Talk:Bra-ket notation

From Wikipedia, the free encyclopedia

You have new messages (last change).

This article is within the scope of WikiProject Physics, which collaborates on articles related to physics.

Physics Portal

This article has been rated as B-Class on the assessment scale.

???

This article has not yet received an importance rating within physics.

Help with this template

This article has been rated but has no comments. If appropriate, please review the article and leave comments here to identify the strengths and weaknesses of the article and what work it will need.

1 Unheadered stuff at top
2 Plane Waves
3 Linear Operators
4 symbol changed midway through
5 C Number Omission
6 Quantum Computing
7 Banach Spaces
8 Riesz representation
9 what does bra-ket notation mean?
10 reason for deleting section
11 H*
12 Notation used by mathematicians
13 Small "errors" / omissions
14 Rewrite?

[edit] Unheadered stuff at top

I have changed the (unfinished) last paragraph about Hermitian operators of the section "Linear Operators" to include some physical meaning and to prevent operators from jumping onto the bras'/kets' label. I think this notation is unsuitable for beginners and not very useful for experts. Y!qtr9f 18:11, 30 November 2006 (UTC)

what does "is dual to" mean?

See Dual space! Y!qtr9f 18:11, 30 November 2006 (UTC)

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? -- Jan Hidders 12:19 Mar 5, 2003 (UTC)

What's wrong with th ordinary angle brackets < and > ? or & lt ; and & gt ; ? Theresa knott

Mozilla renders it fine for me. rang and lang are HTML 4.0 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 (see here). Are you using IE 3? Do 〈 and 〉 (generated from the numeric codes) work for you? -- 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 〈 - Patrick 21:01 Mar 5, 2003 (UTC)

Can you see the characters on http://www.htmlhelp.com/reference/html40/entities/symbols.html ? -- 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. -- 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

([[Image:Del.gif]]). Out of curiosity, what "official policy" are you referring to? -- 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. -- 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 <. - 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. -- Jan Hidders 22:40 Mar 5, 2003 (UTC)

Using <FONT FACE="SYMBOL">& #9001 ; &lang ;</FONT> 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)" - 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? Theresa knott 14:20 Mar 6, 2003 (UTC)

Alternatively, we can use TeX all the time for these brackets, they work fine also. - Patrick 22:06 Mar 5, 2003 (UTC)

In-line TeX is usually discouraged. See Wikipedia:WikiProject_Mathematics. -- 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. - 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 $<\;$ . -- CYD

Perhaps Patrick is referring to the size and not the shape? -- Jan Hidders 10:36 Mar 7, 2003 (UTC)

No, to the shape. In my case 〈 and < have a much smaller angle than $\langle$ . - 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? -- 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 ?Theresa knott 09:30 Mar 11, 2003 (UTC)

Regarding

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,

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 vectors and c a complex number. Then, from hermitian property

<cA|B> = < B|cA>^* ,

but from linear property

< B|cA> = c< B|A>, hence <cA|B > = (c< B|A>)^* ,

hence

<cA| B> = c^*< B|A>^* = c^*<A|B > .

Q.E.D. MathKnight 21:10, 28 Mar 2004 (UTC)

Regarding

<cA|B> = < B|cA>^* ,

This should instead be

<cA|B> = < B|c*A>^* ,

Your proof is therefore erroneous. -- CYD

The passage

<cA|B> = < B|cA>^*

is correct, since we mark D = cA and then from Conjugate Symmetry,

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

and subsitute again we get

<cA|B> = < 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

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

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 b 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 ).

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". 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: $\lang x | A | \psi \rang = \frac{d}{dx} \lang x | \psi \rang = \frac{d}{dx} \psi (x)$ . MathKnight 20:10, 4 Aug 2004 (UTC)

[edit] 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. -- 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, <x|p> is the proper notation. MathKnight 17:30, 16 Oct 2004 (UTC)

[edit] 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. [1] 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. -- CYD

[edit] 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?

[edit] C Number Omission

Hello all: I've dared to put my remarks at the top of the page, as I'm unaware (as yet) of any convention concerning where to post newer remarks, other than below for responses. I'm sure I'll get well smacked around if I'm being presumptious about this.

I've noted that while Dirac's Bra and Ket notation is addressed, there is both a spelling omission and Dirac's "c number", which accounts for the omitted letter referenced.

