![]() | This is an archive of past discussions. Do not edit the contents of this page. If you wish to start a new discussion or revive an old one, please do so on the current talk page. |
Archive 1 | Archive 2 |
In the section "The unit operator" shouldn't the formula be
rather than
? (v and w exchanged in the last part)
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?
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
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
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)
Alternatively, we can use TeX all the time for these brackets, they work fine also. - Patrick 22:06 Mar 5, 2003 (UTC)
Regarding
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
but from linear property
hence
Q.E.D. MathKnight 21:10, 28 Mar 2004 (UTC)
Regarding
This should instead be
Your proof is therefore erroneous. -- CYD
The passage
is correct, since we mark D = cA and then from Conjugate Symmetry,
and subsitute again we get
Notice that c (a scalar) multiplies A.
I can also show it directly, define bra-ket inner product as:
you immidietly see that
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:
The concept of "multiplying inside the ket" is redundant in b reprenstation of the Delta function around x we get
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: .
MathKnight 20:10, 4 Aug 2004 (UTC)
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.
I think the defenition
Operators can also act on bras. Applying the operator A to the bra results in the bra (A), defined as a linear functional on H by the rule
- .
This expression is commonly written as
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)
" \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?
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)
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)
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)
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)
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
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)
or is it more general like this?:
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 or . 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)
In the last section, H* is defined to be the space of linear operators on H. Was "linear functionals" intended? The defined in the article seems to map h into the field, not back into H.
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)
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)
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, 19:30, 12 December 2006
In section 'Linear operators':
Can someone give an example about the outer product? I mean if we let
What will it be for the outer product?
Thanks! Justin545 01:56, 14 October 2007 (UTC)
Is it correct to say that (if we are in a Hilbert space) the bra is the image of the ket under the Riesz isomorphism? I.e. ? Eriatarka ( talk) 15:00, 14 June 2008 (UTC)
What is a "unit ray"? That sounds like a contradiction since a ray, as far as I know, has no size. Xezlec ( talk) 05:02, 17 June 2008 (UTC)
In the following equation:
In the bit which states: Should it not be: ? —Preceding unsigned comment added by Etanol ( talk • contribs) 16:38, 17 March 2009 (UTC)
"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 kets and the continuous linear functionals by bras. Over any vector space without topology, we may also notate the vectors by kets and the linear functionals by bras. In these more general contexts, the bracket does not have the meaning of an inner product, because the Riesz representation theorem does not apply." relevance? non-existent. trivial. i can denote any vector by a ket. i can denote any OBJECT by a ket. —Preceding unsigned comment added by 91.15.144.196 ( talk) 21:18, 2 April 2009 (UTC)
This article, as written, is math rather than physics. The Dirac notation has some nice practical applications as described by Feynman and other authors. However, with comments like
"(cur) (prev) 19:22, 17 October 2009 Sbyrnes321 (talk | contribs) m (20,364 bytes) (Undid revision 320439307 by 128.151.39.155 (talk) silly to describe two examples of a totally ubiquitous thing) (undo)"
this article will most likely remain as a math piece. Sbyrness321 deleted the short applications paragraph because it gave a only few practical examples (not just two) when the article right now has NONE. Very interesting. —Preceding unsigned comment added by 128.151.39.161 ( talk) 20:22, 18 October 2009 (UTC)
3 months ago, DMacks ( talk · contribs) changed the rendering in the lead from the standard LaTeX usage, , to a nowiki style, <Φ|Ψ>, to avoid potential confusion between ( and < for some resolutions. While I agree that can appear similar to at first glance, it is the standard used throughout Wikipedia and most other sources. I don't see anything unusual about the MediaWiki rendering compared to other sources using LaTeX. I think we should use the same style for this lead, especially considering the large number of incoming links from articles that use the LaTeX style. I experimented with another style that renders phi and psi better, , but I ultimately decided to go back to the standard used by the scientific community. — UncleDouggie ( talk) 13:58, 24 January 2011 (UTC)
Why is the differential operator applied to a ket considered to be an abuse of notation? It seems to me like the derivative of a vector is well defined, even in the absence of a chosen basis. Hiiiiiiiiiiiiiiiiiiiii ( talk) 00:57, 30 March 2011 (UTC)
I chased wikipedia pages all over the place to try to understand this stuff. My complaint is rather like one of the othe commenters here - I could't see any connection between hamiltonian spaces and lasers or electrons. I finally had to view the Oxford University lecture videos to "get" something which is probably obvious to everyone else. Here's kinda/sorta what I wish someone had written here on wikipedia for interested amateurs such as myself. If someone could express the following more clearly and correctly, that would be great.
