From 0c6d376741241636ebd5c7565da37334006e1684 Mon Sep 17 00:00:00 2001
From: Davit Potoyan <david.potoyan@gmail.com>
Date: Fri, 11 Oct 2024 11:49:11 -0500
Subject: [PATCH] 'dirac notation'
---
ch03/note04.md | 100 ++++++++++++++++++++++++++++++-------------------
1 file changed, 62 insertions(+), 38 deletions(-)
diff --git a/ch03/note04.md b/ch03/note04.md
index 052799a7..367cd7d0 100644
--- a/ch03/note04.md
+++ b/ch03/note04.md
@@ -73,19 +73,18 @@ $$
### Commutations of operators
-- From linear algebra we know that order of matrix multiplicaiton matters and that $AB\neq BA$ for two matrices $A$ and $B$
-
-- Thus we generally ecpect $\hat{A}\hat{B} \neq \hat{B}\hat{A}$.
-- We can quantify relationship between two operators by computing the **Commutator**:
-
:::{admonition} **Commutator $\hat{A}$ and $\hat{B}$**
:class: important
$${\left[\hat{A},\hat{B}\right]f = \left(\hat{A}\hat{B} - \hat{B}\hat{A}\right)f}$$
:::
-- If the commutator is zero, it means that order in multiplication of operators or matrices can be changed.
-- If the commutator is non-zero, the order matters and can not be changed!
+- From linear algebra we know that order of matrix multiplicaiton matters and that $AB\neq BA$ for two matrices $A$ and $B$
+- Thus we also generally ecpect $\hat{A}\hat{B} \neq \hat{B}\hat{A}$ for any two operators.
+
+- We can quantify relationship between two operators by computing the **Commutator**
+ - If the commutator is zero, it means that order in multiplication of operators or matrices can be changed.
+ - If the commutator is non-zero, the order matters and can not be changed!
:::{admonition} **Example**
@@ -144,23 +143,16 @@ In general, it turns out that for operators $\hat{A}$ and $\hat{B}$ that do not
$${\Delta A\Delta B \ge \frac{1}{2}\left|\left<\left[\hat{A},\hat{B}\right]\right>\right|}$$
-:::{admonition} **Example**
-:class: note
-
-Obtain the position/momentum uncertainty principle
-:::
-
-:::{admonition} **Solution**
-:class: dropdown
-
-Denote $\hat{A} = \hat{x}$ and $\hat{B} = \hat{p}_x$.
+- Let's check this relation on the example of momentum and position operators
+- Denote $\hat{A} = \hat{x}$ and $\hat{B} = \hat{p}_x$.
$$\frac{1}{2}\left|\left<\left[\hat{A},\hat{B}\right]\right>\right| = \frac{1}{2}\left|\left<\left[\hat{x},\hat{p}_x\right]\right>\right| = \frac{1}{2}\left|\left<\frac{\hbar}{i}\right>\right|
= \frac{1}{2}\left|\left<\psi\left|\frac{\hbar}{i}\right|\psi\right>\right| = \frac{1}{2}\left|\frac{\hbar}{i}\underbrace{\left<\psi\left|\psi\right.\right>}_{=1}\right| = \frac{\hbar}{2}$$
$$\Rightarrow \Delta x\Delta p_x \ge \frac{\hbar}{2}$$
-:::
+
+- We find that we can not measure precise values of position or momentum simulatneously.
