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Bra-ket notation is the standard notation used for describing quantum mechanical states. It was invented by Paul Dirac, and is also known as Dirac notation. It is so called because the inner product of two states is denoted by a bracket, ‹φ|ψ›, consisting of a left part, ‹φ|, called the bra, and a right part, |ψ›, called the ket.
Bras and kets
In quantum mechanics, the state of a physical system is identified with a vector in a Hilbert space, H. Each vector is called a ket, and written as
- <math>|\psi\rangle<math>
where ψ is an arbitrary label for the ket. Each element of the dual space of H (i.e. each linear functional from H to the complex numbers C) is known as a bra, and written as
- <math>\langle\phi|<math>
where φ is an arbitrary label for the bra. Applying the bra ‹φ| to the ket |ψ› results in a complex number, called a bra-ket, which we write as the inner product
- <math>\langle\phi|\psi\rangle.<math>
Duals
Every ket |ψ› has a dual bra, written as ‹ψ|, a continuous linear function on H defined as follows:
- <math>\langle\psi|\rho\rangle = ( |\psi\rangle , |\rho\rangle )<math> for all kets <math>|\rho\rangle<math>
where the right hand side ( , ) denotes the inner product given on the Hilbert space. The notation is justified by the Riesz representation theorem, which states that a Hilbert space and its dual space are isometrically isomorphic. Thus, each bra corresponds to exactly one ket, and vice versa.
Properties
Bras and kets can be manipulated in the following ways:
- Given any bra ‹φ|, kets |ψ1› and |ψ2›, and complex numbers c1 and c2, then, since bras are linear functionals,
- <math>\langle\phi|(c_1|\psi_1\rangle + c_2|\psi_2\rangle) = c_1\langle\phi|\psi_1\rangle + c_2\langle\phi|\psi_2\rangle. <math>
- Given any ket |ψ›, bras ‹φ1| and ‹φ2|, and complex numbers c1 and c2, then, by the definition of addition and scalar multiplication of linear functionals,
- <math>(c_1 \langle\phi_1| + c_2 \langle\phi_2|)|\psi\rangle = c_1 \langle\phi_1|\psi\rangle + c_2\langle\phi_2|\psi\rangle. <math>
- <math>
c_1|\psi_1\rangle + c_2|\psi_2\rangle<math> is dual to <math> c_1^* \langle\psi_1| + c_2^* \langle\psi_2|.
<math>
- Given any bra ‹φ| and ket |ψ›, the inner product axiom gives
- <math>\langle\phi|\psi\rangle = \langle\psi|\phi\rangle^*.<math>
Linear operators
If A : H → H is a linear operator, we can apply A to the ket |ψ› to obtain the ket (A|ψ›). The operator also acts on bras: applying the operator A to the bra ‹φ| results in the bra (‹φ|A), defined as a linear functional on H by the rule
- <math>(\langle\phi|A) \; |\psi\rangle = \langle\phi| \; (A|\psi\rangle).<math>
This expression is commonly written as
- <math>\langle\phi|A|\psi\rangle.<math>
A convenient way to define linear operators on H is given by the outer product: if ‹φ| is a bra and |ψ› is a ket, the outer product |φ›‹ψ| denotes the operator which maps the ket |ρ› to the ket |φ›‹ψ|ρ› (here the scalar ‹ψ|ρ› is written to the right of the vector |φ›). One use of the outer product is to construct projection operators. Given a ket |ψ› of norm 1, the orthogonal projection onto the subspace spanned by |ψ› is
- <math>|\psi\rangle\langle\psi|<math>
Composite bras and kets
Two Hilbert spaces V and W may form a third space <math> V \otimes W <math> by a tensor product. If |ψ› is a ket in V and |φ› is a ket in W, the tensor product of the two kets is a ket in <math> V \otimes W <math>. This is written variously as
- <math>|\psi\rangle|\phi\rangle<math> or <math>|\psi\rangle \otimes |\phi\rangle<math> or <math>|\psi \phi\rangle<math>.
Representations in terms of bras and kets
It is sometimes more convenient to work with the projections of state vectors onto a particular basis, rather than the vectors themselves. The reason is that the former are simply complex numbers, and can be formulated in terms of partial differential equations (see, for example, the derivation of the position-basis Schrödinger equation.) This process is very similiar to the use of coordinate vectors in linear algebra.
