# 4 Basis sets and bra-ket algebra#

In this section, bra-ket algebra is described more fully together with basis sets and how they are intimately related to the bra and ket. Some algebra for manipulating the bra and ket is also described.

## 4.1 To each ket $$|\;\rangle$$ belongs a bra $$\langle\;|$$#

In the Dirac notation, a ket is a column and a bra a row vector. The elements of the vector are those of the basis set used. Each ket always has a corresponding bra. The bra is formed as the conjugate transpose of the ket, which means converting column into row and then taking the complex conjugate of each term. Formally the ket is the Hermitian adjoint of the bra, see Chapter 7.4.8. For example, if the ket is

$\begin{split} \displaystyle \qquad\qquad |v\rangle =\begin{bmatrix} a\\ b \\ c \end{bmatrix}\qquad\qquad \text{(27)} \end{split}$

the bra is

$\displaystyle \qquad\qquad \langle v| = \begin{bmatrix}a^*& b^*& c^*\end{bmatrix} \qquad \tag{28}$

where * indicates a complex conjugate. The $$a, b, c$$ are the elements of the basis set being used. However, these are themselves only aliases or references to the actual properties. By manipulating these symbols, problems can be solved so that only at the end of the calculation do numbers have to be used. The bra and ket can be multiplied together in two ways, as an inner product (dot product) to produce a scalar number

$\begin{split}\displaystyle \qquad\qquad \langle A|B\rangle =\begin{bmatrix} a^*_1 & a^*_2 & a^*_3 \cdots\end{bmatrix}\begin{bmatrix} b_1\\b_2\\b_3\\\vdots\end{bmatrix} = \sum_i a^*_ib_i\qquad\qquad \qquad\qquad \text{(29)}\end{split}$

and an outer product which is an operator and produces a matrix.

$\begin{split}\displaystyle \qquad\qquad |B\rangle\langle A|= \begin{bmatrix} b_1\\b_2\\b_3\\\vdots\end{bmatrix}\begin{bmatrix} a^*_1 & a^*_2 & a^*_3 \cdots\end{bmatrix}=\begin{bmatrix}b_1a_1^* & b_1a_2^* & b_1a_3^*\cdots \\ b_2a_1^* & b_2a_2^* &\cdots \\ \vdots & \vdots& \ddots\end{bmatrix} \qquad\qquad \qquad\qquad \text{(30)}\end{split}$

which are the same rules for any matrix met in Chapter 7. The only difference in their usage in quantum mechanics is that the elements of the basis set must form an orthogonal set. If they are also normalised then they form an orthonormal set.

## 4.2 Discrete basis sets#

The bra and ket, equations 27 and 28 have terms $$a,\, b,\, c$$ and these are the elements of the basis set chosen for the calculation. All quantum problems using bra-ket notation must have a basis set defined before the calculation starts. Unlike many calculations using vectors, the basis set does not usually represent spatial coordinates but is instead some other relevant property. The spin state of the proton, which is described next, is an example of this.

The spin state of an atom or nucleus can be described with a spin quantum number $$s$$ and magnetic (azimuthal) quantum numbers $$m$$, which, when $$s = 1/2$$ has values $$+1/2$$ or $$-1/2$$, or, in general, $$m = +s, +s - 1, \cdots -s$$ and thus is a range of values from $$s \to -s$$ each separated from one another by unity. Sometimes the magnetic quantum number is labelled $$m_s$$ or $$m_z$$; the latter because it is the projection of angular momentum onto a unique axis normally labelled $$z$$ when a unidirectional magnetic field is present, Fig. 4. The magnitude(expectation value of the angular momentum is $$\hbar\sqrt{ s(s+1)}$$. Nuclear spin angular momentum is usually labelled with $$I$$ instead of $$s$$. If a magnetic field $$B$$ is along the $$z$$-axis, the energy of the spin depends on its orientation and is $$E = -\gamma m_z\hbar B$$ where $$\gamma$$ is the magnetogyric ratio with units of rad T$${-1}$$ s$$^{-1}$$.

