<aside> <img src="https://prod-files-secure.s3.us-west-2.amazonaws.com/369dfa6b-d4d9-4cf2-a446-e369553b6347/cae7d0f5-acbf-496f-b7f2-f125482c70de/inner_product.png" alt="https://prod-files-secure.s3.us-west-2.amazonaws.com/369dfa6b-d4d9-4cf2-a446-e369553b6347/cae7d0f5-acbf-496f-b7f2-f125482c70de/inner_product.png" width="40px" /> Inner product: $\lang \vec a |\vec b \rang$ is a scalar function (takes 2 vectors and returns a scalar). For a complex vector space $V^N(\mathbb C)$, the inner product has the following properties:
🗒️ Note: thus we get the norm is $||\vec a||^2=\left <\vec a|\vec a \right > \ge 0$. $||\vec a||^2=0$ if $\vec a =\vec 0$
<aside> <img src="https://prod-files-secure.s3.us-west-2.amazonaws.com/369dfa6b-d4d9-4cf2-a446-e369553b6347/2597e894-cb51-44f4-89ec-bdfdaf6d1bf5/Orthogonality.png" alt="https://prod-files-secure.s3.us-west-2.amazonaws.com/369dfa6b-d4d9-4cf2-a446-e369553b6347/2597e894-cb51-44f4-89ec-bdfdaf6d1bf5/Orthogonality.png" width="40px" /> Orthogonality: 2 vectors $\vec a$ $\vec b$ which are elements of the inner product space $V^N$ are orthogonal if $\left < \vec a |\vec b \right >=0$
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<aside> <img src="https://prod-files-secure.s3.us-west-2.amazonaws.com/369dfa6b-d4d9-4cf2-a446-e369553b6347/334e8c57-ad03-4811-be2d-4884459252b1/Orthogonal_basis.png" alt="https://prod-files-secure.s3.us-west-2.amazonaws.com/369dfa6b-d4d9-4cf2-a446-e369553b6347/334e8c57-ad03-4811-be2d-4884459252b1/Orthogonal_basis.png" width="40px" /> Orthogonal basis: A basis set $\{ \hat e_j\}^N_{j=1}$ of the inner product space is said to be orthogonal if $\left < \hat e_i |\hat e_j \right > =A\delta _{ij}$ where $A\in \mathbb C$ and $\delta _{ij}$ is the Kronecker-$\delta$. If $A=1$ then it is orthonormal.
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Consider the set of vectors $\{ \hat e_j\}^N_{j=1}$ which form a complete and orthogonal basis set for the vector space $V^N$. Two vectors $\vec a,\vec b\in V^N$ with representation $\vec a=\sum_j a_j\hat e_j$ and $\vec b=\sum_j b_j \hat e_j$ will have an inner product of the form
$$ \begin{aligned} \left <\vec a |\vec b \right >&=\sum^N_{i,j=1}\overline a_ib_j \\ &=\sum^N_{i,j=1}\overline{a_i}b_{j}=\sum^N_{i,j=1}\overline a_i \delta_{ij} b_j=\sum^N_{i=1} \overline a_i b_i
\end{aligned} $$
🗒️ Note: this leads to the following:
$$ \left < \hat e_k |\vec a\right >=\sum^N_{j=1}a_j \left < \hat e_k|\hat e_j\right >=\sum ^N_{j=1}a_j \delta_{kj}=a_k $$
tells us how much of the vector $\vec a$ is in the direction $\hat e_k$
<aside> <img src="https://prod-files-secure.s3.us-west-2.amazonaws.com/369dfa6b-d4d9-4cf2-a446-e369553b6347/3b5e5812-9eec-4cda-a902-101e74c2c007/Gram-Schmidt_orthogonalization.png" alt="https://prod-files-secure.s3.us-west-2.amazonaws.com/369dfa6b-d4d9-4cf2-a446-e369553b6347/3b5e5812-9eec-4cda-a902-101e74c2c007/Gram-Schmidt_orthogonalization.png" width="40px" /> Gram-Schmidt orthogonalization: Consider a general basis $\{\vec v_j\}^N_{j=1}$ for the inner-product space $V^N.$ We can construct an orthonormal basis $\{\hat e_j\}^N_{j=1}$ using the following procedure:
For any arbitrary starting vector we have $\vec u_1=\vec v_1$
Then we have $\vec u_2=\vec v_2-\left < \vec u_1 |\vec v_2 \right > \vec u_1$ or more generally
$$ \vec u_m= \vec v_m-\sum^{m-1}_{j=1}\left < \vec u_j |\vec v_m\right > \vec u_j $$
The vectors $\{\vec u_j\}^N_{j=1}$ are an orthogonal basis for the vector space $V^N$, but have an orthonormal basis, we normalise the vectors, such that $\hat e_j=\vec u_j/||\vec u_j||$
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💃 Example: Consider the following vector in $\R^3$
$$ \vec v_1=\begin{pmatrix} 1 \\ 1 \\ 1 \end{pmatrix} \quad \vec v_2=\begin{pmatrix} 1 \\ 0 \\ 1 \end{pmatrix} \quad \vec v_3= \begin{pmatrix} 1 \\ 1 \\ 0 \end{pmatrix} $$
Step 1
$$ \vec u_1=\vec v_1=\begin{pmatrix} 1 \\ 1 \\ 1 \end{pmatrix} $$
Step 2
$$ \vec u_2=\vec v_2-\underbrace{\left < \vec u_1 | \vec v_2 \right >}_{2}\vec u_1=\begin{pmatrix} -1 \\ -2 \\ -1 \end{pmatrix} $$
Step 3
$$ \vec u_3=\vec v_3-\underbrace{\left < \vec u_1 | \vec v_3 \right >}{2}\vec u_1-\underbrace{\left < \vec u_2 | \vec v_3 \right >}{-3}\vec u_2=\begin{pmatrix} -4 \\ -7 \\ -5 \end{pmatrix} $$
$\left | \psi \right >$ is a ket and it represents a vector space $V$
💃 Example: given a vector space $V^N$ with arbitrary vector $\vec v=\sum^N_{j=1}v_j \hat e_j$ where $\{ \hat e_j\}^N_{j=1}$ is a basis for $V^N$, in Dirac notation this vector looks like:
$$ \left | \vec v \right >=\sum^N_{j=1}v_j\left |\vec e_j \right > $$
$\left < \psi \right |$ is a bra, it has both addition and scalar multiplication in the same as ket’s.
If $\left | \psi \right >$ is an element of vector space $V$ then $\left < \psi \right |$ is a vector $V^*$ which is called dual space or the dual of $V$
relation between dual space is, the adjoint or Hermitian conjugate
$$ (\left | \psi \right > )^\dagger =\left < \psi \right | \qquad \;(\left < \psi \right | )^\dagger =\left | \psi \right > $$
This map is antilinear
$$ (\alpha \left |\psi\right >+\beta \left |\phi\right >)^\dagger=\overline \alpha \left <\psi\right |+\overline \beta \left <\phi\right | $$
