# Quantum Chemistry/Example 32

## The Wavefunction of a Particle in a One-Dimensional Box

Despite the wavefunction having very little physical interpretation, the solution to the wavefunction of a particle is beneficial in order to determine the probability of such particle existing in a given area. When deriving the equation for the wavefunction of a particle in a one-dimensional box, we first begin by defining the Hamiltonian operators (i.e., kinetic and potential energy) for the given system, and substituting these values into the Schrodinger equation.

In quantum mechanics, the Schrodinger equation is given as,

 The Schrodinger Equation ${\displaystyle {\hat {H}}\psi =E\psi }$ ${\displaystyle \left({\hat {T}}+{\hat {V}}\right)\psi =E\psi }$ 

where ${\displaystyle {\hat {H}}}$ is the Hamiltonian operator, ${\displaystyle {\hat {T}}}$ is the kinetic energy operator, ${\displaystyle {\hat {V}}}$ is the potential energy operator, ${\displaystyle \psi }$ is the wavefuntion, and ${\displaystyle E}$ is the total energy of the system. The first equation is a simplification of the second equation, since ${\displaystyle {\hat {H}}={\hat {T}}+{\hat {V}}}$. Mathematically, the wavefunction ${\displaystyle \psi }$ is an eigenfunction of the Hamiltonian operator ${\displaystyle {\hat {H}}}$ with an eigenvalue of ${\displaystyle E}$.

### Defining the Hamiltonian

For the case of a particle in a one-dimensional box, its potential energy is equal to zero within the constraints of such box (i.e., ${\displaystyle 0\leq x\leq L}$), but the potential energy of the particle is infinite outside of these constraints:

 Potential Energy Range of Particle in 1D Box ${\displaystyle V(x)={\begin{cases}\infty ,&x<0\\0,&0\leq x\leq L\\\infty ,&x>L\end{cases}}}$

Additionally, the kinetic energy operator in the ${\displaystyle x}$ dimension is given as ${\displaystyle {\hat {T}}=-{\frac {{\hslash }^{2}}{2m}}{\frac {{d}^{2}}{{dx}^{2}}}}$. Therefore, since ${\displaystyle V(x)=0}$ inside of the box, the values defined for ${\displaystyle {\hat {T}}}$ and ${\displaystyle {\hat {V}}}$ can be substituted into the Schrodinger equation, which simplifies to ${\displaystyle {\frac {{d}^{2}\psi (x)}{{dx}^{2}}}=-{\frac {2E\cdot m}{{\hslash }^{2}}}\cdot \psi (x)}$.

### Boundary Conditions of the Wavefunction

The boundary conditions must then be identified for a particle in a one-dimensional box. The constraints of the box have been previously defined as the position of the particle, ${\displaystyle x}$, being contained within a box of length ${\displaystyle L}$. This means that the particle cannot exist when ${\displaystyle x=L}$ or when ${\displaystyle x=0}$. The boundary conditions for the value of the wavefunction must therefore be ${\displaystyle \psi (0)=0}$ and ${\displaystyle \psi (L)=0}$.

### General Solution

This is a second-order linear homogenous equation with respect to ${\displaystyle \psi (x)}$, and has a general solution of

${\displaystyle \psi (x)=A\cdot \sin(k\cdot x)+B\cdot \cos(k\cdot x)}$, where ${\displaystyle k={\frac {\sqrt {2\cdot m\cdot E}}{\hslash }}}$

Thus, the solution for the given second-order linear homogenous equation is

${\displaystyle \psi (x)=A\cdot \sin \left({\frac {\sqrt {2\cdot m\cdot E}}{\hslash }}\cdot x\right)+B\cdot \cos \left({\frac {\sqrt {2\cdot m\cdot E}}{\hslash }}\cdot x\right)}$

To determine the specific solution for this wavefunction, the constants ${\displaystyle A}$ and ${\displaystyle B}$ must be found.

### Determining the Unknown Variables

By substituting the first boundary condition that ${\displaystyle \psi (0)=0}$ into the general solution of the wavefunction, the value of ${\displaystyle B}$ may be found.

${\displaystyle \psi (0)=A\cdot \sin \left({\frac {\sqrt {2\cdot m\cdot E}}{\hslash }}\cdot 0\right)+B\cdot \cos \left({\frac {\sqrt {2\cdot m\cdot E}}{\hslash }}\cdot 0\right)}$

${\displaystyle \psi (0)=A\cdot \sin(0)+B\cdot \cos(0)=0}$

${\displaystyle A\cdot 0+B\cdot 1=0}$

${\displaystyle \therefore B=0}$

We now have ${\displaystyle \psi (x)=A\cdot \sin \left({\frac {\sqrt {2\cdot m\cdot E}}{\hslash }}\cdot x\right)}$. The second boundary condition, ${\displaystyle \psi (a)=0}$, may be applied.

${\displaystyle \psi (L)=A\cdot \sin \left({\frac {\sqrt {2\cdot m\cdot E}}{\hslash }}\cdot L\right)=0}$

The value of A cannot be equal to zero, otherwise the value of the wavefunction would be equal to zero at any position of ${\displaystyle x}$. So, ${\displaystyle \sin \left({\frac {\sqrt {2\cdot m\cdot E}}{\hslash }}\cdot L\right)=0}$.