The spin basis set for a spin half particle (electron, $$^1$$H, $$^{13}$$13C, etc.) can be constructed in terms of the two quantum numbers $$s$$ and $$m$$, labelled with two indices as $$(s, m)$$, and is

$\displaystyle (1/2,\, +1/2) (1/2,\, -1/2)$

where $$(1/2, +1/2)$$, is a label describing one spin state, it has no mathematical significance, and the label $$(1/2, -1/2)$$ describes the other. In the case of spin half particles we conventionally reserve symbols $$\alpha$$ to represent the state $$(1/2, +1/2)$$ and $$\beta$$ to represent the state $$(1/2, -1/2)$$ and the basis set is now written as $$(\alpha\,\beta)$$ or as kets $$|\alpha \rangle$$ and $$|\beta\rangle$$ as $$(|\alpha\rangle \;|\beta\rangle)$$. Both notations are common. In this two-dimensional basis, the $$\alpha$$ spin state $$(s = 1/2, m = 1/2)$$ is represented as the column vector in the standard basis set as $$\displaystyle |\alpha\rangle= \begin{bmatrix} 1 \\ 0 \end{bmatrix}$$.

As $$\alpha$$ is a basis element (basis ket) of the standard basis set, it is zero everywhere except for one element representing its position in the basis set and which is given the value of unity. Similarly the $$\beta$$ spin state $$(s = 1/2, m = -1/2)$$ is represented as as $$\displaystyle |\beta\rangle= \begin{bmatrix} 0 \\ 1 \end{bmatrix}$$, and the basis set could be written as

$\begin{split}\displaystyle (\alpha, \beta) \equiv \left(\begin{bmatrix} 1 \\ 0 \end{bmatrix}\begin{bmatrix} 0 \\ 1 \end{bmatrix}\right) \end{split}$

The basis set can be ordered differently; for example, $$\displaystyle (\beta,\alpha, ) \equiv \left(\begin{bmatrix} 0 \\ 1 \end{bmatrix}\begin{bmatrix} 1 \\ 0 \end{bmatrix}\right)$$ would be equally valid, it is only necessary to decide in which order the elements are placed and keep to it. Figure 4. Sketch describing spin quantum numbers and angular momentum for a $$^1$$H nucleus. The angular momentum vector $$\pmb{I}$$ can be at any angle around the cone.

If there are two electrons (or nuclei) each with spin quantum number $$+1/2$$ then we have, depending on spin orientation, a total spin of $$S = 1$$ or $$0$$, loosely speaking these correspond to spins being ‘parallel’ or ‘anti-parallel’ to one another. The complete set of quantum numbers is $$S=0, m=0$$, and $$S=1$$ with $$m=+1,0,-1$$, making four states in all. A minimal basis set therefore has four terms

$\displaystyle (0,0) \;(1,-1) \;(1,0) \;(1,+1)$

and four basis vectors. The vectors can be labelled with the quantum numbers $$(s, m)$$ as below, but any four unique symbols could be used, e.g. $$\xi, \psi, x, y$$. The basis set order is also arbitrary;

$\begin{split}\displaystyle \qquad\qquad|0,0\rangle =\begin{bmatrix}1\\0\\0\\0\end{bmatrix}\quad |1,-1\rangle =\begin{bmatrix}0\\1\\0\\0\end{bmatrix}\quad |1,0\rangle =\begin{bmatrix}0\\0\\1\\0\end{bmatrix}\quad |1,1\rangle =\begin{bmatrix}0\\0\\0\\1\end{bmatrix}\qquad\qquad\qquad\qquad\text{(31)}\end{split}$

It is important to realize that in $$| 0, 0 \rangle$$ for example, the $$0, 0$$ is just a label, it has no other meaning per se but the label tells us that this is state $$(0, 0)$$; it does not enter into the maths. The basis set above is orthonormal as can be seen by calculating $$\langle 0, 0 | 1, -1\rangle$$ or any of the other eleven combinations.

To study d-orbitals a basis of five elements is needed. These can be characterized either by the wavefunction labels, $$z^2, x^2 - y^2$$, etc. or by the orbital angular momentum quantum numbers, $$L$$, which for d-orbitals has two units of angular momentum and $$m$$ quantum numbers which take values from $$-2 \to 2$$ in unit steps as $$2, 1, 0, -1, -2$$. The basis can be written using the names of the orbitals as shown on the left, or by using quantum numbers on the right,

$\displaystyle (\psi_{x^2-y^2}, \psi_{xz}, \psi_{z^2}, \psi_{yz}, \psi_{xy}) \equiv((2, -2)\; (2, -1)\; (2, 0)\; (2, +1)\; (2, +2))$

Either way, the $$z^2$$ wavefunction is $$\psi_{z^2}=\begin{bmatrix} 0\\0\\1\\0\\0 \end{bmatrix}$$ and similarly for the other wavefunctions.

