Bra-ket notation

Bra-ket notation is the standard notation for describing quantum states in the theory of quantum mechanics. It can also be used to denote abstract vector space and linear functionals in pure mathematics. It is so called because the inner product of two states is denoted by a bracket,

\langle\phi|\psi\rangle

, consisting of a left part,

\langle\phi|

, called the bra, and a right part,

|\psi\rangle

, called the ket. The notation was invented by Paul Dirac, and is also known as Dirac notation. == Bras and kets == In quantum mechanics, the state of a physics system is identified with a vector (spatial) in a complex number Hilbert space, ''H''. Each vector is called a "ket", and written as :

|\psi\rangle

where ψ denotes the particular ket. Every ket

|\psi\rangle

has a dual bra, written as :

\langle\psi|

This is a continuous linear functional from ''H'' to the complex numbers C, defined by: :

\langle\psi|\rho\rangle = \bigg( |\psi\rangle \;,\; |\rho\rangle \bigg)

for all kets

|\rho\rangle

where ( , ) denotes the inner product defined 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. This is not always the case however, on page 111 of Quantum Mechanics by Cohen-Tannoudji et. al. it is clarified that there is such a relationship between bras and kets so long as the defining functions used are square integrable. Consider a continuous basis and a Dirac delta function or a sine or cosine wave as a wave function. Such functions are not square integrable and therefore it arises that there are bras that exist with no corresponding ket. The reason this does not hinder quantum mechanics is because all wave functions are square integrable in reality. Incidentally, bra-ket notation can be used even if the vector space is not a Hilbert space. In any Banach space ''B'', the vectors may be notated by bras and the continuous linear functionals by kets. Over any vector space without topology, we may also notate the vectors by bras and the linear functionals by kets. In these more general contexts, the bracket does not have the meaning of an inner product, because the Riesz representation theorem does not apply. Applying the bra

\langle\phi|

to the ket

|\psi\rangle

results in a complex number, called a "bra-ket" or "bracket", which is written as :

\langle\phi|\psi\rangle

. In quantum mechanics, this is the probability amplitude for the state ψ to collapse into the state φ. == Properties == Bras and kets can be manipulated in the following ways: * Given any bra

\langle\phi|

, kets

|\psi_1\rangle

and

|\psi_2\rangle

, and complex numbers ''c''₁ and ''c''₂, then, since bras are ''linear'' functionals, ::

\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.

* Given any ket

|\psi\rangle

, bras

\langle\phi_1|

and

\langle\phi_2|

, and complex numbers ''c''₁ and ''c''₂, then, by the definition of addition and scalar multiplication of linear functionals, ::

\bigg(c_1 \langle\phi_1| + c_2 \langle\phi_2|\bigg) \; |\psi\rangle = c_1 \langle\phi_1|\psi\rangle + c_2\langle\phi_2|\psi\rangle.

* Given any kets

|\psi_1\rangle

and

|\psi_2\rangle

, and complex numbers ''c''₁ and ''c''₂, from the properties of the inner product (with c* denoting the complex conjugate of c), ::

c_1|\psi_1\rangle + c_2|\psi_2\rangle

is dual to

c_1^* \langle\psi_1| + c_2^* \langle\psi_2|.

* Given any bra

\langle\phi|

and ket

|\psi\rangle

, an axiomatic property of the inner product gives ::

\langle\phi|\psi\rangle = \langle\psi|\phi\rangle^*

. == Linear operators == If ''A'' : ''H'' → ''H'' is a linear operator, we can apply ''A'' to the ket

|\psi\rangle

to obtain the ket

(A|\psi\rangle)

. Linear operators are ubiquitous in the theory of quantum mechanics. For example, self-adjoint operator are used to represent observable physical quantities, such as energy or momentum, whereas unitary operator linear operators represent transformative processes such as rotation or the progression of time. Operators can also be viewed as acting on bras ''from the right hand side''. Applying the operator ''A'' to the bra

\langle\phi|

results in the bra

(\langle\phi|A)

, defined as a linear functional on ''H'' by the rule :

\bigg(\langle\phi|A\bigg) \; |\psi\rangle = \langle\phi| \; \bigg(A|\psi\rangle\bigg)

. This expression is commonly written as :

\langle\phi|A|\psi\rangle.

