From our previous post Monopoles-5-dirac-monopoles-part-1, we have for the wavefunction of the momentum operator, $$ -i\hbar \frac{\partial\psi}{\partial{x}} = -i\hbar\,e^{i\beta}\left[\frac{\partial\psi_1}{\partial{x}} + i\frac{\partial\beta}{\partial{x}}\right] = e^{i\beta} \left(-i\hbar\frac{\partial}{\partial{x}} + \hbar{k_x}\right)\psi_1$$
With all three components we get to know that, if the wavefunction $\psi$ satisfies any wave equation for the operator $\hat{p}$, then $\psi_1$ will satisfy the same wave equation for the operator $\hat{p} + \hbar\vec{k}$ Similarly, for the energy operator, $$ H\psi = i\hbar\frac{\partial\psi}{\partial{t}} = i\hbar\left[e^{i\beta}\frac{\partial\psi_1}{\partial{t}} + \psi_1ie^{i\beta}\frac{\partial\beta}{\partial{t}}\right] \\~\\ = e^{i\beta}\left[i\hbar\frac{\partial}{\partial{t}} - \hbar{k_0}\right]\psi_1$$ where the wave function $\psi_1$ will satisfy the wave equation for the operator, $ E - \hbar{k_0}$ The reason we take our wave function in the above given structure helps us to compare with a similar kind of wave equation, which we will encounter in the Gauge transformation of the wave function of free charge in an Electromagnetic field.
Let us look at the Schrodinger equation for a particle with mass "m" and charge "e" in an Electromagnetic field described by its potentials $\vec{A}\,,\,\phi$ as, $$ i\hbar\frac{\partial\psi}{\partial{t}} = H\psi = \frac{1}{2m}\left(\vec{p} - \frac{e\vec{A}}{c}\right)^2\psi + e\phi\psi$$
$$ H\psi = \frac{\vec{p}^2}{2m}\psi -\frac{e}{2mc}\left(\vec{p}\cdot\vec{A} + \vec{A}\cdot\vec{p}\right)\psi +\frac{e^2}{2mc^2}\vec{A}^2\psi +e\phi\psi$$
But wait.. What I have done above is wrong!!
The reason is because I treated the operators as some usual terms in multiplication and did the usual product.
Here is the main point when you deal with operators.
Never work with operators blindly without any function. You can do all operations with operators only after it acts on any function.
Let us try it again by acting on a function.
$$ \left(\vec{p} - \frac{e}{c}\vec{A}\right)^2\psi = \left(\vec{p} - \frac{e}{c}\vec{A}\right)\left(\vec{p\psi}-\frac{e}{c}\vec{A\psi}\right) = \left(\vec{p}^2\psi - \frac{e}{c}\vec{p}\cdot\vec{A\psi}-\frac{e}{c}\vec{A}\cdot\vec{p\psi}+\frac{e^2}{c^2}\vec{A}^2\psi\right)$$
But $$ \left(\vec{p}\cdot\vec{A}\right)\psi = \vec{p}\cdot\vec{A\psi} +\vec{p\psi}\cdot\vec{A} \\\rightarrow\,\,\,\,\,\,\vec{p}\cdot\vec{A\psi} =\left( \vec{p}\cdot\vec{A}\right)\psi - \vec{p\psi}\cdot\vec{A}$$
So, $$\left[\vec{p}^2\psi - \frac{e}{c}\left(\vec{p}\cdot\vec{A}\right)\psi - \frac{e}{c}\vec{A}\cdot\vec{p\psi}-\frac{e}{c}\vec{A}\cdot\vec{p\psi}+\frac{e^2}{c^2}\vec{A}^2\psi\right]$$ Combining the terms and substituting in the Hamiltonian, we get, $$ H\psi = \frac{1}{2m}\left[\vec{p}^2\psi - \left(\frac{e}{c}\vec{p}\cdot\vec{A}\right)\psi - 2 \frac{e}{c}\vec{A}\cdot\vec{p\psi}+\frac{e^2}{c^2}\vec{A}^2\psi\right]$$
or simply, $$H\psi = \frac{\vec{p}^2}{2m}\psi -\frac{e}{2mc}\left(\vec{p}\cdot\vec{A}\right)\psi + \frac{e}{mc}\vec{A}\cdot\vec{p}\psi +\frac{e^2}{2mc^2}\vec{A}^2\psi +e\phi\psi$$
From this, the application of gauge transformation transfers the potential to new values. Accordingly, to maintain the same structure and physical results, the wave function should be transformed into a new wave function.