### Commuting operators and simultaneous measurments
@@ -208,8 +200,11 @@ Note that the commutation relation must apply to all well-behaved functions and
:::{admonition} **Hermitian Matrix**
:class: important
+$$A = A^\dagger$$
+
$$a_{jk} = a^{*}_{kj}$$
+- $A^\dagger$ is called **conjugate transpose of matrix** where one trasposes elements and replaces with complex conjugate elements
:::
@@ -217,15 +212,23 @@ $$a_{jk} = a^{*}_{kj}$$
:::{admonition} **Hermitian Operator**
:class: important
+$$A = A^\dagger$$
+
$$
-{\int {\color{blue} \psi^*_j} {\color{green}\hat{A} \psi_k} d\tau = \int { \color{green} \psi_k} {\color{blue} (\hat{A}\psi_j)^{*} } d\tau}
+\int {\color{blue} \psi^*_j} \hat{A} {\color{green}\psi_k} d\tau = \int { \color{green} \psi_k} \hat{A}^\dagger {\color{blue}\psi_j^{*} } d\tau= \int { \color{green} \psi_k} \hat{A} {\color{blue}\psi_j^{*} } d\tau
$$
+**In Dirac Notation**
+
+$$ \langle j| \hat{A}|k\rangle = \langle k| \hat{A} | j \rangle^{*}$$
:::
-- *Note the symmetry between complex conjugate pair of wavefunctions:* The expression remains the same wether the same operator acts on wavefunction or its complex conjugate pair.
+
+- On the left, $\hat{A}$ acts on $\psi_k$, and the result is integrated against $\psi_j^*$.
+- On the right, $\hat{A}^\dagger$ acts on $\psi_j^*$, and the result is integrated against $\psi_k$.
+- **For Hermitian operators** we have special case when $\hat{A} = \hat{A}^\dagger$ and this equation becomes symmetric.
- In general most matrices/operators in mathematics are not Hermitian. Meaning you get different result when you feed complex conjugate function to the same operator. Some examples are below
:::{admonition} **Example of Hermitian Matrix**
@@ -252,12 +255,15 @@ i & 0 \\
:class: dropdown
- For the first matrix we have $a_{12}=2\neq a^{*}_{21}=3$, non-Hermitian
-- For the second matrix $a_{12}= a^{*}_{21}=0$, Hermitian
+- For the second matrix $a_{11}\neq a^{*}_{11}=0$, non-Hermitian
- For the third matrix $a_{12}=-3i =a^{*}_{21} = (3i)^{*}=-3i$, Hermitian
- For the fourth matrix $a_{12}=2i \neq a^{*}_{21} = (2i)^{*} = -2i$, Hermitian
:::
+- To see that Differentiation operators are Hermitian requires a little more work. - A trick that helps see it is integration by parts where the constant term is zero because wavefunction decays to zero at boundaries (postulate 1, keeping probability finite)!
+
+$$\int \psi_1 d\psi_2 =- \int \psi_2d\psi_1 + \psi_1\psi_2\Big|_{x_{min}}^{x_{max}} =- \int \psi_2d\psi_1$$
:::{admonition} **Example of Hermitian Operator**
:class: note
@@ -271,47 +277,65 @@ Prove that the momentum operator (in one dimension) is Hermitian.
${\int\limits_{-\infty}^{\infty}\psi_j^*(x)\left(-i\hbar\frac{d\psi_k(x)}{dx}\right)dx} = -i\hbar\int\limits_{-\infty}^{\infty}\psi_j^*(x)\frac{d\psi_k(x)}{dx}dx = \\ \overbrace{\int\limits_{-\infty}^{\infty}\psi_k(x)\left(i\hbar\frac{d\psi_j^*(x)}{dx}\right)dx}^{{integration\, by\, parts}}$
$ = {\int\limits_{-\infty}^{\infty}\psi_k(x)\left(-i\hbar\frac{d\psi_j(x)}{dx}\right)^*dx} \Rightarrow \hat{p}_x\textnormal{ is Hermitian}$.
-Note that the wavefunctions approach zero at infinity and thus the boundary term in the integration by parts does not contribute. In 3-D, one would have to use the [Green identities](http://en.wikipedia.org/wiki/Green's_identities).
+
:::
-### Two conseqeuences of Hermitian property
-#### Eigenvalues of Hermitian operator are real
+### Two Consequences of Hermitian Property
-- Note that operators and eigenfunctions may be complex valued; however, **eigenvalues** of quantum mechanical operators **must be real** because they correspond to real values obtained from measurements.
-- By allowing wavefunctions to be complex, it is merely possible to store more information in it (i.e., both the real and imaginary parts or ``density and velocity'')
-- When computing experimental quantities complex conjugate pair of wavefunctions must be combined to yield real values.
