Wave function of a particle in free state

The wave function of a free particle:

Suppose, A particle of mass $m$ is in motion along the x-axis. Suppose no force is acting on the particle so that the potential energy of the particle is constant. For convenience, the constant potential energy is taken to b zero. Therefore, the time-independent Schrodinger equation for a free particle is:

$-\frac{\hbar^{2}}{2m} \frac{d^{2} \psi}{dx^{2}}=E \psi \qquad(1)$

Since the particle is moving freely with zero potential energy, its total energy $E$ is the kinetic energy given by


Where $p_{x}$ is the momentum of the particle which is moving along the x-axis.

$\frac{d^{2} \psi}{dx^{2}} + \frac{2mE}{\hbar^{2}} \psi =0$

Let $k^{2}=\frac{2mE}{\hbar^{2}} \qquad(2)$, Now substitute this value in the above equation that can be written as

$\frac{d^{2} \psi}{dx^{2}} + k^{2} \psi =0 \qquad(3)$

The solution of the above equation $(3)$

$\psi (x) = A e^{ikx} + B e^{-ikx} \qquad(4)$

Here $A$ and $B$ are constants.

The above equation$(4)$ gives the time independent of the wave function. The complete wave function (i.e for both time-dependent and independent ) for a particle is given by

$\psi(x,t)=\psi(x) e^{-i\omega t}$

$\psi(x,t)=\left( A e^{ikx} + B e^{-ikx} \right) e^{-i\omega t}$

$\psi(x,t)=A e^{\left(ikx-i\omega t \right)} + B e^{\left(-ikx -i\omega t \right)} $

$\psi(x,t)= A e^{-i\left(\omega t - kx \right)} + B e^{-i\left(\omega t + kx\right)} \quad(5)$

The above equation $(5)$ represents a continuous plane simple harmonic wave. The first term on the right side of the above equation $(5)$ represents the wave traveling in the positive x-direction, and the second term represents the wave traveling in the negative x-direction. Therefore, the wave function for the motion of a particle in the positive x-direction, we have

$\psi(x,t)= A e^{-i\left(\omega t - kx \right)} \qquad(6)$

The complex conjugate of the above wave function of free particle:

$\psi^{*}(x,t)= A e^{i\left(\omega t - k x \right)} \qquad(7)$

Eigenfunction and Eigen Value of linear momentum operator:

Now the momentum operator $\frac{\hbar}{i} \frac{\partial}{\partial x}$, operating on the equation $(6)$ i.e wave function of a free particle moving along the positive x-axis:

$\frac{\hbar}{i} \frac{\partial \psi(x,t)}{\partial x}= \frac{\hbar}{i} \frac{\partial }{\partial x} \left[ A e^{-i\left(\omega t - k x \right)} \right]$

$\frac{\hbar}{i} \frac{\partial \psi(x,t)}{\partial x}= \frac{\hbar}{i} \left[ A \left( ik \right)e^{-i\left(\omega t - kx \right)} \right]$

$\frac{\hbar}{i} \frac{\partial \psi(x,t)}{\partial x}= \hbar k \left[ A e^{-i\left(\omega t - kx \right)} \right]$

$\frac{\hbar}{i} \frac{\partial \psi(x,t)}{\partial x}= \hbar k \psi(x,t) \qquad(8)$

This equation shows that the wave function $\psi(x,t)$ for the particle is an eigenfunction of the linear momentum operator, and the momentum $p_{x}$ is the eigenvalue of the operator. Hence the momentum remains sharp with the value $p_{x}$

Probability of finding the free particle:

The probability of finding the position a particle in the region between $x$ and $x+dx$ is given by

$P \: dx =\psi(x,t) \psi^{*}(x,t) dx \qquad(8)$

Now substitute the value of $\psi(x,t)$ and $\psi^{*}(x,t)$ from equation $(6)$ and equation $(7)$ in above equation $(8)$, then we get

$P \: dx = A^{2} dx \qquad(9)$

Therefore the probability density $P$ for the position of the particle with the definite value of momentum is constant over the x-axis, i.e., All positions of the particle are equally probable. This conclusion is also obtained from the principle of uncertainty.

According to the interpretation of the wave function, the probability of finding the particle somewhere in space must be equal to $1$. i.e

$\int_{-\infty}^{+\infty} \psi(x,t) \psi^{*}(x,t) dx=1 \qquad(9)$

In this case $\psi(x,t) \psi^{*}(x,t) = A^{2}$ ,

There, the integral on the left side of the equation $(9)$ is infinite. Hence the wave function for the free particle cannot be normalized and $A$ must remain arbitrary. The difficulty arises because we are dealing with an ideal case. In practice, we can not have an absolutely free particle. The particle will always be confined within an enclosure in the laboratory, and hence its position can be determined. This means that its momentum cannot be determined with absolute accuracy.

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