A convenient way to define linear operators on ''H'' is given by the outer product: if

\langle\phi|

is a bra and

|\psi\rangle

is a ket, the outer product :

|\phi\rang \lang \psi|

denotes the rank one operator that maps the ket

|\rho\rangle

to the ket

|\phi\rangle\langle\psi|\rho\rangle

(where

\langle\psi|\rho\rangle

is a scalar multiplying the vector

|\phi\rangle

). One of the uses of the outer product is to construct projection operators. Given a ket

|\psi\rangle

of norm 1, the orthogonal projection onto the Linear subspace spanned by

|\psi\rangle

is :

|\psi\rangle\langle\psi|

== Composite bras and kets == Two Hilbert spaces ''V'' and ''W'' may form a third space

V \otimes W

by a tensor product. In quantum mechanics, this is used for describing composite systems. If a system is composed of two subsystems described by ''V'' and ''W'' respectively, then the Hilbert space of the entire system is the tensor product of the two spaces. (The exception to this is if the subsystems are actually identical particles. In that case, the situation is a little more complicated.) If

|\psi\rangle

is a ket in V and

|\phi\rangle

is a ket in W, the tensor product of the two kets is a ket in

V \otimes W

. This is written variously as :

|\psi\rangle|\phi\rangle

|\psi\rangle \otimes |\phi\rangle

|\psi \phi\rangle

. == Representations in terms of bras and kets == In quantum mechanics, it is often 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 Schr�dinger equation). This process is very similar to the use of coordinates vector in linear algebra. For instance, the Hilbert space of a spin (physics) point particle is spanned by a position basis

\lbrace|\mathbf{x}\rangle\rbrace

, where the label x extends over the set of position vectors. Starting from any ket

|\psi\rangle

in this Hilbert space, we can ''define'' a complex scalar function of x, known as a wavefunction: :

\psi(\mathbf{x}) \equiv \lang \mathbf{x}|\psi\rang

. It is then customary to define linear operators acting on wavefunctions in terms of linear operators acting on kets, by :

A \psi(\mathbf{x}) \equiv \lang \mathbf{x}|A|\psi\rang

. 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 entities: the first acts on wavefunctions, and the second acts on kets. For instance, the momentum operator p has the following form: :

\mathbf{p} \psi(\mathbf{x}) \equiv \lang \mathbf{x} |\mathbf{p}|\psi\rang = - i \hbar \nabla \psi(x)

. One occasionally encounters an expression like :

- i \hbar \nabla |\psi\rang

. 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 differentiating wavefunctions once the expression is projected into the position basis: :

- i \hbar \nabla \lang\mathbf{x}|\psi\rang

. ''For further details, see rigged Hilbert space.'' Quantum mechanics Mathematical notation Bilinear forms