The basis vectors of a ket vector correspond to "classical"states. For instance, in QM the energy of a particle may be one of an infinite series of discrete values - E0, E1, E2 and so on. Each position in the ket corresponds (loosely speaking) to its corresponding energy level. The value psi_1 is the probability amplitude for the system having energy level e1, and so on. Each value is a complex number, and the sum of the moduli of the squares of those numbers ( ie: sum of (psi_i . conj psi_i) ) must be 1.
The key thing, the really important thing, is that the system is not "in" state E0, e1, e2 and so on - its the whole of that ket vector that is its physical state, and its the whole of the ket vector that changes as the system propagates forward in time, or however. For this reason, it often makes sense to express the ket as a function rather than as a list of numbers.
For any physical system, there will be several ways that you can describe it completely. For instance, it's often the case that describing the *position* of a particle comlpetely also describes the *moment* of the particle completely. This means that the position ket can be transformed into a momentum ket by choosing different basis vectors and doing some sort of transform. A takeaway here is that sometimes you want to think of the ket as being "the state of the system" without being specific about how that state is expressed. It's still a meaningful thing to do.
A bra can be thought of as an observation that you would like to make on a ket. < bra | ket > - the projection of the ket on the bra - gives us the physical probability of making that observation, given that physical state. But "observation" is a term-of-art. < bra | ket > sort of corresponds to when you do something to a quantum system. It's an action that you can perform. It has *physical* meaning.
For any ket, we can construct a corresponding bra. This doing < foo | bar >, where we understand foo to be a bra-ified ket, gives us a probability - the probability that if the system is in the state bar, that it will look like it is in state foo should we observe it.
Note that < foo | foo > always sums to 1. The physical meaning of this is that if you know that a system is in some state, then the probability of observing it in that state is 1. Duh. This feature is mathematically equivalent to the rule "modulus squared of all the values in any ket must sum to one".
Wwe can construct a set of basis ket vectors
| 1, 0, 0, 0 ... > | 0, 1, 0, 0 ... > | 0, 0, 1, 0 ... >
The process of converting these into bras and applying them to some ket does the job of inquiring about the value for each position in the ket. It's equivalent to simply reading the square of the modulus of the ket at a position.
A bra or ket does not need to have an infinite number of values. For instance, in a two slit experiment there are two states "photonn goes through the left slit" and "photon goes through the right slit". Another on is spin. Rather than e0, e1, e2, e3 and so on, if you test the spin of a particle you will get one of two possible answers: up or down. It can be represented by a ket with two values | spin up , spin down > . When we take a simple "spin up or down" observation on the system - ie, when we do a < 1, 0 | or <0, 1| on it, then we get a simple probability back. But the system actually has three degrees of freedom. The question "spin left or right?" is effectively a matter of choosing a different set of basis vectors and performing a coordinate transform on the ket to express its spin in terms of those new vectors.
The other thing I would like to know is - what's the story with reducing the size of a ket? When we perform a simple "yes or no" question of a complex system, obviously we are somehow mapping a system with many dimensions ont a systems with fewer. For instance, we could map | e0, e1, e2, e3 ...> onto a two-state | E lt e3 , e ge E3 >. We can't do that with a simple dot-product - we have to use a matrix or something. What do you call that, and how does it fit into the notation?
Paul Murray ( talk) 05:39, 22 June 2011 (UTC)
I intend to re-write, not delete or remove content, but add and adjust content in the beginning, to make this far more approachable. It charges into so much mathematical terminology and notation which simply couldn't be understood by any layperson. This topic can be introduced in analogy with ordinary Euclidean vectors, bases, and the dot product, paralleling with the basis kets, dual vectors and the inner product - simultaneously explaining the generality of the Dirac formalism and its abstract scope beyond the physical, geometric Euclidean vectors from the elementary vector calculus lots of people can understand fairly well, and the reason pointing to the wavefunction article on its powerful application.-- F = q( E + v × B) 11:37, 22 January 2012 (UTC)
I have produced a number of diagrams to accompany the above quote: illustrate the parallels between "usual" vectors and bra-ket vectors, though more will be added soon to illistrate the abstractness - these are intended for the lead.