It can be derived by substituting the new potentials in the Hamiltonian and comparing it with the old Hamiltonian.
The Gauge transformation is given by, $$ \vec{A'} = \vec{A}+\nabla\chi \\~\\ V' = V - \frac{1}{c}\frac{\partial\chi}{\partial{t}}$$ The transformation of the wave function is, $$ \psi' = \psi e^{\frac{ie\chi}{\hbar{c}}}$$
If I make the initial Potentials zero, then $$\vec{A} = 0 \,\,\,\,\rightarrow\,\,\,\, \vec{A'} = \nabla\chi\\ V = 0 \,\,\,\,\rightarrow\,\,\,\, V' = -\frac{1}{c} \frac{\partial\chi}{\partial{t}}$$ which says that if $\psi$ satisfies the Hamiltonian where there is no Electromagnetic field i.e. in free space, then $\psi'$ will satisfy the Hamiltonian with the Electromagnetic potentials given by the above relations.
This is exactly similar to the result we derived at the beginning. Comparing with the corresponding equations we get, $$ \beta = \frac{e\chi}{\hbar{c}}$$ and so, $$ \vec{A'} = \nabla\chi = \frac{\hbar{c}}{e}\left[\frac{\partial\beta}{\partial{x}}\hat{x} +\frac{\partial\beta}{\partial{y}}\hat{y} +\frac{\partial\beta}{\partial{z}}\hat{z}\right] = \frac{\hbar{c}}{e}\vec{k} $$ Similarly, $$ V' = \frac{-\hbar{c}}{e} \frac{1}{c} \frac{\partial\beta}{\partial{t}} = -\frac{\hbar{c}}{e}k_0$$ Thus, our initial wave equation now gets a physical meaning with its corresponding wave equation of a particle in an Electromagnetic field described by our Potentials.
We will try to analyze completely the physical implications imposed by our potentials in the next post.
With all three components we get to know that, if the wavefunction $\psi$ satisfies any wave equation for the operator $\hat{p}$, then $\psi_1$ will satisfy the same wave equation for the operator $\hat{p} + \hbar\vec{k}$ Similarly, for the energy operator, $$ H\psi = i\hbar\frac{\partial\psi}{\partial{t}} = i\hbar\left[e^{i\beta}\frac{\partial\psi_1}{\partial{t}} + \psi_1ie^{i\beta}\frac{\partial\beta}{\partial{t}}\right] \\~\\ = e^{i\beta}\left[i\hbar\frac{\partial}{\partial{t}} - \hbar{k_0}\right]\psi_1$$ where the wave function $\psi_1$ will satisfy the wave equation for the operator, $ E - \hbar{k_0}$ The reason we take our wave function in the above given structure helps us to compare with a similar kind of wave equation, which we will encounter in the Gauge transformation of the wave function of free charge in an Electromagnetic field.
Let us look at the Schrodinger equation for a particle with mass "m" and charge "e" in an Electromagnetic field described by its potentials $\vec{A}\,,\,\phi$ as, $$ i\hbar\frac{\partial\psi}{\partial{t}} = H\psi = \frac{1}{2m}\left(\vec{p} - \frac{e\vec{A}}{c}\right)^2\psi + e\phi\psi$$
$$ H\psi = \frac{\vec{p}^2}{2m}\psi -\frac{e}{2mc}\left(\vec{p}\cdot\vec{A} + \vec{A}\cdot\vec{p}\right)\psi +\frac{e^2}{2mc^2}\vec{A}^2\psi +e\phi\psi$$
But wait.. What I have done above is wrong!!
The reason is because I treated the operators as some usual terms in multiplication and did the usual product.
Here is the main point when you deal with operators.