Bra-ket notation

Should we really use & rang ; and & lang ;? My Mozilla on Windows and IE both don't render it. (For mozilla this is reported as a bug in bugzilla: http://bugzilla.mozilla.org/show_bug.cgi?id=15731 ) Is there actually a browser that does? -- User:Jan Hidders 12:19 Mar 5, 2003 (UTC) What's wrong with th ordinary angle brackets < and > ? or & lt ; and & gt ; ? User:Theresa knott :Mozilla renders it fine for me. rang and lang are HTML 4.0 [http://www.mozilla.org/newlayout/testcases/layout/entities.html character entity references] specifically for "bras" and "kets", so it's more correct to use them. It also looks more legible; < and > make the bras and kets somewhat more difficult to read. :IE 4 should be able to display the characters ([http://www.alanwood.net/unicode/explorer_older.html see here]). Are you using IE 3? Do 〈 and 〉 (generated from the numeric codes) work for you? -- User:CYD I use IE 6. I can see everything on Wikipedia:Special characters and on http://www.unicode.org/iuc/iuc10/x-utf8.html , but neither 〈 nor ⟨ - User:Patrick 21:01 Mar 5, 2003 (UTC) Can you see the characters on http://www.htmlhelp.com/reference/html40/entities/symbols.html ? -- User:CYD : I couldn't. I was also using IE6 under W'98 and there it didn't render. IE6 under XP also doesn't render it. Mozilla 1.2 under W'98 doesn't render it either, but Mozilla 1.3 under Linux and XP do. I'll see what happens if I upgrade to Mozialla 1.3 under W'98. If that works, than I'm happy with lang and rang, although strictly that would not be enough for the official policy on special characters. -- User:Jan Hidders 21:22 Mar 5, 2003 (UTC) ::Is it a problem with the browser or simply the correct font that is missing? : Then maybe we should use an image, similar to what is done for (). Out of curiosity, what "official policy" are you referring to? -- User:CYD :: You could, but images are really the last resort. They don't scale and are sometimes positioned awkwardly by different browsers. So I would suggest using < and >. If you really don't like those then my next choice would be to use lang and rang anyway, just as long as we can tell people that if they want to see the page in its full glory they have to install the latest Mozilla. You might even want to plead for a change of policy on the mailing list. In that case you have my vote. :-) The "official policy" is more or less implicit in the page on special characters. From what I remember from previous discussions on the mailing list the main argument is always that we should keep Wikipedia as accessible as possible and therefore only use special characters if we really need them. -- User:Jan Hidders 21:35 Mar 5, 2003 (UTC) :::I would say that having to use one particular browser is much worse than either an image or a regular <. - User:Patrick 22:11 Mar 5, 2003 (UTC) ::::Absolutely, however, where Mozilla goes so do the browsers that are based on the Gecko rendering engine, and since it follows the standards the KHTML-based browsers (Konquerer et al.) and other open source browsers are usually not far behind and even IE will probably catch up if it has not already. Besides, Mozilla is pretty easy to install these days and availiable on many platforms. -- User:Jan Hidders 22:40 Mar 5, 2003 (UTC) Using & #9001 ; &lang ; I can see them here: 〈 ⟨, in accordance with http://www.alanwood.net/unicode/miscellaneous_technical.html , which says "LEFT-POINTING ANGLE BRACKET (present in WGL4 and in Symbol font)" - User:Patrick 21:58 Mar 5, 2003 (UTC) :Ah I can see them now [IE 5 windows 2000]. Is it safe to assume that symbol is a pretty much unversal font? User:Theresa knott 14:20 Mar 6, 2003 (UTC) Alternatively, we can use TeX all the time for these brackets, they work fine also. - User:Patrick 22:06 Mar 5, 2003 (UTC) : In-line TeX is usually discouraged. See Wikipedia:WikiProject_Mathematics. -- User:Jan Hidders 22:40 Mar 5, 2003 (UTC) ::comparison: the "correct" in-line symbol (?), the ordinary less-than, and the TeX symbol: ⟨<

\langle

. I see more difference between the correct in-line symbol and the TeX symbol than between the correct in-line symbol and the ordinary less-than! So using the ordinary < and > seems best. - User:Patrick 04:32 Mar 6, 2003 (UTC) ::: That's strange. The TeX symbol looks exactly like ⟨ to me. The difference between the correct symbol and an ordinary < looks to me like the difference between

\langle

and

<\;

. -- User:CYD :::: Perhaps Patrick is referring to the size and not the shape? -- User:Jan Hidders 10:36 Mar 7, 2003 (UTC) :::::No, to the shape. In my case ⟨ and < have a much smaller angle than

\langle

. - User:Patrick 12:40 Mar 7, 2003 (UTC) ::::::Yup. I am now looking at it with my IE6 under W'98 and also there the shapes differ. The symbol font, byt the way, works for me in IE6 and Mozilla under W'98, XP and Mozilla under Linux, but seems a bit cumbersome to type. Does anybody know if there is some font that I could install under Windows to see lang and rang? -- User:Jan Hidders 15:51 Mar 7, 2003 (UTC) ::::::: Better to be cumbersome to type than impossible to read. I've changed the page accordingly. Could people check for mistakes please. Can anyone still not read the text ?User:Theresa knott 09:30 Mar 11, 2003 (UTC) ---- Regarding :* Given any ket |ψ›, bras ‹φ₁| and ‹φ₂|, and complex numbers ''c''₁ and ''c''₂, then, by the definition of addition and scalar multiplication of linear functionals, ::

(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.

