Month: March 2021

Inner and Outer Product in Quantum Mechanics

| Cosmoshivani

Alright, This is NOT going to be a very revealing kind of post like was planning but here you go…

In quantum mechanics, the particles are represented as state vectors. These state vectors are part of a state space for a given system or a Hilbert Space.

Let’s take two state vectors, \psi and \phi from a given state space. Now, we can perform some operations on these, just like we do with coordinate vectors.

But to define the inner product of state vectors we’ll have to define a dual space too. Given a basis, we can write these state vectors as matrices. Kets are written as column vectors and they are elements of state space, whereas the Bra vectors are written as row vectors and are elements of the dual space.

So, a ket vector is written as | \psi> and a bra vector as < \psi|. Their inner product is defined as,

< \psi|\psi>

The resulting vector is either a scalar or a complex number.

If the inner product is one then, the vector is said to be normalized and if it’s zero, then they are orthogonal.

Whereas, the outer product results in a matrix or an operator and it’s written like this,

|\psi> < \psi|

I’m not really sure if all the outer products result in actual operators in quantum mechanics. Because in QM, they are mostly Hermitian Operators. I’d really like to think more about it in future. 🤔 Hopefully 😄😄

see ya!

Inner and Outer product in Linear Algebra

| Cosmoshivani

Let’s take two coordinate vectors, \textbf v and \textbf u. Both of them can be represented as column matrices (why not as row matrices?) of order m \times 1 and n \times 1.

The inner product of these two coordinate vectors is, \textbf v^T \cdot \textbf u.

\textbf v = \begin{pmatrix} a_1 \\ a_2 \\ a_3\\...\\a_m \end{pmatrix}

\textbf v^T = \begin{pmatrix} a_1 & a_2 & a_3 & ... & a_m \end{pmatrix}

\textbf u = \begin{pmatrix} b_1 \\ b_2 \\ b_3\\...\\ b_n \end{pmatrix}

\textbf v^T \cdot \textbf u = \begin{pmatrix} a_1 & a_2 & a_3& ... & a_m \end{pmatrix} \cdot \begin{pmatrix} b_1 \\ b_2 \\b_3 \\ ... \\b_n \end{pmatrix}

\textbf v^T \cdot \textbf u = ( a_1 b_1 + a_2 b_2 + a_3 b_3 + ... + a_m b_n)

Notice that it only works if m = n and it always results in a matrix of order 1 \times 1.

The outer product of these two vectors is \textbf v \cdot \textbf u^T.

\textbf v \cdot \textbf u^T = \begin{pmatrix} a_1 \\ a_2 \\ a_3 \\ ... \\ a_m \end{pmatrix} \cdot \begin{pmatrix} b_1 & b_2 & b_3 & ... & b_n \end{pmatrix}

\textbf v \cdot \textbf u^T = \begin{pmatrix} a_1 b_1 & a_1 b_2 & a_2 b_3 &...&a_1 b_n\\a_2 b_1 & a_2 b_2 & a_2 b_3 &...& a_2 b_n\\a_3 b_1 & a_3 b_2 & a_3 b_3 &...&a_3 b_n\\...& ... & ... & ... & ... \\a_m b_1 & a_m b_2 & a_m b_3 & ... & a_m b_n \end{pmatrix}

If you have a column vector of order m \times 1 and a row vector of order 1 \times n respectively, their outer product will be a matrix of order m \times n. Here m need not be equal to n.

The significance of inner product in quantum mechanics becomes clear right away, but what about the outer product?

To be continued in the next article: Inner and Outer Product in Quantum Mechanics