Opinions? -- Maschen ( talk) 17:51, 22 January 2012 (UTC)
I would naturally disagree, but thought as much someone would say this... user:F=q(E+v^B) you were 100% completley wrong, which disapoints me after all that excitement... =( The Cartesian_and_ket_vector_components_projection fig is to show that the shadow length is not equal to the inner product, just the projection of magnitudes. All else as suggested will be modified/annihilated now that its all wrong... yet another failure (suprise suprise). Actually Steve, in a sense out of us three you are "the boss", given that you know all this inside out, and we are still learning, which is why i'll do what you reccomend...-- Maschen ( talk) 20:13, 23 January 2012 (UTC)
All previous statements have been unified into here. The current diagrams in the article are no more (and never will).
-- Maschen ( talk) 21:17, 23 January 2012 (UTC)
Right now, the article is a slight mess so I’ll have a go at making it more coherent. I think the new images are fine Maschen - they do match Steve's recommendations above, so by all means add them when you see fit. -- F = q( E + v × B) 22:25, 25 January 2012 (UTC)
I don't understand the equation:
I can't get it to work. For example, let's say A is (1,1,1) is Cartesian coordinates, (sqrt(3),54°, 45°) in spherical coordinates. What direction does point? I would have guessed, parallel to A, so (1/sqrt(3), 1/sqrt(3), 1/sqrt(3)) in Cartesian coordinates. But no, that's not right, because then with no other terms. So what is ? What are the other e 's? -- Steve ( talk) 15:54, 27 January 2012 (UTC)
Yes, it is indeed incorrect, and I didn't even think to check that:
gives approx. 54.73561(...)° for x = y = z = 1... which is really bad... =(
I've seen that representation for spherical polars before, so decided to included it. (Most charming of you to say lazy - but then I am sometimes!...) -- F = q( E + v × B) 23:41, 29 January 2012 (UTC)
In section Notation used by mathematicians, it was quite amusing/odd to read:
I don't remember reading that from recently when the page was re-written, because that section was left alone and almost everything else it was re-written... Anyway is there any reason for embarrassment? F = q(E+v×B) ⇄ ∑ici 18:23, 6 June 2012 (UTC)
See {{ langle}} and {{ rangle}}. Hope people like the change. F = q(E+v×B) ⇄ ∑ici 16:44, 12 June 2012 (UTC)
How do you say <A|B> when speaking? Is it standardized, or are there both lengthy and brief ways to read it? This should be included in the lede, just like the pronunciation of bra and ket. 130.60.6.54 ( talk) 20:25, 10 June 2013 (UTC)
Another question is the pronunciation of ''. The article currently states it as /brɑː/ (brah), but is it more logical that it is /bræ/ (bra, short 'a'), as in the pronunciation of the first syllable of bracket? I've never heard it spoken, so I don't know the answer. — Loadmaster ( talk) 17:34, 18 December 2015 (UTC)
In the section introducing vector spaces over ℂ I amended the phrase 'though the coordinates and vectors are now all complex-valued' to read instead simply 'though the coordinates are now all complex-valued'. Although I see what the original author meant, it's not necessary for the vectors in a vector space over ℂ to in any way 'contain' complex numbers - it's simply necessary that the vectors (whatever they are) to be able to be scaled by elements in ℂ. For example, you can give ℝ2 the structure of a vector space over ℂ [when you scale (x,y) by a+ib you end up with (xa-yb,xb+ya)] but in this case the vectors (elements of ℝ2) don't 'contain' complex numbers. Felix116 ( talk) 21:31, 14 October 2013 (UTC)
I reluctantly reverted the edit of User:131.231.242.221 today, who complained: "The characters "〈" and "〉" appear on my computer as rectangles with little numbers in. Is there an alternative?" Indeed, older systems lack one or both pairs of html characters. Every few months, there a complaint of this sort somewhere or elsewhere in WP. There is also the alternate set of ⟨ψ|z⟩, I strongly suspect equally invisible/missing-character'd to him/her. Supplanting with <.> (lessthan-greater than) as a workaround gives the text an amateur look, and TeXing in text, gives that distinctly ransom note look in those very systems. I have no good ideas, not even on how to advise this user to reset his/her preferrences to render formulas better. Cuzkatzimhut ( talk) 11:17, 20 October 2013 (UTC)
If it helps I saw bra-ket symbol issue viewing the page with google chrome browser, but not with firefox or mirosoft internet explorer. — Preceding unsigned comment added by 67.249.102.221 ( talk) 21:30, 16 November 2013 (UTC)
This section has a number of errors (or simplifications to the point of excessive inaccuracy). A R3 point is NOT a vector. You do not add points. You do not multiply points. You don't take their dot (inner) products. A vector might be a entity with only magnitude and direction, but it is common in physics to also locate the vector in 2 or 3-d space. (ie. there are two distinct technical meanings for the term, (I'm not sure if vector fields can be used as equivalent to a located vector?)). Only vectors originating from the origin can be described with three scalars in 3-d space (n scalars in n-space). That is only vectors which only have magnitude and direction (not position) are the subject of this section.