As far as I know, it should be c₁^* and c₂^* since an inner product over the complex field is sesqi-linear in first component, or hermitian. Proof: let A and B be Vector (spatial)s and c a complex number. Then, from hermitian property : = < B|cA>^* , but from linear property : < B|cA> = c< B|A>, hence = (c< B|A>)^* , hence : = c^*< B|A>^* = c^* . Q.E.D. User:MathKnight 21:10, 28 Mar 2004 (UTC) Regarding : = < B|cA>^* , This should instead be : = < B|c*A>^* , Your proof is therefore erroneous. -- User:CYD The passage : = < B|cA>^* is correct, since we mark D = cA and then from [http://en.wikipedia.org/wiki/Inner_product#Definitions Conjugate Symmetry], :

\lang D | B \rang = { \lang B | D \rang }^*

and subsitute again we get : = < B|cA>^* Notice that ''c'' (a scalar) multiplies A. I can also show it directly, define bra-ket inner product as: :

\lang F | G \rang = \int{ F^* \cdot G dx }

you immidietly see that :

\lang c \cdot F | G \rang = \int{ (cF)^* \cdot G dx } = c^* \int{ F^* \cdot G dx } = c^* \cdot \lang F | G \rang

User:MathKnight 09:44, 15 May 2004 (UTC) Okay, I see the confusion now. Well, since the relevant line in the article clearly refers to the addition and scalar multiplication of linear functionals, your original objection is obviously invalid. In any case, the issue has already addressed; look at the third point of that section: ::

c_1|\psi_1\rangle + c_2|\psi_2\rangle \;\;
\hbox{is dual to} \;\;
c_1^* \langle\psi_1| + c_2^* \langle\psi_2|.

The concept of "multiplying inside the ket" is redundant in bra-ket notation. If the article is to talk about that, it should do so in a careful way in order to avoid confusion. -- User:CYD ---- In the first sentence of the "Linear Operators" section, does anyone see :''A'' : ''H'' → ''H'' with a square instead of a right arrow? I can see the right arrow if written in the Village Pump section, but not in this article (nor in some of the math-related articles). –User:Matthew Low 23:35, 12 Jun 2004 (UTC) == braket-notation + cross-product == This primarily is a question: Can somebody give the braket form of the cross product: a vector x b vector = c vector? Thanks. *Have you tried asking the Wikipedia:Reference_desk? User:Matthew Low–User:Matthew Low 22:44, 15 Jun 2004 (UTC) Since the statevectors aren't spatial vectors (i.e. (x,y,z) ) but rather an abstract entities representing quantum states, your question isn't defined well. User:MathKnight 10:51, 16 Jun 2004 (UTC) == differential operators == Just some thoughts regarding: For example, if ''A = d/dx'', then we are differenting |ψ› when we calculate '' ‹φ|A|ψ› ''. What's ''x'' here? If it parameterizes |ψ⟩ then shouldn't the ket be written |ψ(''x'')⟩? In any case, it might be an idea to use something other than ''x'' in order to prevent confusion with position based Schroedinger operators which act on the wavefunction ⟨''x''|ψ⟩. Besides this, differential operators are usually understood to act to the right, whilst QM operators may act either way. === Answer: === Well, in QM it is not correct to say that ''A'' is ''d/dx'' but rather than ''A=d/dx'' on place-representation. When taking A in place-representation, than the ket '' |ψ> '' is also represent as a function ''x''. We might just as well say that ''A = p * i/(hbar)'' and then |&psil> will be represent as a function of ''p'' (momentum).
The formal mathimatical way to prove the above statement is as follows:
Lets suppose

A = p = \frac{\hbar}{i} \frac{d}{dx}

(i.e. A is the momentum operator, which is an hermitian operator. For simplicity of writing we'll assume that ''hbar = 1''.
In order to obtain the differential form of ''A'' (we prove the statement above) 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 plain waves are eigenvalues of the momentum operator

\lang x |x \rang = c e^{ i p x }

, ( ''c'' is normalization factor and his exact value is not realy important right now), we know that: :

\lang x | A | 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

and hence we indeed reaffirmed (found) A's differential form. For a general function, recall that

\lang x | \psi \rang = \psi (x)

and since

\lang x | A | x \rang = \frac{1}{i} \frac{d}{dx} c e^{ i p x }

the problem in the ''x'' basis (position representation) is done by something similiar to linear basis ,

\int{| x \rang \lang x | \ dx} = Id

(identity operator, and hence don't change the equation), :

A | \psi \rang = \int { A | x \rang \lang x | \psi \rang \ dx }

\lang x | A | \psi \rang = \int{  \lang x | A | x \rang \psi(x) \ dx } = \int{ \left( c \frac{1}{i} \frac{d}{dx} e^{ i p x } \right)^{*} \psi(x) \ dx }

and with integration by parts we shall yield :

\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 }

Now, recalling the the left term (left to the dot) in the integral is actually a reprenstation of the Delta function around ''x'' we get :

\lang x | A | \psi \rang = \frac{1}{i} \psi ' (x)

as desired.

