Dirac notation

The Dirac notation , and bra-ket notation is, in quantum mechanics, a notation for quantum mechanical states . The notation goes back to Paul Dirac . Introduced by him also name bra-ket notation is a play on words with the English term for a clamp ( bracket ). In the Bra-Ket notation, a state is characterized exclusively by its quantum numbers .

In Bra-Ket notation, the vectors of a vector space are also written outside of a scalar product with an angle bracket as a ket . Each ket corresponds to a bra that belongs to the dual space , i.e. represents a linear mapping of the underlying body , and vice versa. The result of the operation of a bra on a ket is written, which establishes the connection with the conventional notation of the scalar product. ${\displaystyle V}$ ${\displaystyle |v\rangle }$ ${\displaystyle |v\rangle }$ ${\displaystyle \langle v|\,,}$ ${\displaystyle V^{*}}$ ${\displaystyle V}$ ${\displaystyle K}$ ${\displaystyle \langle v|}$ ${\displaystyle |w\rangle }$ ${\displaystyle \langle v|w\rangle }$

The notation is used in physics, regardless of whether it is a question of vectors of a vector space or functions in a Hilbert space . The mathematical justification for the Bra-Ket notation results from the theorem of Fréchet-Riesz , which F. Riesz and M. Fréchet independently proved in 1907. Among other things, it states that a Hilbert space and its topological dual space are isometrically isomorphic to one another. In our context: For every ket there is the corresponding bra , and vice versa. ${\displaystyle |v\rangle }$ ${\displaystyle \langle v|}$

presentation

Let be a vector of a complex -dimensional vector space . The Ket expression can be used as column vector with complex elements ( ) shown are: ${\displaystyle v}$ ${\displaystyle m}$ ${\displaystyle \left(v\in \mathbb {C} ^{m}\right)}$ ${\displaystyle \left|v\right\rangle }$ ${\displaystyle v_{n}}$ ${\displaystyle v_{n}\in \mathbb {C} }$

${\displaystyle \left|v\right\rangle \doteq {\begin{pmatrix}v_{1}\\v_{2}\\v_{3}\\\vdots \\v_{m}\end{pmatrix}}}$

It is important that and the associated column vector are not the same mathematical object and therefore no equal sign may be used. This is particularly clear from the fact that the Bra-Ket notation is independent of the choice of a base , while the representation by means of coordinate vectors requires the choice of a base. Instead, it should be made clear that this is the representation of . This can be through the use of characters such as , , done etc. ${\displaystyle \left|v\right\rangle }$ ${\displaystyle (v_{1},v_{2},v_{3},\dotsc ,v_{m})^{T}}$ ${\displaystyle (v_{1},v_{2},v_{3},\dotsc ,v_{m})^{T}}$ ${\displaystyle \left|v\right\rangle }$ ${\displaystyle \Rightarrow }$ ${\displaystyle \doteq }$ ${\displaystyle \leftrightarrow }$

The Bra expression can therefore be represented as a row vector with the conjugate values: ${\displaystyle \left\langle v\right|}$

${\displaystyle \left\langle v\right|\doteq {\begin{pmatrix}v_{1}^{*}&v_{2}^{*}&v_{3}^{*}&\dotso &v_{m}^{*}\end{pmatrix}}}$

Examples

Particles with spin

Using the notation , an electron in the 1s state can be referred to as the spin up of the hydrogen atom . ${\displaystyle |1s,{\uparrow }\rangle }$

photon

The polarization state of a photon can be given as the superposition of two basic states (vertically polarized) and (horizontally polarized): ${\displaystyle |V\rangle }$ ${\displaystyle |H\rangle }$

${\displaystyle |\gamma \rangle =\alpha |V\rangle +\beta |H\rangle }$,

in which

${\displaystyle \alpha ,\beta \in \mathbb {C} }$

and

${\displaystyle \alpha ^{*}\alpha +\beta ^{*}\beta =|\alpha |^{2}+|\beta |^{2}=1}$

System of several bosons

Let a number of bosons be given, each with a certain momentum . The state can be mapped compactly using the Dirac notation: ${\displaystyle n}$ ${\displaystyle q_{k}}$ ${\displaystyle p_{k}=k{\frac {2\pi }{L}}}$