ALSO: The coordinates are NOT "the number of basis vectors in each direction". Who wrote this??!! 216.96.79.61 ( talk) 22:19, 21 January 2014 (UTC)
The initial paragraphs of this section need a rewrite.
kets and the vectors that represent kets are not equal! A ket is related to a vector that represents the ket, like a poster of a star is related to the star. A "coordinate vector" needs the reference to a basis, whereas the ket does not. See Modern Quantum Mechanics Revised Revision, Sakurai, p. 20. Therefore the equal sing between a ket and a vector needs to be changed to another symbol since the mathematical objects on both sides are not equal. For example instead of something like should be used (with like Sakurai for "is represented by" or like Shankar) -- Biggerj1 ( talk) 12:27, 2 February 2014 (UTC)
Biggerj1 (
talk)
08:07, 11 May 2014 (UTC)
Any idea why the kets are showing up as square boxes? Having the issue on multiple computers. Sam Walton ( talk) 14:38, 2 June 2014 (UTC)
In the article it was mentioned:
Shouldn't it have been
![]() | This page is an archive of past discussions. Do not edit the contents of this page. If you wish to start a new discussion or revive an old one, please do so on the current talk page. |
![]() | This is an archive of past discussions. Do not edit the contents of this page. If you wish to start a new discussion or revive an old one, please do so on the current talk page. |
Archive 1 | Archive 2 |
In the section "The unit operator" shouldn't the formula be
rather than
? (v and w exchanged in the last part)
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?
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
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
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)
Alternatively, we can use TeX all the time for these brackets, they work fine also. - Patrick 22:06 Mar 5, 2003 (UTC)
Regarding
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
but from linear property
hence
Q.E.D. MathKnight 21:10, 28 Mar 2004 (UTC)
Regarding
This should instead be
Your proof is therefore erroneous. -- CYD
The passage
is correct, since we mark D = cA and then from Conjugate Symmetry,
and subsitute again we get
Notice that c (a scalar) multiplies A.
I can also show it directly, define bra-ket inner product as:
you immidietly see that
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:
The concept of "multiplying inside the ket" is redundant in b reprenstation of the Delta function around x we get
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: .
MathKnight 20:10, 4 Aug 2004 (UTC)
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.
I think the defenition
Operators can also act on bras. Applying the operator A to the bra results in the bra (A), defined as a linear functional on H by the rule
- .
This expression is commonly written as
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)
" \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?
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)
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)
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)
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)
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
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)
or is it more general like this?:
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 or . 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)
In the last section, H* is defined to be the space of linear operators on H. Was "linear functionals" intended? The defined in the article seems to map h into the field, not back into H.
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)
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)
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, 19:30, 12 December 2006
In section 'Linear operators':
Can someone give an example about the outer product? I mean if we let
What will it be for the outer product?