A state such as $$\Psi=a\psi_{xy}+b\psi_{xz} = \begin{bmatrix} 0\\a\\0\\0\\b \end{bmatrix}$$.

## 4.3 Inner products $$\langle\alpha |\beta\rangle$$#

The dot or inner product of the two vectors is always a number, i.e. a scalar. Using the normalized and orthogonal (spin) states $$\alpha,\beta$$

$\begin{split}\displaystyle \langle\alpha |\alpha\rangle =\begin{bmatrix}1& 0\end{bmatrix}\begin{bmatrix}1\\ 0\end{bmatrix}=1 \quad \langle\beta |\beta\rangle =\begin{bmatrix}0& 1\end{bmatrix}\begin{bmatrix}0\\ 1\end{bmatrix}=1\quad \langle\alpha |\beta\rangle =\langle\beta |\alpha\rangle =0\end{split}$

There is a somewhat peculiar situation with respect to spin angular momentum as opposed to orbital angular momentum, in that we do not know what the equations describing spin angular momentum are because they are not functions of space. However, this is, in fact, unimportant, because everything can be derived by simply defining the symbols for these states.

Consider the state

$\displaystyle |\varphi\rangle = a|\alpha\rangle + b|\beta\rangle$

which is the linear combination of the $$\alpha$$ and $$\beta$$ basis spin states where $$a$$ and $$b$$ are numbers, often complex numbers. This new state can be represented as the column vector $$\displaystyle |\varphi\rangle =\begin{bmatrix}a\\b\end{bmatrix}$$. To normalize the function $$\varphi$$, the same procedure as for any vector is followed. The inner or dot product $$\langle\varphi |\varphi\rangle$$ is calculated

$\begin{split} \displaystyle \langle\varphi |\varphi\rangle=\begin{bmatrix}a^* &b^*\end{bmatrix} \begin{bmatrix} a\\b\end{bmatrix} = a^*a+b^*b\end{split}$

and the normalization equation defined as $$N^2\langle\varphi |\varphi\rangle = 1$$ with normalization constant $$N$$. Hence $$N^2 = a*a + b*b$$ and normalized wavefunction is

$\displaystyle |\varphi\rangle =\frac{a|\alpha\rangle+b|\beta\rangle}{\sqrt{a^*a+b^*b} }$

If the constants are real numbers, the complex conjugates are not needed. If there are equal amounts of $$a$$ and $$b$$ in $$\varphi$$, then

$\displaystyle |\varphi\rangle =\frac{|\alpha\rangle+|\beta\rangle}{\sqrt{2} }$

To find out how much of state $$\alpha$$ is in the linear combination state $$\varphi$$, this is probed with the basis state $$\alpha$$. The calculation is

$\begin{split} \displaystyle \langle\alpha |\varphi\rangle=\frac{1}{\sqrt{a^*a+b^*b}} \begin{bmatrix}1 &0\end{bmatrix} \begin{bmatrix}a\\b \end{bmatrix} =\frac{a}{\sqrt{a^*a+b^*b}}\end{split}$

If the coefficients $$a$$ and $$b$$ are real and normalized to unity when forming $$\varphi$$ so that $$a^2 +b^2 =1$$, then

$\displaystyle \langle \alpha|\varphi\rangle=a\langle \alpha|\alpha\rangle+b\langle \alpha|\beta\rangle=a\qquad\tag{32}$

but, in either case, this result is interpreted to be the probability amplitude for the state $$\varphi$$ to collapse into the state $$\alpha$$ and the probability of this happening is $$\langle|\varphi^2$$. Similarly, acting on $$\varphi$$ with $$\beta$$ gives the probability amplitude $$\langle\beta|\varphi\rangle =b$$. As the product $$\langle \alpha|\varphi\rangle$$ evaluates to the coefficient $$a$$, we can write the odd looking equation

$\displaystyle |\varphi\rangle=\langle\alpha|\varphi\rangle |\alpha\rangle + \langle\beta|\varphi\rangle |\beta\rangle$

Expanding this out produces

$\begin{split}\displaystyle |\varphi\rangle= \begin{bmatrix}1 & 0 \end{bmatrix}\begin{bmatrix}a\\b \end{bmatrix}\begin{bmatrix}1\\0 \end{bmatrix}+\begin{bmatrix}0 & 1 \end{bmatrix}\begin{bmatrix}a\\b \end{bmatrix}\begin{bmatrix}0\\1 \end{bmatrix} =a|\alpha\rangle +b|\beta\rangle\end{split}$