-- **Proof** Let $\psi$ be an eigenfunction of $\hat{A}$ with eigenvalue $a$. Choose $\psi_j = \psi_k = \psi$. Then we can write the result of the left and right hand side of hermitian condition:
+#### Eigenvalues of Hermitian Operators Are Real
+
+- Operators and eigenfunctions in quantum mechanics may be complex valued; however, **eigenvalues** of quantum mechanical operators **must be real** because they correspond to the real values obtained from measurements.
+- By allowing wavefunctions to be complex, it is possible to store more information (i.e., both the real and imaginary parts, or "density and velocity").
+- When computing experimental quantities, the complex conjugate pair of wavefunctions must be combined to yield real values.
+
+- **Proof**: Let $\psi$ be an eigenfunction of $\hat{A}$ with eigenvalue $a$. Choose $\psi_j = \psi_k = \psi$. Then we can write the result of the left-hand and right-hand sides of the Hermitian condition:
+
+$$
+\int \psi^* \hat{A} \psi \, d\tau = a
+$$
-$$\int\psi^*\hat{A}\psi d\tau = a$$
+$$
+\int \psi \left(\hat{A} \psi\right)^* \, d\tau = a^*
+$$
-$$\int\psi\left(\hat{A}\psi\right)^*d\tau = a^*$$
+- Since the operator is Hermitian, this leads to the conclusion:
-$$a = a^*$$
+$$
+a = a^*
+$$
+- Thus, the eigenvalue $a$ must be real.
-#### Eigenfunction of Hermitian operator are orthogonal
+#### Eigenfunctions of Hermitian Operators Are Orthogonal
-The Hermitian property can also be used to show that the eigenfunctions ($\psi_j$ and $\psi_k$), which have different eigenvalues (i.e., $a_j$ and $a_k$ with $a_j \ne a_k$; ``non-degenerate''), are orthogonal to each other:
+The Hermitian property can also be used to show that eigenfunctions $\psi_j$ and $\psi_k$, corresponding to different eigenvalues $a_j$ and $a_k$ (with $a_j \neq a_k$, i.e., "non-degenerate"), are orthogonal to each other:
$$
-{\textnormal{LHS: }\int\psi_j^*\hat{A}\psi_kd\tau = \int\psi_j^*a_k\psi_kd\tau = a_k\int\psi_j^*\psi_kd\tau}
+\textnormal{LHS: } \int \psi_j^* \hat{A} \psi_k \, d\tau = \int \psi_j^* a_k \psi_k \, d\tau = a_k \int \psi_j^* \psi_k \, d\tau
$$
$$
-{\textnormal{RHS: }\int\psi_k\left(\hat{A}\psi_j\right)^*d\tau = \int\psi_k\left(a_j\psi_j\right)^*d\tau = a_j\int\psi_j^*\psi_kd\tau}
+\textnormal{RHS: } \int \psi_k \left(\hat{A} \psi_j \right)^* \, d\tau = \int \psi_k \left(a_j \psi_j \right)^* \, d\tau = a_j \int \psi_j^* \psi_k \, d\tau
$$
+Since the operator is Hermitian, we require that LHS = RHS. This results in:
-Here Hermiticity requires LHS = RHS. If $a_j \ne a_k$, then we are dealing with:
+$$
+\left(a_k - a_j \right) \int \psi_j^* \psi_k \, d\tau = 0
+$$
+
+If $a_j \neq a_k$, then we have:
$$
-{{\left(a_k - a_j\right)}{\ne 0}\int\psi^*_j\psi_kd\tau = 0}
+\int \psi_j^* \psi_k \, d\tau = 0
$$
-Note that if $a_j = a_k$, meaning that the values are [degenerate](http://en.wikipedia.org/wiki/Degenerate_energy_level), this result does not hold.
+This shows that $\psi_j$ and $\psi_k$ are orthogonal.
+
+- **Note**: If $a_j = a_k$, meaning the eigenvalues are degenerate, this result does not hold. Degeneracy refers to eigenstates having the same eigenvalue, and in that case, orthogonality may not apply without further specification.