pick either the sine or the cosine.

Answer 4

Since, the case here is solved using Schrodinger's equation (a second order differential), we arrive at that particular solution theoretically as follows.

In the case of Particle in a box, $\mathbf V(x) = 0$, for a particle inside the box. And hence, the Schrodinger equation becomes, $$\frac{d^{2}\psi}{dx^2} + \mathbf {k^2}\psi (x) = 0$$ where, $$\mathbf k = \sqrt{\frac{2mE}{\hbar^2}}$$

Now, the general solution to such differential equation is given by, $$\psi (x) = C_1e^{ikx} + C_2e^{-ikx}$$

This can be further be written into, $$\psi (x) = C_1sin(kx) + C_2cos(kx)$$ (By using Euler's identity)

Now, if you're asking why can we write the solution of the equation in this fashion, then...

The general form of Second order differential equation is, $$y(x)'' + k^2y(x) = 0$$ the roots of this equation is given as:

$y(x) = C_1{y_1(x)} + C_2{y_2(x)}$ (by using the principle of superposition)

Any function $y_1$ and $y_2$ can satisfy the differential equation if it forms the general solution to the equation. We find that function by taking some trial function and substituting back in the main equation as one of the ways to find the solution.

So why we take exponential/trigonometric functions as trials? Here is an answer, https://math.stackexchange.com/questions/856476/why-do-we-chose-exponential-function-as-a-trial-solution-for-second-order-linear

To conclude, We saw that the solutions at which we arrived are all theoretically derived from mathematical equations. So, no, it is not experimentally derived.