The sine function is periodic, therefore, there are multiple values of ${\displaystyle x}$ at which ${\displaystyle \sin(x)=0}$. The values of x in which this is true are integer multiples of ${\displaystyle \pi }$. Thus, ${\displaystyle {\frac {\sqrt {2\cdot m\cdot E}}{\hslash }}\cdot L=n\pi }$. This can be further simplified by noting that the sine function is symmetric, therefore, the negative value of some ${\displaystyle n}$ will have the same wavefunction as its positive value.

Additionally, substituting ${\displaystyle n=0}$ into our above equation would result in a value of 0 for the wavefunction at every position.

${\displaystyle \psi (x)=A\cdot \sin(0\cdot \pi \cdot x)=A\cdot \sin(0)=0}$ Therefore, ${\displaystyle n\geq 1}$.

We can then plug ${\displaystyle {\frac {\sqrt {2\cdot m\cdot E}}{\hslash }}={\frac {n\pi }{L}}}$ back into our wavefunction equation, which gives ${\displaystyle \psi (x)=A\cdot \sin \left({\frac {n\pi x}{L}}\right)}$

The second property of a wavefunction is that it is normalized, meaning that the integral of the square of the wavefunction (i.e., the probability distribution) over all space must be equal to 1. We can now apply this to the above function to determine the constant value of ${\displaystyle A}$.

${\displaystyle \int _{0}^{L}(\psi (x))^{2}dx=1}$

${\displaystyle \int _{0}^{L}A^{2}\sin ^{2}\left({\frac {n\pi x}{L}}\right)dx=1}$

${\displaystyle A^{2}\int _{0}^{L}\sin ^{2}\left({\frac {n\pi x}{L}}\right)dx=1}$

The tabulated solution to this integral gives a constant value for ${\displaystyle A}$ as

${\displaystyle A={\sqrt {\frac {2}{L}}}}$

Therefore, the finalized specific solution of the wavefunction for a particle in a one-dimensional box is as follows.

${\displaystyle \psi (x)={\sqrt {\frac {2}{L}}}\sin \left({\frac {n\pi x}{L}}\right)}$, where ${\displaystyle n=1,2,3...}$

## Example

The wavefunction of an electrion moving along a platinum wire with length ${\displaystyle a}$ is given by ${\displaystyle \psi (x)=B(x^{3}-3x^{2}+2x)}$, where ${\displaystyle x}$ is the position of the electron. Outside the boundaries of this wire, the value of the wavefunction is zero (${\displaystyle \psi (x)=0}$). Determine the equation of the wavefunction of this electron by calculating ${\displaystyle B}$, the normalization constant, in terms of ${\displaystyle a}$. Is this normalized wavefunction a valid wavefunction? Explain why or why not.

As previously mentioned, one of the properties of a wavefunction is that it must be normalized. When normalizing the wavefunction, a normalization constant appears. Therefore, to determine the normalization constant ${\displaystyle B}$ for this wavefunction, we must integrate over all space and set it equal to 1.

${\displaystyle \int _{0}^{L}(\psi (x))^{2}dx=1}$

In this case, the length of the wire is an arbitrary constant ${\displaystyle a}$. Thus, ${\displaystyle L=a}$. Additionally, the equation for the wavefunction is ${\displaystyle B(x^{3}-3x^{2}+2x)}$, which can be substituted in for ${\displaystyle \psi (x)}$.

${\displaystyle \int _{0}^{a}(B(x^{3}-3x^{2}+2x))^{2}dx=1}$

${\displaystyle B^{2}\int _{0}^{a}x^{6}+9x^{4}+4x^{2}dx=1}$

${\displaystyle B^{2}\left[{\frac {x^{7}}{7}}+{\frac {9x^{5}}{5}}+{\frac {4x^{3}}{3}}\right]_{0}^{a}=1}$

${\displaystyle B^{2}\left({\frac {a^{7}}{7}}+{\frac {9a^{5}}{5}}+{\frac {4a^{3}}{3}}\right)=1}$

${\displaystyle B^{2}={\frac {1}{\frac {15a^{7}+329a^{3}}{105}}}}$

${\displaystyle \therefore B={\sqrt {\frac {105}{15a^{7}+329a^{3}}}}}$

To ensure that this is a valid wavefunction, one of two properties have been satisfied: the function is normalized. We must now ensure that the boundary conditions are valid as well (i.e., ${\displaystyle \psi (0)=0}$ and ${\displaystyle \psi (a)=0}$).

${\displaystyle \psi (0)={\sqrt {\frac {105}{15a^{7}+329a^{3}}}}\cdot 0=0}$

The first boundary condition is valid.

${\displaystyle \psi (a)={\sqrt {\frac {105}{15a^{7}+329a^{3}}}}(a^{3}-3a^{2}+2a)\neq 0}$

The value of the wavefunction at ${\displaystyle x=a}$ will never be equal to zero, since the length of the wire is always a nonzero value. Thus, because this wavefunction disobeys the second property of ${\displaystyle \psi (0)=0}$ and ${\displaystyle \psi (a)=0}$, it is not considered a valid wavefunction.