Thanks! Justin545 01:56, 14 October 2007 (UTC)
Is it correct to say that (if we are in a Hilbert space) the bra is the image of the ket under the Riesz isomorphism? I.e. ? Eriatarka ( talk) 15:00, 14 June 2008 (UTC)
What is a "unit ray"? That sounds like a contradiction since a ray, as far as I know, has no size. Xezlec ( talk) 05:02, 17 June 2008 (UTC)
In the following equation:
In the bit which states: Should it not be: ? —Preceding unsigned comment added by Etanol ( talk • contribs) 16:38, 17 March 2009 (UTC)
"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 kets and the continuous linear functionals by bras. Over any vector space without topology, we may also notate the vectors by kets and the linear functionals by bras. In these more general contexts, the bracket does not have the meaning of an inner product, because the Riesz representation theorem does not apply." relevance? non-existent. trivial. i can denote any vector by a ket. i can denote any OBJECT by a ket. —Preceding unsigned comment added by 91.15.144.196 ( talk) 21:18, 2 April 2009 (UTC)
This article, as written, is math rather than physics. The Dirac notation has some nice practical applications as described by Feynman and other authors. However, with comments like
"(cur) (prev) 19:22, 17 October 2009 Sbyrnes321 (talk | contribs) m (20,364 bytes) (Undid revision 320439307 by 128.151.39.155 (talk) silly to describe two examples of a totally ubiquitous thing) (undo)"
this article will most likely remain as a math piece. Sbyrness321 deleted the short applications paragraph because it gave a only few practical examples (not just two) when the article right now has NONE. Very interesting. —Preceding unsigned comment added by 128.151.39.161 ( talk) 20:22, 18 October 2009 (UTC)
3 months ago, DMacks ( talk · contribs) changed the rendering in the lead from the standard LaTeX usage, , to a nowiki style, <Φ|Ψ>, to avoid potential confusion between ( and < for some resolutions. While I agree that can appear similar to at first glance, it is the standard used throughout Wikipedia and most other sources. I don't see anything unusual about the MediaWiki rendering compared to other sources using LaTeX. I think we should use the same style for this lead, especially considering the large number of incoming links from articles that use the LaTeX style. I experimented with another style that renders phi and psi better, , but I ultimately decided to go back to the standard used by the scientific community. — UncleDouggie ( talk) 13:58, 24 January 2011 (UTC)
Why is the differential operator applied to a ket considered to be an abuse of notation? It seems to me like the derivative of a vector is well defined, even in the absence of a chosen basis. Hiiiiiiiiiiiiiiiiiiiii ( talk) 00:57, 30 March 2011 (UTC)
I chased wikipedia pages all over the place to try to understand this stuff. My complaint is rather like one of the othe commenters here - I could't see any connection between hamiltonian spaces and lasers or electrons. I finally had to view the Oxford University lecture videos to "get" something which is probably obvious to everyone else. Here's kinda/sorta what I wish someone had written here on wikipedia for interested amateurs such as myself. If someone could express the following more clearly and correctly, that would be great.
The basis vectors of a ket vector correspond to "classical"states. For instance, in QM the energy of a particle may be one of an infinite series of discrete values - E0, E1, E2 and so on. Each position in the ket corresponds (loosely speaking) to its corresponding energy level. The value psi_1 is the probability amplitude for the system having energy level e1, and so on. Each value is a complex number, and the sum of the moduli of the squares of those numbers ( ie: sum of (psi_i . conj psi_i) ) must be 1.
The key thing, the really important thing, is that the system is not "in" state E0, e1, e2 and so on - its the whole of that ket vector that is its physical state, and its the whole of the ket vector that changes as the system propagates forward in time, or however. For this reason, it often makes sense to express the ket as a function rather than as a list of numbers.
For any physical system, there will be several ways that you can describe it completely. For instance, it's often the case that describing the *position* of a particle comlpetely also describes the *moment* of the particle completely. This means that the position ket can be transformed into a momentum ket by choosing different basis vectors and doing some sort of transform. A takeaway here is that sometimes you want to think of the ket as being "the state of the system" without being specific about how that state is expressed. It's still a meaningful thing to do.
A bra can be thought of as an observation that you would like to make on a ket. < bra | ket > - the projection of the ket on the bra - gives us the physical probability of making that observation, given that physical state. But "observation" is a term-of-art. < bra | ket > sort of corresponds to when you do something to a quantum system. It's an action that you can perform. It has *physical* meaning.
For any ket, we can construct a corresponding bra. This doing < foo | bar >, where we understand foo to be a bra-ified ket, gives us a probability - the probability that if the system is in the state bar, that it will look like it is in state foo should we observe it.
Note that < foo | foo > always sums to 1. The physical meaning of this is that if you know that a system is in some state, then the probability of observing it in that state is 1. Duh. This feature is mathematically equivalent to the rule "modulus squared of all the values in any ket must sum to one".