A complete bra and ket notation discussion actually should reference a number c which goes between the bra and ket.

I'll include more about this when I access my copy of Dirac's 4th edition, which at the moment is probably in my storage unit at home. Doug Reiss 15:08 January 27, 2006 (EDST)

[edit] Quantum Computing

The beginning of the article says: "It has recently become popular in quantum computing." Has it ever not been popular in quantum computing? It seemed to be popular in 1999, which I don't consider "recent". I don't remember reading any quantum computer papers without bra-ket notation. A5 19:52, 27 February 2006 (UTC)

[edit] Banach Spaces

I have a question about these statements:

1. "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. This does not hinder quantum mechanics because all physically realistic wave functions are square integrable."

What exactly is the bra here? Shouldn't it say "therefore it arises that there are kets without any bra"?

2. "Bra-ket notation can be used even if the vector space is not a Hilbert space."

Is the space mentioned in the statement 1 a Hilbert space? It shouldn't be, because otherwise the Riesz theorem would hold, right? Is it a Banach space? If it is, shouldn't it be moved to the following paragraph with statement 2 as a (useful) example? A5 20:41, 27 February 2006 (UTC)

So far as I know, the two statements should be equivalent, but I'm no expert.
I believe that the current ordering is attempting a transition from the physically motivated to theory and spaces. It's a wiki so be bold if you don't like it as is. — Laura Scudder ☎ 21:12, 27 February 2006 (UTC)

You can use a rigged Hilbert space for the first thing. You can use any reflexive Banach space for the second thing. Charles Matthews 16:19, 3 March 2006 (UTC)

yeah, how often does one find a piece of literature on Banach spaces, or functional analysis in general for that matter, that uses the bra ket notation? never. it's essentially a physical notation, and as such perhaps lend itself more to physical interpretation. doesn't really pay to try to pin down its rigorous meaning. although Charles is right in what he said. In the finite dim case, of coure, rigorization can be done trivially. Mct mht 20:05, 22 April 2006 (UTC)

The statement quoted is typical physical language, can be rigorized using spectral theory for self adjoint operators on Hilbert spaces. Or, like Charles said, rigged Hilbert spaces. Mct mht 20:13, 22 April 2006 (UTC)

[edit] Riesz representation

small mistake in article. the Riesz representation is not quite an isomorphism; it's conjugate linear. i didn't change it because, IMHO, it's rather pointless to justify the notation rigorously. one does not use a piece of notation to speak of "continuous basis" on a separable Hilbert space, and use that same notation to discuss, say, Hahn-Banach. Mct mht

[edit] what does bra-ket notation mean?

Can bra-ket notation be converted to normal mathematics? I'm pretty sure it can, because Schrödinger's equation can be expressed both with bra-ket notation and derivatives. Fresheneesz 10:46, 25 April 2006 (UTC)

From what I can glean from a pervious discussion on this page: does this pattern describe how bra-ket notation is always used?

$\lang independant \,\, variable | operator | function \rang = operator \lang independant \,\, variable | \psi \rang = operator \psi (independant \,\, variable )$

or is it more general like this?:

$\lang independant \,\, variable | a | function \rang = a * \lang independant \,\, variable | \psi \rang = a * \psi (independant \,\, variable )$

??? Fresheneesz 20:48, 25 April 2006 (UTC)

I think the most concrete example is

$\lang \psi | \hat O | \psi \rang = \int_{-\infty}^{+\infty} \psi^* \hat O \psi \, dx$

- mako 22:53, 25 April 2006 (UTC)

Should we put that somewhere in the introduction, because in this article, its very unclear how bra-ket notation is translated into something meaningful. This article is rather on the properties of manipulating and using the notation. 68.6.112.70 02:14, 26 April 2006 (UTC)

Right now I think you're supposed to get that from the third equation when it says $\lang \phi | \psi \rang$ is an inner product, but it could be made more explicit. The problem with trying to tie it in to wavefunction mechanics with examples is that you don't want to accidentally imply that bra-ket notation is just some kind of shorthand for doing wavefunctions in some particular representations. The above example, for instance, technically does this middle step with the identity matrix $| x \rang \lang x |$ :