Two kets in the basis $$(x, y, z, \cdots)$$,

$\begin{split}\displaystyle |A\rangle=a_1|x\rangle + a_2|y\rangle + a_3|z\rangle+\cdots + \\ |B\rangle=b_1|x\rangle + b_2|y\rangle + b_3|z\rangle+\cdots +\end{split}$

form the dot or inner product as do any single column or row matrices or vectors

$\begin{split}\displaystyle \langle A|B\rangle = \begin{bmatrix} a^*_1\; a^*_2\;\cdots\end{bmatrix}\begin{bmatrix}b_1\\b_2\\\vdots\end{bmatrix}=\sum_i a^*_ib_i \end{split}$

If a continuous basis set is being used with a function such as $$R(r)$$ or any wavefunction $$\psi(x)$$ then the inner product becomes an integral. The basis set elements are so closely spaced that the summation of the finite basis is replaced by the integral, and then we have the familiar formula

$\displaystyle \langle\varphi|\psi\rangle =\int \varphi(x)\psi(x)dx$

## 4.4 Outer products:$$|\alpha\rangle\langle \beta|$$ and projection operators $$|\alpha\rangle\langle\alpha|$$ are matrices#

Objects of the form $$|\alpha\rangle\langle \alpha|$$, $$|\alpha\rangle\langle \beta|$$ or $$|\psi_{z^2}\rangle\langle \psi_{xy}|$$ and so forth are always square matrices and correspond to operators. This means that when they are placed to the left of a ket they operate on this and a new ket is formed. Using the definitions above

$\begin{split}\displaystyle|\alpha\rangle\langle\beta|=\begin{bmatrix} 1 \\ 0 \end{bmatrix} \begin{bmatrix} 1 & 0 \end{bmatrix} = \begin{bmatrix} 0 & 1\\ 0 & 0\end{bmatrix}\end{split}$
$\begin{split}\displaystyle|\psi_{z^2}\rangle\langle\psi_{xy}|=\begin{bmatrix}0 \\0 \\ 1 \\ 0 \\0 \end{bmatrix} \begin{bmatrix}0 & 0 & 0 & 1 & 0\end{bmatrix} = \begin{bmatrix} 0 & 0 & 0 & 0 & 0\\ 0 & 0 & 0 & 0 & 0\\ 0 & 0 & 0 & 1 & 0 \\0 & 0 & 0 & 0 & 0\\ 0 & 0 & 0 & 0 & 0 \end{bmatrix}\end{split}$

Operating on the ket

$|\varphi\rangle = a|\alpha\rangle + b|\beta\rangle$

with $$|\alpha\rangle\langle\beta|$$ produces the new ket, $$b|\alpha\rangle$$,

$\displaystyle (|\alpha\rangle\langle\beta|)|\varphi\rangle =|\alpha \rangle\;\langle\beta |\varphi\rangle= \langle\beta|\varphi\rangle|\alpha\rangle=b|\alpha\rangle$

and the last step is made by identifying with $$|\alpha\rangle$$. The result of $$|\alpha\rangle \langle \beta|$$ acting on the ket $$|\varphi\rangle$$, is to produce another ket which is $$b|\alpha\rangle$$, and hence, $$|\alpha\rangle\langle \beta|$$ is an operator. The $$2 \times 2$$ matrix produced is zero except for just one place; hence, it will extract just one value. The alternative calculation can be done by multiplying the two right-hand matrices first, operating with $$\langle \beta|$$ on $$|\varphi\rangle$$ and then multiplying the answer by $$|\alpha\rangle$$,

$\begin{split}\displaystyle |\alpha\rangle\left(\langle\beta| \varphi \rangle \right) = \begin{bmatrix}1\\0\end{bmatrix}\left(\begin{bmatrix} 0&1\end{bmatrix} \begin{bmatrix}a\\b\end{bmatrix}\right)=\begin{bmatrix}1\\0\end{bmatrix}b=b|\alpha\rangle\end{split}$

The operator $$|\alpha\rangle\langle\alpha|$$ will project $$a|\alpha\rangle$$ out of the ket $$|\varphi\rangle$$, and is therefore called a projection operator. The calculation is

$\begin{split}\displaystyle |\alpha\rangle\langle\alpha||\varphi\rangle=\begin{bmatrix}1& 0\\0 & 0\end{bmatrix}|\varphi\rangle=\begin{bmatrix}1& 0\\0 & 0\end{bmatrix}\begin{bmatrix}a\\b\end{bmatrix} = a|\alpha\rangle\end{split}$