Wwe can construct a set of basis ket vectors
| 1, 0, 0, 0 ... > | 0, 1, 0, 0 ... > | 0, 0, 1, 0 ... >
The process of converting these into bras and applying them to some ket does the job of inquiring about the value for each position in the ket. It's equivalent to simply reading the square of the modulus of the ket at a position.
A bra or ket does not need to have an infinite number of values. For instance, in a two slit experiment there are two states "photonn goes through the left slit" and "photon goes through the right slit". Another on is spin. Rather than e0, e1, e2, e3 and so on, if you test the spin of a particle you will get one of two possible answers: up or down. It can be represented by a ket with two values | spin up , spin down > . When we take a simple "spin up or down" observation on the system - ie, when we do a < 1, 0 | or <0, 1| on it, then we get a simple probability back. But the system actually has three degrees of freedom. The question "spin left or right?" is effectively a matter of choosing a different set of basis vectors and performing a coordinate transform on the ket to express its spin in terms of those new vectors.
The other thing I would like to know is - what's the story with reducing the size of a ket? When we perform a simple "yes or no" question of a complex system, obviously we are somehow mapping a system with many dimensions ont a systems with fewer. For instance, we could map | e0, e1, e2, e3 ...> onto a two-state | E lt e3 , e ge E3 >. We can't do that with a simple dot-product - we have to use a matrix or something. What do you call that, and how does it fit into the notation?
Paul Murray ( talk) 05:39, 22 June 2011 (UTC)
I intend to re-write, not delete or remove content, but add and adjust content in the beginning, to make this far more approachable. It charges into so much mathematical terminology and notation which simply couldn't be understood by any layperson. This topic can be introduced in analogy with ordinary Euclidean vectors, bases, and the dot product, paralleling with the basis kets, dual vectors and the inner product - simultaneously explaining the generality of the Dirac formalism and its abstract scope beyond the physical, geometric Euclidean vectors from the elementary vector calculus lots of people can understand fairly well, and the reason pointing to the wavefunction article on its powerful application.-- F = q( E + v × B) 11:37, 22 January 2012 (UTC)
I have produced a number of diagrams to accompany the above quote: illustrate the parallels between "usual" vectors and bra-ket vectors, though more will be added soon to illistrate the abstractness - these are intended for the lead.
Opinions? -- Maschen ( talk) 17:51, 22 January 2012 (UTC)
I would naturally disagree, but thought as much someone would say this... user:F=q(E+v^B) you were 100% completley wrong, which disapoints me after all that excitement... =( The Cartesian_and_ket_vector_components_projection fig is to show that the shadow length is not equal to the inner product, just the projection of magnitudes. All else as suggested will be modified/annihilated now that its all wrong... yet another failure (suprise suprise). Actually Steve, in a sense out of us three you are "the boss", given that you know all this inside out, and we are still learning, which is why i'll do what you reccomend...-- Maschen ( talk) 20:13, 23 January 2012 (UTC)
All previous statements have been unified into here. The current diagrams in the article are no more (and never will).
