Quantum Mechanical Operators

Operator →

An operator is defined as a mathematical term that is used in the operation of a function so that this function may or may not be transformed into another function.

Operators of Quantum Mechanics →

There are the following quantum mechanical operators which are used in the wave function of particles:-

  • Momentum Operator

  • Kinetic Energy Operator

  • Total Energy Operator (Hamiltonian Operator)

  • Total Energy Operator in terms of the differential with respect to time


  • Momentum Operator →

    The wave function for a free particle moving along the position $x$-direction is

    $\psi(x,t)=A e^{\frac{i}{\hbar}(P_{x}x-Et)}$

    Differentiate the above equation with respect to $x$ then we get

    $\frac{\partial \psi}{\partial x}= A e^{\frac{i}{\hbar}(P_{x}x-Et)} \frac{i}{\hbar} P_{x} $

    $\frac{\partial \psi}{\partial x}= \psi \frac{i}{\hbar} P_{x} $

    $ P_{x} \psi = \frac{\hbar}{i}\frac{\partial \psi}{\partial x}$

    $ P_{x} = \frac{\hbar}{i}\frac{\partial}{\partial x}$

    For three dimensional:-

    $\overrightarrow{P}= \frac{\hbar}{i} \overrightarrow{\nabla}$

    Kinetic Energy Operator →

    We know that the momentum operator

    $ P_{x} \psi = \frac{\hbar}{i}\frac{\partial \psi}{\partial x} \qquad(1)$

    Differentiate the above equation $(1)$ with respect to $x$ then we get

    $ P_{x} \frac{\partial \psi}{\partial x} = \frac{\hbar}{i}\frac{\partial^{2} \psi}{\partial x^{2}} \qquad(2)$

    Now substitute the value of $\frac{\partial \psi}{\partial x}$ from equation $(1)$ to equation $(2)$

    $\frac{\hbar}{i}\frac{\partial^{2} \psi}{\partial x^{2}} = P_{x} \frac{i}{\hbar} P_{x} \psi$

    $\frac{\hbar^{2}}{i^{2}}\frac{\partial^{2} \psi}{\partial x^{2}} = P_{x}^{2} \psi$

    $ -\hbar^{2}\frac{\partial^{2} \psi}{\partial x^{2}} = P_{x}^{2} \psi \qquad (\because i^{2}=-1)$

    $ -\frac{\hbar^{2}}{2m}\frac{\partial^{2} \psi}{\partial x^{2}} = \frac{P_{x}^{2}}{2m} \psi \qquad {3}$

    $ -\frac{\hbar^{2}}{2m}\frac{\partial^{2} \psi}{\partial x^{2}} = K \psi \qquad (\because \frac{P_{x}^{2}}{2m} = K)$

    $ K \psi = -\frac{\hbar^{2}}{2m}\frac{\partial^{2} \psi}{\partial x^{2}} $

    $ K = -\frac{\hbar^{2}}{2m}\frac{\partial^{2} }{\partial x^{2}} $

    For three dimensions:-

    $ K = -\frac{\hbar^{2}}{2m}\nabla^{2} $

    Total Energy Operator (Hamiltonian Operator) →

    The total energy of the particle moving along $x4-aix is given by

    $E=\frac{P_{x}^{2}}{2m} + V(x) \qquad(1)$

    Where V(x) → Potential Energy

    We know that the kinetic energy operator

    $ K = -\frac{\hbar^{2}}{2m}\frac{\partial^{2} }{\partial x^{2}} $

    $ \frac{P_{x}^{2}}{2m} = -\frac{\hbar^{2}}{2m}\frac{\partial^{2} }{\partial x^{2}} \qquad (\because K=\frac{P_{x}^{2}}{2m})$

    Now substitute the value of $ \frac{P_{x}^{2}}{2m}$ in equation$(1)$

    $E= -\frac{\hbar^{2}}{2m}\frac{\partial^{2} }{\partial x^{2}} + V(x)$

    Multiply $\psi$ on the both side of above equation

    $E \psi= -\frac{\hbar^{2}}{2m}\frac{\partial^{2} \psi }{\partial x^{2}} + V(x) \psi$

    $E \psi= \left [ -\frac{\hbar^{2}}{2m}\frac{\partial^{2} }{\partial x^{2}} + V(x) \right ] \psi$

    $E \psi= \hat{H} \psi$

    So the total energy operator

    $ \hat{H} = \left [ -\frac{\hbar^{2}}{2m}\frac{\partial^{2} }{\partial x^{2}} + V(x) \right ] $

    For three dimensions:-

    $\hat{H} = \left [ -\frac{\hbar^{2}}{2m}\nabla^{2} + V(x) \right ] $

    The total energy operator is denoted by $\hat{H}$ and called the Hamiltonian Operator.


    Total Energy Operator in terms of the differential with respect to time →

    We know that the wave function

    $\psi= A e^{\frac{i}{\hbar}}\left( P_{x}x - Et \right)$

    Differentiate the above equation $(1)$ with respect to $t$ then we get

    $\frac{\partial \psi}{\partial t}= A e^{\frac{i}{\hbar}(P_{x} x -Et)} \frac{i}{\hbar} (-E) $

    $\frac{\partial \psi}{\partial t}= - \frac{i}{\hbar} E \psi $

    $E \psi= -\frac{\hbar}{i} \frac{\partial \psi}{\partial t}$

    $E \psi= i^{2} \frac{\hbar}{i} \frac{\partial \psi}{\partial t} \qquad (\because i^{2}=-1)$

    $E \psi= i \hbar \frac{\partial \psi}{\partial t}$

    This energy operator is denoted by $E$ so

    $E = i \hbar \frac{\partial }{\partial t}$

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