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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.

Formula

Suppose there are two time-independent one-dimensional wave functions $u_{1}$, $u_{2}$. Let’s assume that $u_{1}$ and $u_{2}$ are normalized and not orthogonal to each other. Then, the following wave function $u$ is a normalized wave function orthogonal to $u_{1}$.

$$ \begin{align*} u &= \dfrac { \displaystyle \left(- \int u_{1}^{\ast} u_{2} dx \right)u_{1} + u_{2}}{\displaystyle \sqrt{ 1-\left| \int u_{1}^{\ast} u_{2} dx \right|^{2}}} \\ &= \dfrac { -\braket{u_{1} | u_{2}} u_{1} + u_{2}}{\sqrt{ 1-\left| \braket{u_{1} | u_{2}} \right|^{2}}} \\ \end{align*} $$

Explanation

Since the goal is to find a new eigenfunction $u$ that is orthogonal to $u_{1}$ from $u_{1}$ and $u_{2}$, let’s denote it first as $u = c_{1} u_{1} + c_{2}u_{2}$. Now, by using the condition that $u$ is orthogonal to $u_{1}$ and the normalization condition of $u$, $c_{1}$ and $c_{2}$ can be obtained.

  • Part 1. $u$ and $u_{1}$ are orthogonal.

    Since $u$ is orthogonal to $u_{1}$, the following holds true.

    $$ \begin{align*} \int uu_{1}^{\ast} dx &= \int \left( c_{1} u_{1}u_{1}^{\ast} + c_{2}u_{2}u_{1}^{\ast}\right) dx \\ &= \int c_{1}u_{1} u_{1}^{\ast}dx + \int c_{2}u_{2}u_{1}^{\ast} dx \\ &= c_{1} + c_{2}\int u_{1}^{\ast}u_{2} dx \\ &= 0 \end{align*} $$

    Thus, we obtain the following.

    $$ c_{1}=-c_{2} \int u_{1}^{\ast}u_{2} dx \tag{1} $$

  • Part 2. $u$ is a normalized function.

    Since $u$ is a normalized eigenfunction, the following holds true.

    $$ \begin{align*} \int uu^{\ast} dx &= \int (c_{1}u_{1}+c_{2}u_{2})(c_{1}u_{1}^{\ast}+c_{2}u_{2}^{\ast})dx \\ &= (c_{1})^{2} \int u_{1}u_{1}^{\ast} dx + (c_{2})^{2} \int u_{2}u_{2}^{\ast} dx + c_{1}c_{2} \left( \int u_{1}^{\ast}u_{2} dx + \int u_{1}u_{2}^{\ast}dx \right) \\ &= (c_{1})^{2} + (c_{2})^{2} + c_{1}c_{2} \left( \int u_{1}^{\ast}u_{2} dx + \int u_{1}u_{2}^{\ast}dx \right) \\ &=1 \end{align*} $$

    Substituting $(1)$ in the last equation yields the following.

    $$ (c_{2})^{2} \left( \int u_{1}^{\ast} u_{2} dx \right)^{2} + (c_{2})^{2} -(c_{2})^{2} \left( \int u_{1}^{\ast} u_{2} dx\right)^{2} -(c_{2})^{2} \int u_{1}^{\ast} u_{2} dx \int u_{1}u_{2}^{\ast} dx =1 $$

    $$ \implies (c_{2})^{2} -(c_{2})^{2} \int u_{1}^{\ast} u_{2} dx \int u_{1}u_{2}^{\ast} dx =1 $$

    $$ \implies (c_{2})^{2} \left( 1- \left| \int u_{1}^{\ast}u_{2} dx \right |^{2} \right)=1 $$

    $$ \implies c_{2}=\dfrac{1}{\displaystyle \sqrt{1- \left|\int u_{1}^{\ast}u_{2} dx \right|^{2}}} $$

    The sign of $c_{2}$ chosen to be $+$ is simply for convenience; $-$ could also be chosen. Substituting this into $(1)$ yields $c_{1}$.

    $$ c_{1} = \dfrac{\displaystyle -\int u_{1}^{\ast}u_{2} dx }{\displaystyle \sqrt{1- \left|\int u_{1}^{\ast}u_{2} dx \right|^{2}}} $$

Thus, the function $u$, which is orthogonal to $u_{1}$ and normalized, is as follows.

$$ u = \dfrac {\displaystyle \left(- \int u_{1}^{\ast} u_{2} dx \right)u_{1} + u_{2}}{\displaystyle \sqrt{ 1-\left| \int u_{1}^{\ast} u_{2} dx \right|^{2}}} $$