-- Maschen ( talk) 21:17, 23 January 2012 (UTC)
Right now, the article is a slight mess so I’ll have a go at making it more coherent. I think the new images are fine Maschen - they do match Steve's recommendations above, so by all means add them when you see fit. -- F = q( E + v × B) 22:25, 25 January 2012 (UTC)
I don't understand the equation:
I can't get it to work. For example, let's say A is (1,1,1) is Cartesian coordinates, (sqrt(3),54°, 45°) in spherical coordinates. What direction does point? I would have guessed, parallel to A, so (1/sqrt(3), 1/sqrt(3), 1/sqrt(3)) in Cartesian coordinates. But no, that's not right, because then with no other terms. So what is ? What are the other e 's? -- Steve ( talk) 15:54, 27 January 2012 (UTC)
Yes, it is indeed incorrect, and I didn't even think to check that:
gives approx. 54.73561(...)° for x = y = z = 1... which is really bad... =(
I've seen that representation for spherical polars before, so decided to included it. (Most charming of you to say lazy - but then I am sometimes!...) -- F = q( E + v × B) 23:41, 29 January 2012 (UTC)
In section Notation used by mathematicians, it was quite amusing/odd to read:
I don't remember reading that from recently when the page was re-written, because that section was left alone and almost everything else it was re-written... Anyway is there any reason for embarrassment? F = q(E+v×B) ⇄ ∑ici 18:23, 6 June 2012 (UTC)
See {{ langle}} and {{ rangle}}. Hope people like the change. F = q(E+v×B) ⇄ ∑ici 16:44, 12 June 2012 (UTC)
How do you say <A|B> when speaking? Is it standardized, or are there both lengthy and brief ways to read it? This should be included in the lede, just like the pronunciation of bra and ket. 130.60.6.54 ( talk) 20:25, 10 June 2013 (UTC)
Another question is the pronunciation of ''. The article currently states it as /brɑː/ (brah), but is it more logical that it is /bræ/ (bra, short 'a'), as in the pronunciation of the first syllable of bracket? I've never heard it spoken, so I don't know the answer. — Loadmaster ( talk) 17:34, 18 December 2015 (UTC)
In the section introducing vector spaces over ℂ I amended the phrase 'though the coordinates and vectors are now all complex-valued' to read instead simply 'though the coordinates are now all complex-valued'. Although I see what the original author meant, it's not necessary for the vectors in a vector space over ℂ to in any way 'contain' complex numbers - it's simply necessary that the vectors (whatever they are) to be able to be scaled by elements in ℂ. For example, you can give ℝ2 the structure of a vector space over ℂ [when you scale (x,y) by a+ib you end up with (xa-yb,xb+ya)] but in this case the vectors (elements of ℝ2) don't 'contain' complex numbers. Felix116 ( talk) 21:31, 14 October 2013 (UTC)
I reluctantly reverted the edit of User:131.231.242.221 today, who complained: "The characters "〈" and "〉" appear on my computer as rectangles with little numbers in. Is there an alternative?" Indeed, older systems lack one or both pairs of html characters. Every few months, there a complaint of this sort somewhere or elsewhere in WP. There is also the alternate set of ⟨ψ|z⟩, I strongly suspect equally invisible/missing-character'd to him/her. Supplanting with <.> (lessthan-greater than) as a workaround gives the text an amateur look, and TeXing in text, gives that distinctly ransom note look in those very systems. I have no good ideas, not even on how to advise this user to reset his/her preferrences to render formulas better. Cuzkatzimhut ( talk) 11:17, 20 October 2013 (UTC)
If it helps I saw bra-ket symbol issue viewing the page with google chrome browser, but not with firefox or mirosoft internet explorer. — Preceding unsigned comment added by 67.249.102.221 ( talk) 21:30, 16 November 2013 (UTC)
This section has a number of errors (or simplifications to the point of excessive inaccuracy). A R3 point is NOT a vector. You do not add points. You do not multiply points. You don't take their dot (inner) products. A vector might be a entity with only magnitude and direction, but it is common in physics to also locate the vector in 2 or 3-d space. (ie. there are two distinct technical meanings for the term, (I'm not sure if vector fields can be used as equivalent to a located vector?)). Only vectors originating from the origin can be described with three scalars in 3-d space (n scalars in n-space). That is only vectors which only have magnitude and direction (not position) are the subject of this section.
ALSO: The coordinates are NOT "the number of basis vectors in each direction". Who wrote this??!! 216.96.79.61 ( talk) 22:19, 21 January 2014 (UTC)
The initial paragraphs of this section need a rewrite.
kets and the vectors that represent kets are not equal! A ket is related to a vector that represents the ket, like a poster of a star is related to the star. A "coordinate vector" needs the reference to a basis, whereas the ket does not. See Modern Quantum Mechanics Revised Revision, Sakurai, p. 20. Therefore the equal sing between a ket and a vector needs to be changed to another symbol since the mathematical objects on both sides are not equal. For example instead of something like should be used (with like Sakurai for "is represented by" or like Shankar) -- Biggerj1 ( talk) 12:27, 2 February 2014 (UTC)
Biggerj1 (
talk)
08:07, 11 May 2014 (UTC)
Any idea why the kets are showing up as square boxes? Having the issue on multiple computers. Sam Walton ( talk) 14:38, 2 June 2014 (UTC)
In the article it was mentioned:
Shouldn't it have been
![]() | This page is an archive of past discussions. Do not edit the contents of this page. If you wish to start a new discussion or revive an old one, please do so on the current talk page. |