Dirac Kets and Operators

State Vector

The physcial state of a system can be described with the state vector \(\left|\Psi\right>\). A wavefunction, $$\Psi(x)=\sum^\infty_{n=0}c_n\phi_n(x),$$ using dirac notation is expressed as $$\left|\Psi\right>=\sum^\infty_{n=0}c_n\left|e_n\right>.$$ The \(\left|e_n\right>\) are the basis vectors that span the Hilbert space. We can represent the \(\left|\Psi\right>\) in terms of column vectors $$\left|\Psi\right> = c_1\left|e_1\right> + c_2\left|e_2\right> + ... + c_N\left|e_N\right> = \begin{bmatrix}c_1 \\ c_2 \\.\\.\\. \\ c_N\end{bmatrix},$$ where $$\left|e_1\right>=\begin{bmatrix}1\\0\\0\\.\\.\\.\\0\end{bmatrix}, \left|e_2\right>=\begin{bmatrix}0\\1\\0\\.\\.\\.\\0\end{bmatrix},...$$

Continuous Basis States

Let \(\left| q \right>\) be a continuous state. The orthonormality relation can be given by $$\left< q' \middle| q \right> = \delta(q - q')$$ And the completness can be achieved by $$\int^\infty_{-\infty} \left| q \right>\left< q\right| dq = I.$$ Then, a ket can be represented by \begin{align} \left| \psi \right> &= \int \psi(q) \left| q \right> dq \\ 1 &= \left<\psi\middle|\psi\right>\\ &= \int \psi^*(q')\psi(q)\left< q' \middle| q\right> dq dq' \\ &= \int \psi^*(q')\psi(q)\delta(q-q')dqdq'\\ &= \int |\psi(q)|^2 dq \end{align}

Inner Product

The inner product between ket \(\left|A\right>=\sum_{i} a_i\left|e_i\right>\) and ket \(\left|B\right>=\sum_{i} b_i\left|e_i\right>\) is defined as $$\left< A \middle| B\right> = \sum_i a^*_ib_i,$$ where \(\left< A \right| \) is called a bar. A bar maps a ket, a vector, to a scalar, so it is a one-form, i.e. a \(\begin{pmatrix}0\\1\end{pmatrix}\) tensor. In terms of the inner products between their basis vectors, the inner product between \(\left|A\right>\) and \(\left|B\right>\) is given by \begin{align} \left< A \middle| B \right> &= \left(=\sum_{i} a_i\left|e_i\right>\right)\left(\sum_{j} b_j\left|e_j\right>\right) \\ &= \sum_i\sum_j a_i^*b_j\left< e_i \middle| e_j\right>\\ &= \sum_i\sum_j a_i^*b_j \delta_{ij}\\ &= \sum_i a_i^*b_i, \end{align} which made use of the orthonormality of basis vectors.

Outer Product

The outer product between ket \(\left|A\right>=\sum_{i} a_i\left|e_i\right>\) and ket \(\left|B\right>=\sum_{i} b_i\left|e_i\right>\) is defined as $$\left| A \right> \left< B \right| = \begin{bmatrix} a_1\\a_2\\.\\.\\.\\a_N \end{bmatrix}\begin{bmatrix}b_1^* & b_2^* & ... &b^*_3\end{bmatrix}.$$ It gives a \(N\times N\) matrix with components \(a_ib^*_j\) at the \(i\)th row and \(j\)th column. THe matrix maps two bars to a scalar. Thus the outer product is a \(\begin{pmatrix}2\\0\end{pmatrix}\) tensor.

Change of Basis

The completeness of the basis gives us $$\sum_n\left| e_n \right> \left< e_n \right| = I$$ Let \(\{\left| e_n\right>\}\) be a set of old basis. We would like to change to a new basis \(\{\left| f_n\right>\}\).

For a vector, \begin{align} \left|\Psi\right> &= \sum_n c_n \left| e_n \right> \\ &= I \sum_n c_n \left| e_n \right> \\ &= \sum_k \left| f_k \right> \left< f_k \right| \sum_n c_n \left| e_n \right> \\ &= \sum_k\left(\sum_n c_n\left< f_k \middle| e_n \right> \right)\left| f_k \right> \\ &= \sum_k b_k \left| f_k \right>, \end{align} where $$b_k = \sum_n c_n \left< f_k \middle| e_n \right>.$$

For a matrix, \begin{align} M &= \sum_p\sum_q m_{pq} \left| a_p \right> \left< a_q \right| \\ &= \sum_p \sum_q \sum_n \left| b_n \right> \left< b_n \middle| a_p\right> m_{qp} \left< a_q \right| \sum_k \left| b_k \right> \left< b_k \right|\\ &= \sum_n \sum_k \lambda_{nk} \left| b_n \right> \left< b_k \right|, \\ \end{align} where $$\lambda_{nk} = \sum_{pq} \left< b_n \middle| a_p \right> m_{pq} \left< a_q \middle| b_k \right>$$