The calculation is long and exausting, but since many useful operators are just combination of ''x'' and ''p = (h/i) * d/dx'' the common way is to skip the calculation just solving differential equation for the ket, where we search the ket expressed as a function of ''x'' (it later can represent as a function of ''p'' by the Fourier transform ). User:MathKnight 09:47, 3 Aug 2004 (UTC) Revision: I want later to enter this answer to the article as an explanation about "bra-ket and concrete representations". User:MathKnight 15:18, 3 Aug 2004 (UTC) Hi MK, thanks for your explanation. I suspected the statement may have been refering to a basis representation of the operator, but am concerned that this subtlety might be overlooked by readers not previously familiar with the material. The simple fact remains that ''d/dx''|ψ⟩ is identically zero, ''unless'' you understand that ''d/dx'' is a label meant to imply that ''A'' is the operator defined so that ''⟨x|A|ψ⟩=d/dx ⟨x|ψ⟩''. You are abeslutly right, therefore I think that this article should handle the representation issue. I think we can rely on the explaination here with few edits such as general intro for the representation problem. I want to add it to the article and I'll work on it. Also, I get white squres in in-text math-symbols (such as ''⟨'' = ''⟨''. I prefer to work with LaTex for math, such as the previous thing you wrote:

\lang x | A | \psi \rang = \frac{d}{dx} \lang x | \psi \rang = \frac{d}{dx} \psi (x)

. User:MathKnight 20:10, 4 Aug 2004 (UTC) == Plane Waves == I'm not sure if this is just a different notation, but I believe the "plane waves" eigenstates of the momentum operator are denoted ⟨x|p⟩ rather than ⟨x|x⟩ which would be some strange delta function concotion. :You're right, of course. In any case, I've deleted the section on the momentum eigenstates, because it's not relevant to the article, which is about bra-ket notation not wavefunctions. -- User:CYD :You're right. The confusion was created by abuse of notation, since |x> is a reserved notation of a eigenstate of ''x'' operator. You are right, is the proper notation. User:MathKnight 17:30, 16 Oct 2004 (UTC) == Linear Operators == I think the defenition

Operators can also act on bras. Applying the operator ''A'' to the bra $\langle\phi|$ results in the bra ( $\langle\phi|$ ''A''), defined as a linear functional on ''H'' by the rule : $(\langle\phi|A) \; |\psi\rangle = \langle\phi| \; (A|\psi\rangle)$ . This expression is commonly written as : $\langle\phi|A|\psi\rangle.$

is a little bit misleading and need clarification. For some reason it ignores the conjugation that A must go through when it swtiches sides in the bra-kets. [http://en.wikipedia.org/w/wiki.phtml?title=Bra-ket_notation&oldid=5003286#Linear_operators] User:MathKnight 10:52, 27 Nov 2004 (UTC) :The statement is obviously correct. I think you're getting confused by the "multiplication inside a ket" issue again. -- User:CYD == symbol changed midway through == " \langle\phi|\psi\rangle. In quantum mechanics, this is the probability amplitude for the state ψ to collapse into the state φ." Is that right, or did some ψ's get changed to phi's at the end of this section? :It is currently correct as is. <φ|ψ> is the projection (linear algebra) of state |ψ> onto state |φ>, which can be interpreted as the probability for |ψ> to change into |φ>. --User:Laurascudder | User talk:Laurascudder 21:26, 16 May 2005 (UTC)

See other meanings of words starting from letter:

B

BA | BC | BD | BE | BF | BG | BH | BI | BJ | BK | BL | BM | BN | BO | BP | BR | BS | BT | BU | BW | BX | BY | BZ |

Words begining with Bra-ket_notation:

Bra-ket_notation
Bra-ket_notation