Bra-ket Notation and Wavefunction
- Quantum Mechanics 1.2
from 「Modern Quantum Mechanics」: Sakurai, J. J.
Bras & Kets, the most basic and important notation in quantum mechanics.
Bras & kets
A physical system is represented by a state vector $\ket{\Psi}$, an element of $L^2$ complex Hilbert space $\mathcal{H}$. (In fact, it doesn’t have to be $L^2$. See $C^*$ formalism who wants exactness.) We denote the dual (covector) of the state vector as $\bra{\Psi}$. Psi is mainly used for the vectors.
$\ket{\Psi}$ is called a ket, and $\bra{\Psi}$ is called a bra. This nomenclature is derived from bra-c-ket〈 〉. $^\dagger$ denotes the hermitian conjugate.
- $ \ket{\Psi} \in \mathcal{H},\; \bra{\Psi} \in \mathcal{H}^* $
- $ \bra{\Psi}^\dagger=\ket{\Psi},\; \ket{\Psi}^\dagger=\bra{\Psi} $
An inner product is expressed as follows, where $^\ast$ denotes the complex conjugate.
- $ \brkt{\phi}{\psi} = \brkt{\psi}{\phi}^\ast $
- $ \brkt{\phi}{c_1\psi_1+c_2\psi_2} = c_1\brkt{\phi}{\psi_1}+c_2\brkt{\phi}{\psi_2} $
- $ \brkt{c_1\phi_1+c_2\phi_2}{\psi} = c_1^\ast \brkt{\phi_1}{\psi}+c_2^\ast \brkt{\phi_2}{\psi} $
In addition, normalized state vectors satisfy the following property:
- $ \brkt{\psi}{\psi}=1 $
Operators
Since we can think of ket as a vector and bra as a co-vector, a linear operator can be considered a square matrix. The hat $\hat{\;\;}$ is often used to indicate an operator. (Regarding an operator as a matrix is not correct since the vectors are infinite-dimensional, but in quantum physics, it doesn’t matter.)
- $ \hat{A}\ket{\Psi} = \ket{\hat{A}\Psi},\; \bra{\Psi}\hat{A} = \bra{\hat{A}^\dagger\Psi} $
- $ \brktop{\phi}{\hat{A}}{\psi} \coloneqq \left(\bra{\phi}\hat{A}\right)\ket{\psi} = \bra{\phi}\left(\hat{A}\ket{\psi}\right) $
If the same state vector appears on both the bra and ket sides,
- $ \langle\hat{A}\rangle \coloneqq \brktop{\psi}{\hat{A}}{\psi} $
then this expression gives the expectation value, or mean or average value, of the observable represented by operator $\hat{A}$ for the physical system in the state $\ket{\psi}$.
A convenient way to define linear operators on a Hilbert space is given by the outer product :
- $ \ket{\phi}\bra{\psi} $
One of the uses of the outer product is to construct projection operators. Given a ket $\ket{\psi}$ of norm $1$, the orthogonal projection onto the subspace spanned by $\ket{\psi}$ is :
- $ \ket{\psi}\bra{\psi} $
Basis
State vectors are elements of infinite-dimensional Hilbert space, which is a vector space. This implies that there exists a basis, which spans every state vectors. We can write this using projection operators.
- $ \dps \ket{\psi} = \sum_{i\in\N} \left( \ket{e_i} \bra{e_i} \right) \ket{\psi} = \sum_{i\in\N} \brkt{e_i}{\psi} \ket{e_i} $
- $ \dps \sum_{i\in\N} \ket{e_i} \bra{e_i} = \mathbb{I} $
Position space and wavefunction
Any vectors that generate the Hilbert space can be a basis. Let’s take a look at position space first. This label is the eigenvalue of the position operator acting on such a basis state, $ \hat{\b{r}}\ket{\b{r}} = \b{r}\ket{\b{r}}$.
The $\ket{\b{r}}$ are the basis vectors, which are orthonormal so their inner product is a delta function;
- $ \brkt{\b{r}\rq}{\b{r}} = \delta^3(\b{r}\rq-\b{r}) $
Thus, starting from any ket $\ket{\Psi}$ in this Hilbert space, one may define a complex scalar function $\Psi(\b{r})$, known as a wavefunction :
- $ \Psi(\b{r}) \coloneqq \brkt{\b{r}}{\Psi} $
- $ \dps \ket{\Psi} = \int \d^3\b{r}\; \Psi(\b{r})\ket{\b{r}} $
- $ \hat{A}(\b{r})\Psi(\b{r}) = \brktop{\b{r}}{\hat{A}}{\Psi} $
Let’s re-write the properties of state vectors with wavefunctions.
- $ \dps \brkt{\phi}{\psi} = \int \phi(\b{r})^\ast \psi(\b{r}) \,\d^3\b{r} $
- $ \dps \brkt{\psi}{\psi} = \int | \psi(\b{r}) |^2 \,\d^3\b{r} \enspace$ ($1$ if normalized)
We’ll talk about momentum space later.