Never work with operators blindly without any function. You can do all operations with operators only after it acts on any function.
Let us try it again by acting on a function.
$$ \left(\vec{p} - \frac{e}{c}\vec{A}\right)^2\psi = \left(\vec{p} - \frac{e}{c}\vec{A}\right)\left(\vec{p\psi}-\frac{e}{c}\vec{A\psi}\right) = \left(\vec{p}^2\psi - \frac{e}{c}\vec{p}\cdot\vec{A\psi}-\frac{e}{c}\vec{A}\cdot\vec{p\psi}+\frac{e^2}{c^2}\vec{A}^2\psi\right)$$
But $$ \left(\vec{p}\cdot\vec{A}\right)\psi = \vec{p}\cdot\vec{A\psi} +\vec{p\psi}\cdot\vec{A} \\\rightarrow\,\,\,\,\,\,\vec{p}\cdot\vec{A\psi} =\left( \vec{p}\cdot\vec{A}\right)\psi - \vec{p\psi}\cdot\vec{A}$$
So, $$\left[\vec{p}^2\psi - \frac{e}{c}\left(\vec{p}\cdot\vec{A}\right)\psi - \frac{e}{c}\vec{A}\cdot\vec{p\psi}-\frac{e}{c}\vec{A}\cdot\vec{p\psi}+\frac{e^2}{c^2}\vec{A}^2\psi\right]$$ Combining the terms and substituting in the Hamiltonian, we get, $$ H\psi = \frac{1}{2m}\left[\vec{p}^2\psi - \left(\frac{e}{c}\vec{p}\cdot\vec{A}\right)\psi - 2 \frac{e}{c}\vec{A}\cdot\vec{p\psi}+\frac{e^2}{c^2}\vec{A}^2\psi\right]$$
or simply, $$H\psi = \frac{\vec{p}^2}{2m}\psi -\frac{e}{2mc}\left(\vec{p}\cdot\vec{A}\right)\psi + \frac{e}{mc}\vec{A}\cdot\vec{p}\psi +\frac{e^2}{2mc^2}\vec{A}^2\psi +e\phi\psi$$
From this, the application of gauge transformation transfers the potential to new values. Accordingly, to maintain the same structure and physical results, the wave function should be transformed into a new wave function.
It can be derived by substituting the new potentials in the Hamiltonian and comparing it with the old Hamiltonian.
The Gauge transformation is given by, $$ \vec{A'} = \vec{A}+\nabla\chi \\~\\ V' = V - \frac{1}{c}\frac{\partial\chi}{\partial{t}}$$ The transformation of the wave function is, $$ \psi' = \psi e^{\frac{ie\chi}{\hbar{c}}}$$
If I make the initial Potentials zero, then $$\vec{A} = 0 \,\,\,\,\rightarrow\,\,\,\, \vec{A'} = \nabla\chi\\ V = 0 \,\,\,\,\rightarrow\,\,\,\, V' = -\frac{1}{c} \frac{\partial\chi}{\partial{t}}$$ which says that if $\psi$ satisfies the Hamiltonian where there is no Electromagnetic field i.e. in free space, then $\psi'$ will satisfy the Hamiltonian with the Electromagnetic potentials given by the above relations.
This is exactly similar to the result we derived at the beginning. Comparing with the corresponding equations we get, $$ \beta = \frac{e\chi}{\hbar{c}}$$ and so, $$ \vec{A'} = \nabla\chi = \frac{\hbar{c}}{e}\left[\frac{\partial\beta}{\partial{x}}\hat{x} +\frac{\partial\beta}{\partial{y}}\hat{y} +\frac{\partial\beta}{\partial{z}}\hat{z}\right] = \frac{\hbar{c}}{e}\vec{k} $$ Similarly, $$ V' = \frac{-\hbar{c}}{e} \frac{1}{c} \frac{\partial\beta}{\partial{t}} = -\frac{\hbar{c}}{e}k_0$$ Thus, our initial wave equation now gets a physical meaning with its corresponding wave equation of a particle in an Electromagnetic field described by our Potentials.
We will try to analyze completely the physical implications imposed by our potentials in the next post.
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