Gram-Schmidt Orthogonalization Process in Quantum Mechanics
📂Quantum Mechanics Gram-Schmidt Orthogonalization Process in Quantum Mechanics Definition The Gram-Schmidt orthogonalization procedure is a method for creating an orthogonal set from vectors that are not orthogonal to each other.
Suppose there are two time-independent one-dimensional wave functions u 1 u_{1} u 1 , u 2 u_{2} u 2 . Let’s assume that u 1 u_{1} u 1 and u 2 u_{2} u 2 are normalized and not orthogonal to each other. Then, the following wave function u u u is a normalized wave function orthogonal to u 1 u_{1} u 1 .
u = ( − ∫ u 1 ∗ u 2 d x ) u 1 + u 2 1 − ∣ ∫ u 1 ∗ u 2 d x ∣ 2 = − ⟨ u 1 ∣ u 2 ⟩ u 1 + u 2 1 − ∣ ⟨ u 1 ∣ u 2 ⟩ ∣ 2
\begin{align*}
u &= \dfrac { \displaystyle \left(- \int u_{1}^{\ast} u_{2} dx \right)u_{1} + u_{2}}{\displaystyle \sqrt{ 1-\left| \int u_{1}^{\ast} u_{2} dx \right|^{2}}} \\
&= \dfrac { -\braket{u_{1} | u_{2}} u_{1} + u_{2}}{\sqrt{ 1-\left| \braket{u_{1} | u_{2}} \right|^{2}}} \\
\end{align*}
u = 1 − ∫ u 1 ∗ u 2 d x 2 ( − ∫ u 1 ∗ u 2 d x ) u 1 + u 2 = 1 − ∣ ⟨ u 1 ∣ u 2 ⟩ ∣ 2 − ⟨ u 1 ∣ u 2 ⟩ u 1 + u 2
Explanation Since the goal is to find a new eigenfunction u u u that is orthogonal to u 1 u_{1} u 1 from u 1 u_{1} u 1 and u 2 u_{2} u 2 , let’s denote it first as u = c 1 u 1 + c 2 u 2 u = c_{1} u_{1} + c_{2}u_{2} u = c 1 u 1 + c 2 u 2 . Now, by using the condition that u u u is orthogonal to u 1 u_{1} u 1 and the normalization condition of u u u , c 1 c_{1} c 1 and c 2 c_{2} c 2 can be obtained.
Part 1. u u u and u 1 u_{1} u 1 are orthogonal.
Since u u u is orthogonal to u 1 u_{1} u 1 , the following holds true.
∫ u u 1 ∗ d x = ∫ ( c 1 u 1 u 1 ∗ + c 2 u 2 u 1 ∗ ) d x = ∫ c 1 u 1 u 1 ∗ d x + ∫ c 2 u 2 u 1 ∗ d x = c 1 + c 2 ∫ u 1 ∗ u 2 d x = 0
\begin{align*}
\int uu_{1}^{\ast} dx &= \int \left( c_{1} u_{1}u_{1}^{\ast} + c_{2}u_{2}u_{1}^{\ast}\right) dx \\
&= \int c_{1}u_{1} u_{1}^{\ast}dx + \int c_{2}u_{2}u_{1}^{\ast} dx \\
&= c_{1} + c_{2}\int u_{1}^{\ast}u_{2} dx \\
&= 0
\end{align*}
∫ u u 1 ∗ d x = ∫ ( c 1 u 1 u 1 ∗ + c 2 u 2 u 1 ∗ ) d x = ∫ c 1 u 1 u 1 ∗ d x + ∫ c 2 u 2 u 1 ∗ d x = c 1 + c 2 ∫ u 1 ∗ u 2 d x = 0
Thus, we obtain the following.
c 1 = − c 2 ∫ u 1 ∗ u 2 d x (1)
c_{1}=-c_{2} \int u_{1}^{\ast}u_{2} dx \tag{1}
c 1 = − c 2 ∫ u 1 ∗ u 2 d x ( 1 )
Part 2. u u u is a normalized function.
Since u u u is a normalized eigenfunction, the following holds true.
