Question

please help to solve this

Answer 1

Particle in a Box

The system we have here is known as Particle in a one-dimensional box in Quantum Mechanics.

The potential is defined in a simple way:

$\sf \hat{V}=\begin{cases} 0 & \textsf{for 0 \leq x \leq a} \\ \infty & \textsf{otherwise} \end{cases}$

We apply the Schrödinger's Equation to this system.

The Equation, in one-dimension is:

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

However, for systems like this, we are usually not concerned about the Time Dependence. Since we just want the eigenfunctions and energy eigenvalues, we may use the Time Independent Schrödinger's Equation, which is just:

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

Now, the Kinetic Energy term of the Hamiltonian $\sf \hat{H}$ is a double partial derivative with respect to x.

Let us see what the eigenfunctions can be. We have something like:

$\displaystyle \sf \frac{\partial^2 \psi(x)}{\partial x^2} \propto \psi(x)$

Clearly, The eigenfunction can be an exponential function, sinusoidal function, or maybe a combination of different exponential or sinusoidal functions.

So we may assume $\sf \psi(x)$ to be

$\psi(x)=Ae^{ikx}+Be^{-ikx} \\ \\ OR \\ \\ \psi(x)=A\sin kx +B\sin kx$

Both satisfy $\frac{\partial^2 \psi(x)}{\partial x^2} \propto \psi(x)$

Let us take the combination of sine and cosine functions and work with it. We have some constraints on the potential, and so we have constraints on the Wave Function too.

We have assumed:

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

where A, B, k are constants.

We have two boundary conditions.

1) At x=0, $\psi=0$

So,

$\psi(0)=0 \\ \\ \implies A\sin 0 + B \cos 0 = 0 \\ \\ \implies 0+B=0 \\ \\ \implies B=0$

Hence now our wavefunction becomes:

$\sf \psi(x) = A\sin kx$

2) At x=a, $\psi=0$

So

$\sf \displaystyle \psi(a)=0\\ \\ \implies A\sin ka=0 \\ \\ \implies \sin ka=0 \\ \\ \implies ka=n\pi \\ \\ \implies k=\frac{n\pi}{a}\qquad n\in \mathbb{Z}$

Appliying these boundary conditions, we get the wavefunction as:

$\sf \psi(x) = \A\sin \left( \frac{n\pi x}{a} \right)$

We can obtain the value of A as well. We need to normalise the wave function.

$\sf\displaystyle \int\limits_0^a\psi^*(x)\ \psi(x)\ dx=1\\\\\\\implies\int\limits_0^a \left(A\sin \frac{n\pi x}{a}\right)^2dx=1\\\\\\\implies \frac{A^2}{2}\int\limits_0^a\left(1-\cos \frac{2n\pi x}{a}\right)dx=1\\\\\\\implies\frac{A^2}{2}\left[x-\frac{a}{2n\pi}\sin \frac{2n\pi x}{a}\right]_0^a=1\\\\\\\implies \frac{A^2}{2}(a)=1\\\\\\\implies \boxed{\sf A=\sqrt{\frac{2}{a}}}$

Hence, the normalised eigenfunction is:

$\Large \boxed{\sf \psi(x)=\sqrt{\frac{2}{a}}\sin \left(\frac{n\pi x}{a}\right)}$

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Coming to the Energy Eigenvalue. For that we use the Time Independent Schrödinger's Equation.

We will use the eigenfunction as $\psi(x)=A\sin\frac{n\pi x}{a}$

$\displaystyle \sf \hat{H}\psi(x)=E\psi(x)\\\\\\\implies -\frac{\hbar^2}{2m}\frac{\partial^2}{\partial x^2} A\sin \frac{n\pi x}{a}=E\psi(x)\\\\\\\implies \frac{n^2\pi^2\hbar^2}{2ma^2}\ A\sin \frac{n\pi x}{a}=E\psi(x)\\\\\\\implies \frac{n^2\cancel{\pi^2}}{2ma^2}\ \frac{h^2}{4\cancel{\pi^2}}\ \bcancel{\psi(x)}=E\bcancel{\psi(x)}\\\\\\\\\implies \huge \boxed{\sf E_n=\frac{n^2h^2}{8ma^2}}$

The above is the expression for the Energy eigenvalue.

Hence, we derived the eigenfunction and energy eigenvalue for the particle in a one-dimensional box system.