We begin as one would expect with Schrodinger's equation in one dimension; which is
\( -\frac{\hbar}{2m}\frac{d^2 \psi }{dx^2} + V \psi = \frac{i \hbar \partial \psi}{\partial t} \) . If we solve the solve this using separation of variable we get a term that is only time dependent and the other that is only spatially dependent. The steps to this abound everywhere so I shall not go into it (It would not be in the spirit of this blog). But we could introduce what is called the Hamiltonian operator and rewrite the time dependent Schrodinger's equation as follows:
\begin{equation}
H\psi = i \hbar \frac{\partial \psi}{\partial t}
\end{equation}
where H = \(-\frac{-\hbar}{2m}\frac{ d^2 }{dx^2} + V \).
The spatially dependent part would look like this; (If we write in terms of our newly defined Hamilton, H)
\begin{equation}
H\psi = E \psi
\end{equation}
Notice that E(energy) is just a number and this looks like some eigenvalue problem which in fact it is. So if we have a system with different energy levels , these would correspond to different eigenfunctions. So in that case our eigenvalue problem would look like this:
\begin{equation}
H\psi_n = E_n \psi_n
\end{equation}
The next statement is very important and will be stated without proof. The \(\psi_n \) live in a vector space that physicist refer to as the Hilbert Space, not only do they live in it but also can form a basis set for it (after of course they have been normalized). In other words if we have some solution to the schrodinger's equation, we can express it as a linear combination of these \(\psi_n \). That is to say that:
\begin{equation}
\Phi = \sum_n c_n \psi_n
\end{equation}
where \( \Phi(t) \) is a solution to Schrodinger's equation.
It is here that the student may think his or herself, " well that's neat, what is the next topic?" Pose for a while and contemplate this, we have a infinite dimensional vector space, spanned by these \( \psi_n \). A little but deep thought could occur to one: why represent the vector by writing the tedious sum expressed in the above equation? Could we not just use the \(c_n \) place them in a column vector (which in this case would be infinite dimensional) and then our operators would be matrices rather than differential operators. After all matrices act upon column vectors. This is the key insight to Dirac notation. Our quantum state will no longer be written by some function but instead it will be some column vector, with the elements in it the \(c_n\).
In other words or symbols:
\begin{equation}
\Phi = \begin{pmatrix} c_1\\ c_2 \\ c_3 \\ \vdots \end{pmatrix}
\end{equation}
NOTE: If we are in an infinite dimensional space we have to worry about convergence but at the most we live in a textbook world everything is well behaved. The next step is to see the transition from differential operators to matrices explicitly.
\( -\frac{\hbar}{2m}\frac{d^2 \psi }{dx^2} + V \psi = \frac{i \hbar \partial \psi}{\partial t} \) . If we solve the solve this using separation of variable we get a term that is only time dependent and the other that is only spatially dependent. The steps to this abound everywhere so I shall not go into it (It would not be in the spirit of this blog). But we could introduce what is called the Hamiltonian operator and rewrite the time dependent Schrodinger's equation as follows:
\begin{equation}
H\psi = i \hbar \frac{\partial \psi}{\partial t}
\end{equation}
where H = \(-\frac{-\hbar}{2m}\frac{ d^2 }{dx^2} + V \).
The spatially dependent part would look like this; (If we write in terms of our newly defined Hamilton, H)
\begin{equation}
H\psi = E \psi
\end{equation}
Notice that E(energy) is just a number and this looks like some eigenvalue problem which in fact it is. So if we have a system with different energy levels , these would correspond to different eigenfunctions. So in that case our eigenvalue problem would look like this:
\begin{equation}
H\psi_n = E_n \psi_n
\end{equation}
The next statement is very important and will be stated without proof. The \(\psi_n \) live in a vector space that physicist refer to as the Hilbert Space, not only do they live in it but also can form a basis set for it (after of course they have been normalized). In other words if we have some solution to the schrodinger's equation, we can express it as a linear combination of these \(\psi_n \). That is to say that:
\begin{equation}
\Phi = \sum_n c_n \psi_n
\end{equation}
where \( \Phi(t) \) is a solution to Schrodinger's equation.
It is here that the student may think his or herself, " well that's neat, what is the next topic?" Pose for a while and contemplate this, we have a infinite dimensional vector space, spanned by these \( \psi_n \). A little but deep thought could occur to one: why represent the vector by writing the tedious sum expressed in the above equation? Could we not just use the \(c_n \) place them in a column vector (which in this case would be infinite dimensional) and then our operators would be matrices rather than differential operators. After all matrices act upon column vectors. This is the key insight to Dirac notation. Our quantum state will no longer be written by some function but instead it will be some column vector, with the elements in it the \(c_n\).
In other words or symbols:
\begin{equation}
\Phi = \begin{pmatrix} c_1\\ c_2 \\ c_3 \\ \vdots \end{pmatrix}
\end{equation}
NOTE: If we are in an infinite dimensional space we have to worry about convergence but at the most we live in a textbook world everything is well behaved. The next step is to see the transition from differential operators to matrices explicitly.
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