∫ u u ∗ d x = ∫ ( c 1 u 1 + c 2 u 2 ) ( c 1 u 1 ∗ + c 2 u 2 ∗ ) d x = ( c 1 ) 2 ∫ u 1 u 1 ∗ d x + ( c 2 ) 2 ∫ u 2 u 2 ∗ d x + c 1 c 2 ( ∫ u 1 ∗ u 2 d x + ∫ u 1 u 2 ∗ d x ) = ( c 1 ) 2 + ( c 2 ) 2 + c 1 c 2 ( ∫ u 1 ∗ u 2 d x + ∫ u 1 u 2 ∗ d x ) = 1
\begin{align*}
\int uu^{\ast} dx &= \int (c_{1}u_{1}+c_{2}u_{2})(c_{1}u_{1}^{\ast}+c_{2}u_{2}^{\ast})dx \\
&= (c_{1})^{2} \int u_{1}u_{1}^{\ast} dx + (c_{2})^{2} \int u_{2}u_{2}^{\ast} dx + c_{1}c_{2} \left( \int u_{1}^{\ast}u_{2} dx + \int u_{1}u_{2}^{\ast}dx \right) \\
&= (c_{1})^{2} + (c_{2})^{2} + c_{1}c_{2} \left( \int u_{1}^{\ast}u_{2} dx + \int u_{1}u_{2}^{\ast}dx \right) \\
&=1
\end{align*}
∫ u u ∗ d x = ∫ ( c 1 u 1 + c 2 u 2 ) ( c 1 u 1 ∗ + c 2 u 2 ∗ ) d x = ( c 1 ) 2 ∫ u 1 u 1 ∗ d x + ( c 2 ) 2 ∫ u 2 u 2 ∗ d x + c 1 c 2 ( ∫ u 1 ∗ u 2 d x + ∫ u 1 u 2 ∗ d x ) = ( c 1 ) 2 + ( c 2 ) 2 + c 1 c 2 ( ∫ u 1 ∗ u 2 d x + ∫ u 1 u 2 ∗ d x ) = 1
Substituting ( 1 ) (1) ( 1 ) in the last equation yields the following.
( c 2 ) 2 ( ∫ u 1 ∗ u 2 d x ) 2 + ( c 2 ) 2 − ( c 2 ) 2 ( ∫ u 1 ∗ u 2 d x ) 2 − ( c 2 ) 2 ∫ u 1 ∗ u 2 d x ∫ u 1 u 2 ∗ d x = 1
(c_{2})^{2} \left( \int u_{1}^{\ast} u_{2} dx \right)^{2} + (c_{2})^{2} -(c_{2})^{2} \left( \int u_{1}^{\ast} u_{2} dx\right)^{2} -(c_{2})^{2} \int u_{1}^{\ast} u_{2} dx \int u_{1}u_{2}^{\ast} dx =1
( c 2 ) 2 ( ∫ u 1 ∗ u 2 d x ) 2 + ( c 2 ) 2 − ( c 2 ) 2 ( ∫ u 1 ∗ u 2 d x ) 2 − ( c 2 ) 2 ∫ u 1 ∗ u 2 d x ∫ u 1 u 2 ∗ d x = 1
⟹ ( c 2 ) 2 − ( c 2 ) 2 ∫ u 1 ∗ u 2 d x ∫ u 1 u 2 ∗ d x = 1
\implies (c_{2})^{2} -(c_{2})^{2} \int u_{1}^{\ast} u_{2} dx \int u_{1}u_{2}^{\ast} dx =1
⟹ ( c 2 ) 2 − ( c 2 ) 2 ∫ u 1 ∗ u 2 d x ∫ u 1 u 2 ∗ d x = 1
⟹ ( c 2 ) 2 ( 1 − ∣ ∫ u 1 ∗ u 2 d x ∣ 2 ) = 1
\implies (c_{2})^{2} \left( 1- \left| \int u_{1}^{\ast}u_{2} dx \right |^{2} \right)=1
⟹ ( c 2 ) 2 ( 1 − ∫ u 1 ∗ u 2 d x 2 ) = 1
⟹ c 2 = 1 1 − ∣ ∫ u 1 ∗ u 2 d x ∣ 2
\implies c_{2}=\dfrac{1}{\displaystyle \sqrt{1- \left|\int u_{1}^{\ast}u_{2} dx \right|^{2}}}
⟹ c 2 = 1 − ∫ u 1 ∗ u 2 d x 2 1
The sign of c 2 c_{2} c 2 chosen to be + + + is simply for convenience; − - − could also be chosen. Substituting this into ( 1 ) (1) ( 1 ) yields c 1 c_{1} c 1 .
c 1 = − ∫ u 1 ∗ u 2 d x 1 − ∣ ∫ u 1 ∗ u 2 d x ∣ 2
c_{1} = \dfrac{\displaystyle -\int u_{1}^{\ast}u_{2} dx }{\displaystyle \sqrt{1- \left|\int u_{1}^{\ast}u_{2} dx \right|^{2}}}
c 1 = 1 − ∫ u 1 ∗ u 2 d x 2 − ∫ u 1 ∗ u 2 d x
Thus, the function u u u , which is orthogonal to u 1 u_{1} u 1 and normalized, is as follows.
u = ( − ∫ u 1 ∗ u 2 d x ) u 1 + u 2 1 − ∣ ∫ u 1 ∗ u 2 d x ∣ 2
u = \dfrac {\displaystyle \left(- \int u_{1}^{\ast} u_{2} dx \right)u_{1} + u_{2}}{\displaystyle \sqrt{ 1-\left| \int u_{1}^{\ast} u_{2} dx \right|^{2}}}
u = 1 − ∫ u 1 ∗ u 2 d x 2 ( − ∫ u 1 ∗ u 2 d x ) u 1 + u 2
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