${\displaystyle n}$	State vector	Occupation number representation	Explanation
0	${\displaystyle \left|0\right\rangle }$	${\displaystyle \left|00\right\rangle }$	0 particles are in state 1. 0 particles are in state 2.
1	${\displaystyle \left|q_{1}\right\rangle }$	${\displaystyle \left|10\right\rangle }$	1 particle is in state 1. 0 particles are in state 2.
1	${\displaystyle \left|q_{2}\right\rangle }$	${\displaystyle \left|01\right\rangle }$	0 particles are in state 1. 1 particle is in state 2.
2	${\displaystyle \left|q_{1},q_{1}\right\rangle }$	${\displaystyle \left|20\right\rangle }$	2 particles are in state 1. 0 particles are in state 2.
2	${\displaystyle \left|q_{1},q_{2}\right\rangle =\left|q_{2},q_{1}\right\rangle }$	${\displaystyle \left|11\right\rangle }$	1 particle is in state 1. 1 particle is in state 2.
2	${\displaystyle \left|q_{2},q_{2}\right\rangle }$	${\displaystyle \left|02\right\rangle }$	0 particles are in state 1. 2 particles are in state 2.
3	${\displaystyle \left|q_{1},q_{1},q_{1}\right\rangle }$	${\displaystyle \left|30\right\rangle }$	3 particles are in state 1. 0 particles are in state 2.
3	${\displaystyle \left|q_{1},q_{1},q_{2}\right\rangle =\left|q_{1},q_{2},q_{1}\right\rangle =\left|q_{2},q_{1},q_{1}\right\rangle }$	${\displaystyle \left|21\right\rangle }$	2 particles are in state 1. 1 particle is in state 2.
${\displaystyle \vdots }$	${\displaystyle \vdots }$	${\displaystyle \vdots }$	${\displaystyle \vdots }$

Scalar product

The scalar product of a Bra with a Ket is written in Bra-Ket notation as: ${\displaystyle \langle \phi |}$ ${\displaystyle |\psi \rangle }$

${\displaystyle \langle \phi |\psi \rangle :=\left(\langle \phi |\right)\cdot \left(|\psi \rangle \right)}$

This can be seen as the application of the bras to the ket . ${\displaystyle \langle \phi |}$ ${\displaystyle |\psi \rangle }$

For complex numbers and the following applies: ${\displaystyle c_{1}}$ ${\displaystyle c_{2}}$

${\displaystyle \langle \phi |\;{\bigg (}c_{1}|\psi _{1}\rangle +c_{2}|\psi _{2}\rangle {\bigg )}=c_{1}\langle \phi |\psi _{1}\rangle +c_{2}\langle \phi |\psi _{2}\rangle }$ (Linearity)

Due to the duality relationship, the following also applies:

${\displaystyle \langle \psi |\phi \rangle =\langle \phi |\psi \rangle ^{*}}$ (complex conjugation)

Tensor product

The tensor product of a ket with a bra is written as ${\displaystyle |\phi \rangle }$ ${\displaystyle \langle \psi |}$

${\displaystyle \phi \otimes \psi \ \ =:\ \ |\phi \rangle \langle \psi |}$

In the case of ordinary vectors, the tensor product corresponds to a matrix.

For a complete orthonormal basis the operation leads ${\displaystyle \{|1\rangle ,|2\rangle ,\dotsc ,|N\rangle \}}$

${\displaystyle |1\rangle \langle 1||\psi \rangle =\langle 1|\psi \rangle |1\rangle =c_{1}|1\rangle }$

a projection onto the base state . This defines the projection operator on the subspace of the state : ${\displaystyle |1\rangle }$ ${\displaystyle |1\rangle }$

${\displaystyle |1\rangle \langle 1|}$

A particularly important application of the multiplication of Ket by Bra is the unit operator, which is the sum of the projection operators to ${\displaystyle I}$

${\displaystyle I=\sum _{n=1}^{N}|n\rangle \langle n|.}$

(In infinite-dimensional Hilbert spaces, the Limes has to be considered with a discrete basis .) ${\displaystyle N\to \infty }$

This “representation of the unit operator” is particularly important because it can be used to develop any state in any basis . ${\displaystyle |a\rangle }$

An example of a basic development by inserting the one :

${\displaystyle |a\rangle =I|a\rangle =\sum _{n=1}^{N}|n\rangle \underbrace {\langle n|a\rangle } _{=:\alpha _{n}}=\sum _{n=1}^{N}\alpha _{n}|n\rangle }$

This is the representation of the state chain in the base by inserting the one . ${\displaystyle |a\rangle }$ ${\displaystyle n}$

That this always works is a direct consequence of the completeness of the Hilbert space , in which the states, i.e. the kets , 'live'.

For a continuous basis, an integral must be formed instead of the sum. For example, one obtains for the spatial space the sum over the spatial continuum and thus the unit operator as an integral over the whole : ${\displaystyle \mathbb {R} ^{3}}$

${\displaystyle I=\sum _{\text{kont. Basis}}|{\vec {x}}\rangle \langle {\vec {x}}|=\int _{\mathbb {R} ^{3}}\,\,\mathrm {d} ^{3}\!x\,|{\vec {x}}\rangle \langle {\vec {x}}|}$

Of course, a basic development is also possible with such a continuous basis, which usually leads to a Fourier integral . Technically, this is not a development according to basis vectors of the Hilbert space, as there can be no continuum of pairwise orthogonal vectors in the considered separable spaces : vectors of this type rather form a mathematically non-trivial extension of the considered Hilbert space, and they are therefore called sometimes also “improper vectors” because like the delta function or like monochromatic plane waves they cannot be square-integrated. (The concept of orthogonality must also be generalized here by using delta functions instead of the otherwise usual Kronecker symbols.) ${\displaystyle |{\vec {x}}\rangle }$ ${\displaystyle \delta _{i,j}}$

If you pay attention to these details in calculations, which basically only amount to the “recipes”    and , the basic development remains a useful analogy. ${\displaystyle \sum _{i}\to \int _{\mathbb {R} ^{3}}\,\,\mathrm {d} ^{3}\!x\,}$ ${\displaystyle \delta _{i,j}\to \delta (x_{i}-x_{j})}$

Representations in quantum mechanics

In quantum mechanics, one often works with projections of state vectors onto a certain basis instead of with the state vectors themselves.

