Question

The time-independent Schrödinger equation for a nonrelativistic free particle of mass \(m\) is obtained from the energy relationship \(E=p^{2} /(2 m)\) by replacing \(E\) and \(p\) with appropriate derivative operators, as suggested by the de Broglie relations. Using this procedure, derive a quantum wave equation for a relativistic particle of mass \(m,\) for which the energy relation is \(E^{2}-p^{2} c^{2}=m^{2} c^{4},\) without taking any square root of this relation.

Short answer

In summary, the wave equation for a relativistic particle derived from the given energy relation without taking the square root is: \( \frac{\partial^2}{\partial t^2} \psi(\vec{r},t) + c^2 \nabla^2 \psi(\vec{r},t) = -\frac{m^2c^4}{\hbar^2} \psi(\vec{r},t) \) where \(\psi(\vec{r},t)\) is the wave function of the particle, \(\nabla^2\) is the Laplacian operator, and \(\hbar\) is the reduced Planck constant. This wave equation takes into account the relativistic effects by incorporating the energy and momentum operators from the de Broglie relations.

Step-by-step solution

Recall the de Broglie relations and the nonrelativistic Schrödinger equation

The de Broglie relations for a particle are given by: \( E = \hbar \omega \) \( p = \hbar k \) Here, \(\hbar\) is the reduced Planck constant, \(\omega\) is the angular frequency, and \(k\) is the wave number. Based on these relations, the nonrelativistic Schrödinger equation with the energy relationship \(E = \frac{p^2}{2m}\) can be written as: \( -\frac{\hbar^2}{2m} \nabla^2 \psi(\vec{r},t) = i\hbar \frac{\partial \psi(\vec{r},t)}{\partial t} \) where \(\psi(\vec{r},t)\) is the wave function of the particle, and \(\nabla^2\) is the Laplacian operator.

Analyze the given energy relation and form the equation

The energy relation for a relativistic particle is given by: \( E^2 - p^2c^2 = m^2c^4 \) Now, we need to replace the energy \(E\) and momentum \(p\) terms with their derivative operators from the de Broglie relations, as in the nonrelativistic case. We have: \( E \to i\hbar \frac{\partial}{\partial t} \) \( p \to -i\hbar \nabla \) Substitute these operators in the energy relation: \( \left( i\hbar \frac{\partial}{\partial t} \right)^2 - ( -i\hbar \nabla )^2c^2 = m^2c^4 \)

Simplify the equation and obtain the wave equation for a relativistic particle

Now simplify the equation replacing \(E\) and \(p\) with their respective operators from the previous step: \( \left( i\hbar \frac{\partial}{\partial t} \right)^2 - ( -i\hbar \nabla )^2c^2 = m^2c^4 \) Simplifying further, we obtain: \( -\hbar^2 \frac{\partial^2}{\partial t^2} \psi(\vec{r},t) -\hbar^2c^2 \nabla^2 \psi(\vec{r},t) = m^2c^4 \psi(\vec{r},t) \) Divide by \(-\hbar^2\): \( \frac{\partial^2}{\partial t^2} \psi(\vec{r},t) + c^2 \nabla^2 \psi(\vec{r},t) = -\frac{m^2c^4}{\hbar^2} \psi(\vec{r},t) \) This is the wave equation for a relativistic particle without taking the square root of the energy relation.