For instance, the Hilbert space of a zero-spin point particle is spanned by a position basis {|x>}, where the label x extends over the set of position vectors. See rigged Hilbert space. Starting from any ket |ψ> in this Hilbert space, one can define the wavefunction
- <math>\psi(\mathbf{x}) \equiv \lang \mathbf{x}|\psi\rang<math>
It is then customary to define linear operators acting on wavefunctions in terms of linear operators acting on kets, by
- <math>A \psi(\mathbf{x}) \equiv \lang \mathbf{x}|A|\psi\rang <math>
Although the operator A on the left hand side of this equation is, by convention, labelled in the same way as the operator on the right hand side, it should be borne in mind that the two are conceptually different (though closely related) entities: the first acts on wavefunctions, and the second acts on kets. For instance, the momentum operator p has the following form:
- <math>\mathbf{p} \psi(\mathbf{x}) \equiv \lang \mathbf{x} |\mathbf{p}|\psi\rang = - i \hbar \nabla \psi(x) <math>
One occasionally encounters an expression like
- <math> - i \hbar \nabla |\psi\rang<math>
This is something of an abuse of notation, though a fairly common one. The differential operator must be understood to be an abstract operator, acting on kets, that has the effect of differenting wavefunctions once the expression is projected into the position basis.
Example: representing the momentum operator in the place basis
The momentum operator is an hermitian operator. For simplicity of writing we'll assume that <math>\hbar = 1<math>.
In order to obtain the differential form of p we shall transform the problem from a general vector space problem into a problem in a vector space spanned by the x (position) basis.
Since p is hermitian we shall find his eigenfunctions and use them to build a differential equation that will yield us how A operates in the basis of eigenfunctions.
Now, since plane waves are eigenvalues of the momentum operator <math> \lang x |x \rang = c e^{ i p x } <math> (where c is a normalization factor whose exact value is not really important right now), we know that:
- <math> \lang x | p | x \rang = p \lang x | x\rang = p c e^{ i p x } = \frac{1}{i} \frac{d}{dx} c e^{ i p x } = \frac{1}{i} \frac{d}{dx} \lang x | x \rang <math>
and hence we indeed reaffirmed (found) p 's differential form.
For a general function, recall that <math> \lang x | \psi \rang = \psi (x) <math> and since <math> \lang x | p | x \rang = \frac{1}{i} \frac{d}{dx} c e^{ i p x } <math> the problem in the x basis (position representation) is done by something similiar to a basis transformation matrix , <math> \int{| x \rang \lang x | \ dx} = Id <math> (identity operator, and hence don't change the equation),
- <math> p | \psi \rang = \int { A | x \rang \lang x | \psi \rang \ dx } <math> or <math> \lang x | p | \psi \rang = \int{ \lang x | p | x \rang \psi(x) \ dx } = \int{ \left( c \frac{1}{i} \frac{d}{dx} e^{ i p x } \right)^{*} \psi(x) \ dx } <math>
and with integration by parts we shall yield
- <math> \int{ \left( c \frac{1}{i} \frac{d}{dx} e^{ i p x} \right) ^{*} \psi(x) \ dx } = \int{ c e^{ - i p x} \cdot \frac{1}{i} \frac{d}{dx} \psi(x) \ dx }<math>
Now, recalling that the term left of the dot in the integral is actually a representation of the Delta function around x we get
- <math> \lang x | p | \psi \rang = \frac{1}{i} \psi ' (x) <math>
as desired. To be more accurate, we need to put back Planck's constant and that can be done from Dimensional analysis calculations, giving
- <math> \lang x | p | \psi \rang = \frac{\hbar}{i} \frac{d}{dx} \lang x | \psi \rang = \frac{\hbar}{i} \frac{d}{dx} \psi (x) <math>
Please note that x and p were used in several different contexts - as operators, as bra (x) or as eigenvalue number (p). It might be a little confusing.
The calculation is long and exhausting, but since many useful operators are just combinations of x and p = (h/i) * d/dx, the common practice is to skip the calculation and just solve differential equation for the ket, where we seek the ket expressed as a function of x (it can later be represented as a function of p by the Fourier transform ).
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