$\lang \psi | \hat O | \psi \rang = \lang \psi | x \rang \lang x | \hat O | x \rang \lang x | \psi \rang = \int_{-\infty}^{+\infty} \psi^*(x) \hat O_x \psi(x) \, dx$

There's nothing special about doing the inner product in position space (could just as easily use any other identity), but that could easily be misunderstood. — Laura Scudder ☎ 05:24, 26 April 2006 (UTC)

Also, so far as being meaningful, there's some interesting situations where with bra-ket notation you can find out a lot of useful information about a system for which you can't solve for the wavefunctions, so it's a useful standalone technique. I think the article could convey that better. — Laura Scudder ☎ 05:37, 26 April 2006 (UTC)

So can you correct/fill-in-the-blanks for me here?:

$\lang a(x) | \hat O | b(x) \rang = \int_{-\infty}^{+\infty} a(x)^* \hat O_x b(x) \, dx$

where

a (x)

is a function of x.

$\hat O$ is ______?

b (x)

is a second function of x.

Fresheneesz 06:11, 30 April 2006 (UTC)

$\hat O$ is an operator (something like $\frac{d}{dx}$ ) Trewornan 03:41, 2 June 2006 (UTC)

To be picky, I usually prefer not to write $\lang a(x) |$ as it's not perfectly clear what's meant. The bra $\lang a |$ is a state that's not in any particular representation, while the projection of that bra onto a basis set is for instance $\lang x | a \rang = a(x)$ . I would usually interpret $\lang a(x) |$ then as $\lang a | x \rang \lang x |$ .

But to answer your question about $\hat{O}$ , in my equations above I meant the operator in no particular representation. So for instance, in that notation I wouldn't be able to write down what the 1-D momentum operator $\hat{p}$ was because it will be different in position and momentum spaces and I haven't yet determined what I'm working in. What I do know how to write down is $\lang x | \hat p | x \rang = - i \hbar \frac{\partial}{\partial x}$ . Projecting the abstract onto a representation basis set is what allows me to write things down in a traditional way.

So a concrete example of what this means:

$\lang \phi | \hat{p} | \psi \rang = \lang \phi | x \rang \lang x | \hat p | x \rang \lang x | \psi \rang = \int \phi^*(x) \left( - i \hbar \frac{\partial}{\partial x} \right) \psi(x) dx$

or equivalently

$\lang \phi | \hat{p} | \psi \rang = \lang \phi | p \rang \lang p | \hat p | p \rang \lang p | \psi \rang = \int \phi^*(p) p \psi(p) dp$

which would be a much easier calculations in some situtations. — Laura Scudder ☎ 18:20, 30 April 2006 (UTC)

Laura's explanation is awesome! It would do everybody a great favor is someone included it on the page's text. I read the page like 3 times over a 6 months period and didn't understand anything; with this explantion I got it (well, kinda, at least I got what the ket thing is about) :) --70.137.159.152 02:32, 25 July 2006 (UTC)

[edit] reason for deleting section

section on comparison with "long-S" notatation has been deleted. first there is no such thing as a "long-S notation". a ket is an element of a Hilbert space, whether the Hilbert space $\mathbb{C}^n$ or $L 2$ . the inner product is just the inner product on those spaces. to say it is so by convention, as claimed in the section is awkward and unnecessary. it's not the notation that changes, but the underlying Hilbert space. while awkwardness is not sufficient for deletion, incorrectness is. Mct mht 21:25, 29 April 2006 (UTC)

More precisely, since the state space is ~~almost~~ always

L 2 (μ)

for some measure

μ

. it is the measure that changes, not the notation. Mct mht 21:29, 29 April 2006 (UTC)

Well firstly, there is such a thing as "long-S" notation, it was invented by Leibnitz (see Integral). Secondly, as it stands this article is opaque to anybody who doesn't already have an understanding of the notation, which defeats the point. However if my contribution offends your pedantry so be it - I really can't be bothered. Trewornan

i have no objection for further elucidation in article, as long as what is said is correct. the very claim that the notation actually changes shows conceptual deficiency. citing irrelevant material from high school calculus texts doesn't really do it. Mct mht 17:09, 30 April 2006 (UTC)

I agree with Mct mht, this article serves 100% *no* purpose for me, or any users I can imagine would want to look at the page. It *needs* explanation as to what it *means*. Laura Scudder gave some examples aboves, but I still find it impossible to generalize that into how to actually interpret bra-ket notation. Fresheneesz 07:32, 6 May 2006 (UTC)

