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}$
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$ P_{x} = \frac{\hbar}{i}\frac{\partial}{\partial x}$
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For three dimensional:-
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$\overrightarrow{P}= \frac{\hbar}{i} \overrightarrow{\nabla}$
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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}} $
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$ K = -\frac{\hbar^{2}}{2m}\frac{\partial^{2} }{\partial x^{2}} $
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For three dimensions:-
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$ K = -\frac{\hbar^{2}}{2m}\nabla^{2} $
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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
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$ \hat{H} = \left [ -\frac{\hbar^{2}}{2m}\frac{\partial^{2} }{\partial x^{2}} + V(x) \right ] $
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For three dimensions:-
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$\hat{H} = \left [ -\frac{\hbar^{2}}{2m}\nabla^{2} + V(x) \right ] $
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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
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$E = i \hbar \frac{\partial }{\partial t}$
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