The projection onto a certain base is called representation . One advantage of this is that the wave functions obtained in this way are complex numbers for which the formalism of quantum mechanics can be written as a partial differential equation .

• Representation in the location space basis ( location representation ):

Let be an eigenstate of the position operator with the property ${\displaystyle |{\vec {x}}\rangle }$ ${\displaystyle {\hat {x}}}$

${\displaystyle {\hat {x}}|{\vec {x}}\rangle ={\vec {x}}|{\vec {x}}\rangle }$.

The wave function results from projection as ${\displaystyle \psi ({\vec {x}})}$

${\displaystyle \psi ({\vec {x}})=\langle {\vec {x}}|\psi \rangle }$

The scalar product is

${\displaystyle \langle \psi _{1}|\psi _{2}\rangle \ =\int _{\mathbb {R} ^{3}({\vec {x}})}\langle \psi _{1}|{\vec {x}}\rangle \langle {\vec {x}}|\psi _{2}\rangle \,\mathrm {d} ^{3}\!x=\int _{\mathbb {R} ^{3}({\vec {x}})}\psi _{1}({\vec {x}})^{*}\,\psi _{2}({\vec {x}})\,\mathrm {d} ^{3}\!x}$

• Representation in the momentum space basis ( momentum representation ):

Let be an eigenstate of the momentum operator with the property ${\displaystyle |{\vec {p}}\rangle }$ ${\displaystyle {\hat {p}}}$

${\displaystyle {\hat {p}}|{\vec {p}}\rangle ={\vec {p}}|{\vec {p}}\rangle }$.

The wave function results from projection as ${\displaystyle \psi ({\vec {p}})}$

${\displaystyle \psi ({\vec {p}})=\langle {\vec {p}}|\psi \rangle \,\,\left(\equiv \,\int {\frac {e^{-i{\vec {p}}\cdot {\vec {x}}}}{(2\pi )^{3/2}}}\,\psi ({\vec {x}})\,\mathrm {d} ^{3}\!x\right)}$

The scalar product is now the same as before

${\displaystyle \langle \psi _{1}|\psi _{2}\rangle \ =\int _{\mathbb {R} ^{3}({\vec {p}})}\langle \psi _{1}|{\vec {p}}\rangle \langle {\vec {p}}|\psi _{2}\rangle \,\mathrm {d} ^{3}\!p=\int _{\mathbb {R} ^{3}({\vec {p}})}\psi _{1}({\vec {p}})^{*}\,\psi _{2}({\vec {p}})\,\mathrm {d} ^{3}\!p}$

In general, scalar products are invariant for any change of base . Examples are the transitions (“change of representation”) from a complete set of eigenvectors and / or improper eigenvectors of self-adjoint operators of the system to the other, e.g. B. the transition from one matrix system to another or the transition from a matrix representation to a position or impulse representation.

• Matrix elements of an invariant defined “measured variable” with an assigned operator that depends on the base used are the same in all bases, although the operators themselves generally have different representations. This is how one calculates in the location representation ${\displaystyle {\hat {A}}\,,}$

${\displaystyle \langle \psi _{1}|{\hat {A}}|\psi _{2}\rangle \ =\iint _{\mathbb {R} ^{3}({\vec {x}})\,\mathbb {R} ^{3}({\vec {x}}')}\langle \psi _{1}|{\vec {x}}\rangle \langle {\vec {x}}|{\hat {A}}|{\vec {x}}'\rangle \langle {\vec {x}}'|\psi _{2}\rangle \,\mathrm {d} ^{3}\!x\,\mathrm {d} ^{3}\!x'\ }$ ${\displaystyle =\iint _{\mathbb {R} ^{3}({\vec {x}})\,\mathbb {R} ^{3}({\vec {x}}')}\psi _{1}({\vec {x}})^{*}{\hat {A}}({\vec {x}},\,{\vec {x}}')\psi _{2}({\vec {x}}')\,\mathrm {d} ^{3}\!x\,\mathrm {d} ^{3}\!x'}$

The diagonal elements, i.e. those with , are at the same time the expectation values ​​of the operator in the respective states. ${\displaystyle |\psi _{1}\rangle =|\psi _{2}\rangle }$

Individual evidence

1. ↑ Modern Quantum Mechanics Revised Revision, Sakurai, p. 20
2. ↑ Modern Quantum Mechanics Revised Revision, Sakurai
3. ↑ Principles of Quantum Mechanics, R. Shankar, Springer London, Limited, 2012