It's actually not that complicated - if you've got a function like f(x)=g(x)+ih(x) and its conjugate f*(x)=g(x)-ih(x) in bra-ket notation you write f(x) as |f> and f*(x) as <f| the trick bit is that when you write both together like this <f|f> you mean the integral $\int f \cdot f^* dx$ in general (particularly when dealing with quantum mechanics) that's all there is to it. The problem certain individuals have is that the same notation is used for vector spaces as for function spaces (actually that's "conceptually deficient") so if you were talking about vectors like the x and y axes rather than functions the product <x|y> isn't an integral like above just a standard vector product. It's the easy way to explain but it's not absolutely correct in some ways.Trewornan

[edit] H*

In the last section, H* is defined to be the space of linear operators on H. Was "linear functionals" intended? The $φ g$ defined in the article seems to map h into the field, not back into H.

[edit] Notation used by mathematicians

Whoever wrote this -- THANK YOU. I think it would be a great idea to have a math translation of all the physics talk found across many of the quantum mechanics articles. I have not seen quantum physics before and I spent an hour pondering about what exactly is happening (and why certain notations are chosen) as I slowed chewed through the early sections. However, since I do have some math background, the last section resolved the entire article in a breeze -- and I wondered about whether I was abusing substances in the previous hour ... --yiliu60 01:40, 6 October 2006 (UTC)

actually that stuff is not quite right, e.g. 1. as pointed out above by unsigned, the dual space H* is not the same as the bounded linear operators on H, 2. H is actually conjugate isomorphic to H*, etc. Mct mht 09:40, 17 October 2006 (UTC)

That's ironic, as I'm an undergrad physics student and I don't have the faintest idea what this article is trying to tell me. Nobody who teaches our course uses much in the way of this notation until later on, so I don't have a clue what it's for, even though i'm pretty good with QM. 86.135.99.158 12:26, 10 December 2006 (UTC)

[edit] Small "errors" / omissions

In the Bras and kets section the bras and kets are written in (a1, a2, ...) respresentation. This is correct for finite and countable-dimensional vectors, but seems a little awkward for functions in a Hilbert space. Perhaps a remark should be added that this notation is an example.

Also, I'm missing the unit operator |ei><ei| (in the basis {|ei>}) (which is important probably, though perhaps not necessarily a property of the bra-ket notation but more an example of bra-ket notation applied to a mathematical identity (namely, that the sum over e_i* e_i is unity) and the equivalence between for example <v|w>, <v|e><e|w>, <v|e><e|e><e|e><e|w> and <v||||w> (which may not be important at all but still, it's a nice notational trick). 213.84.168.13 20:54, 6 November 2006 (UTC)

That's a pretty neat trick, and (yet) another example of how e turns up everywhere. Please be bold and edit the page to include this - you'll do a better job than me :-D --h2g2bob 21:23, 6 November 2006 (UTC)

I've been thinking about this, and I think it would work for any base, ie:

| a i b > < a i b | = 1

For any real a and b, as you end up with

(a i b)(a i b) * = (a i b)(a - i b) = a i b / a i b = 1

I suppose it's no surprise that the answer is real when squaring complex variables, but the result is very nice. I think this (with exponentials) is probably a quite useful trick. --h2g2bob 13:19, 7 November 2006 (UTC)

I heavily edited the first section for the new information to fit in, removing a (in my opinion useless and bad-quality) reference. Regarding the unit operator: I wanted to fit it in the Properties section at first, but it became quite a lot of text to fit in the enumeration so I split it into a separate section. Please review. CompuChip 14:08, 11 November 2006 (UTC) (formerly 213.84.168.13)

[edit] Rewrite?

There are some basic errors and yet there is some really good stuff here. Some of the errors are so nearly correct that fixing them would be difficult without just deleting them. I think it would be great if someone started over, and avoided reading too much into the bra-ket notation. Also, it's difficult to address the bra-ket notation without considering inner product spaces and bilinear forms. The article should be more explicit about the relation to those things.Mathchem271828

Retrieved from "http://en.wikipedia.org../../../b/r/a/Talk%7EBra-ket_notation_a298.html"

Category: B-Class physics articles