Question

What are the boundary conditions of wavefunction particle in a box?

Answer 1

To obtain this result, we note the boundary condi- tions, that the wave function vanish on the left and right boundaries. The left boundary condition, x=0 says ψ(0

Answer 2

Particle in a One-Dimensional Box

Particle in a 1-D box is a simple yet important concept in learning Quantum Mechanics.

Schrödinger's Equation for a one-dimensional system is given by:

$\displaystyle i\hbar \frac{\partial \Psi(x,t)}{\partial t}=\left(-\frac{\hbar^2}{2m} \frac{\partial^2}{\partial x^2}+\hat{V}}\right) \Psi(x,t)$

We solve this by taking $\Psi(x,t) = \psi(x).f(t)$. We will only require the Time Independent Part.

The Time Independent Schrödinger Equation is:

$\hat{H} \psi(x) = E \psi(x)$

We are given a Particle in a box. Let us assume the length of the box is L.

The Potential for such a system is defined by:

$\hat{V} = \begin{cases} 0 & \textsf{for 0\textless x\textless L} \\ \infty & \textsf{otherwise} \end{cases}$

This simply means that The Particle has zero potential inside the box. And infinite potential at the boundaries and outside.

This is to state that the particle cannot escape the box. It also cannot exist at the boundaries.

When we consider 0<x<L, we have $\hat{V} = 0$. From the time independent equation, we see:

$\displaystyle \hat{H} \psi(x)= E \psi(x) \\ \\ \implies -\frac{\hbar^2}{2m}\frac{\partial^2\psi(x)}{\partial x^2} = E \psi(x)$

$\psi(x)$ is an eigenfunction of the Hamiltonian Operator. Double Derivative gives back the same function.

So possible solutions for $\psi(x)$ include $\sin kx$, $\cos kx$, $e^{ikx}$, etc.

We will take a combination of cosine and sine functions, as follows:

$\psi(x) = C \cos kx + D \sin kx$

C and D can be complex numbers as well. We will now apply the boundary conditions for the particle.

-> The Particle cannot exist at the boundary defined by x=0

So we have

$\psi(0) = 0 \\ \\ \implies C\cos 0 + D \sin 0 = 0 \\ \\ \implies C + 0 = 0 \\ \\ \implies \bf C = 0$

Our Wave Function now reduces to:

$\psi(x) = D \sin kx$

We now apply the second boundary condition:

-> Particle also cannot exist at the boundary defined by x=L

We have:

$\psi(L) = 0 \\ \\ \implies D\sin kL = 0 \\ \\ \implies \sin kL=0 \\ \\ \implies kL = n\pi \\ \\ \implies \bf k=\dfrac{n\boldsymbol{\pi}}{L}$

Hence, Finally, after applying the boundary conditions, we have our wave function as

$\bf \boldsymbol{\psi}(x) = D \boldsymbol{\sin} \left( \dfrac{n\boldsymbol{\pi}}{L} x \right)$

Thus, The Boundary Conditions for a Particle in a One-Dimensional Box are that the particle cannot exist at the boundaries defined by x=0 and x=L.

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The following part is extra, and just for additional info.

We can apply the normalisation condition, to get the value of D:

$\displaystyle \int\limits_0^L \psi^*\psi \, dx = 1 \\ \\ \\ \implies D^2 \int\limits_0^L \sin^2 \left( \frac{n\pi}{L} x\right) dx = 1 \\ \\ \\ \implies D^2 \int\limits_0^L \frac{\left(1-\cos \left(\frac{2n\pi x}{L} \right)\right)}{2} \, dx = 1\\ \\ \\ \implies \frac{D^2}{2} \left[x-\frac{L}{2n\pi} \sin\left(\frac{2n\pi x}{L}\right)\right]\limits_0^L = 1 \\ \\ \\ \implies \frac{D^2L}{2}=1 \\ \\ \\ \implies \boxed{\bf D=\sqrt{\frac{2}{L}}}$

And Hence the Final Wave Function becomes:

$\boxed{\bf \boldsymbol{\psi}(x)=\sqrt{\frac{2}{L}} \, \boldsymbol{\sin} \left( \frac{n\boldsymbol{\pi} x}{L}\right)}$

Also, we can derive the energy by using the relation:

$\displaystyle \hat{H} \psi(x) = E \psi(x) \\ \\ \\ \implies -\frac{\hbar^2}{2m} \frac{\partial^2 \psi(x)}{\partial x^2} = E \psi (x) \\ \\ \\ \implies -\frac{\hbar^2}{2m} \frac{\partial^2}{\partial x^2} \left(D \sin\left(\frac{n\pi x}{L}\right)\right)=E \psi(x) \\ \\ \\ \implies -\frac{\hbar^2}{2m} \left(\frac{n\pi}{L}\right)^2 \left( -D \sin\left(\frac{n\pi x}{L}\right)\right) = E\psi(x) \\ \\ \\ Put \: \hbar=\frac{h}{2\pi} \\ \\ \\ \implies \frac{n^2h^2}{8mL^2} \psi(x)=E\psi(x)$

$\implies \boxed{\bf E_n = \frac{n^2h^2}{8mL^2}}$

This is the expression for the Energy for the particle.

So, this was all about Particle in a One Dimensional Box.