Talk: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? -- 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 (http://www.mozilla.org/newlayout/testcases/layout/entities.html) 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 (http://www.alanwood.net/unicode/explorer_older.html)). 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 Missing image
Del.gif
Image:Del.gif

([[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: ⟨<<math>\langle<math>. 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 <math>\langle<math> and <math><\;<math>. -- 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 <math>\langle<math>. - 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,

<math>(c_1 \langle\phi_1| + c_2 \langle\phi_2|)|\psi\rangle = c_1 \langle\phi_1|\psi\rangle + c_2\langle\phi_2|\psi\rangle. <math>

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 (http://en.wikipedia.org/wiki/Inner_product#Definitions),

<math> \lang D | B \rang = { \lang B | D \rang }^* <math>

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:

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

you immidietly see that

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

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:

<math>

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

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. -- 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). –Matt 23:35, 12 Jun 2004 (UTC)

Contents

1 braket-notation + cross-product 2 differential operators 2.1 Answer: 3 Plane Waves 4 Linear Operators 5 symbol changed midway through

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 Reference desk? –Matt 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. 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 <math> A = p = \frac{\hbar}{i} \frac{d}{dx} <math> (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 <math> \lang x |x \rang = c e^{ i p x } <math> , ( c is normalization factor and his exact value is not realy important right now), we know that:

<math> \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 <math>

and hence we indeed reaffirmed (found) A's differential form.

For a general function, recall that <math> \lang x | \psi \rang = \psi (x) <math> and since <math> \lang x | A | x \rang = \frac{1}{i} \frac{d}{dx} c e^{ i p x } <math> the problem in the x basis (position representation) is done by something similiar to basis transformation matrix , <math> \int{| x \rang \lang x | \ dx} = Id <math> (identity operator, and hence don't change the equation),

<math> A | \psi \rang = \int { A | x \rang \lang x | \psi \rang \ dx } <math> or <math> \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 } <math>

and with integration by parts we shall yield

<math> \int{ \left( c \frac{1}{i} \frac{d}{dx} e^{ i p x} \right) ^{*} \psi(x) \ dx } = \int{ c e^{ - i p x} \cdot \frac{1}{i} \frac{d}{dx} \psi(x) \ dx }<math>

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

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

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: <math> \lang x | A | \psi \rang = \frac{d}{dx} \lang x | \psi \rang = \frac{d}{dx} \psi (x) <math> . 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. -- 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)

Linear Operators

I think the defenition

Operators can also act on bras. Applying the operator A to the bra <math>\langle\phi|<math> results in the bra (<math>\langle\phi|<math>A), defined as a linear functional on H by the rule

<math>(\langle\phi|A) \; |\psi\rangle = \langle\phi| \; (A|\psi\rangle)<math>.

This expression is commonly written as

<math>\langle\phi|A|\psi\rangle.<math>

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] (http://en.wikipedia.org/w/wiki.phtml?title=Bra-ket_notation&oldid=5003286#Linear_operators) 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

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 of state |ψ> onto state |φ>, which can be interpreted as the probability for |ψ> to change into |φ>. --Laura Scudder | Talk 21:26, 16 May 2005 (UTC)