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I've gotten stuck with a bit of dirac notation calculation?

  • Thread starter jeebs
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  • #1
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sorry if this looks ugly but I couldn't find out how to write out bras and kets on the Latex thing.

I have these inner products
<f|g> = i<x|(AB - A<B> - <A>B + <A><B>)|x>
and
<g|f> = -i<x|(BA - B<A> - <B>A + <A><B>)|x>
where |x> is some arbitrary ket and A and B do not commute.

I'm trying to follow a derivation of the uncertainty principle that has a step which involves getting
<f|g> + <g|f> = i<x|[A,B]|x>,
but I'm not sure how to add the two things together.

I tried to see if it was allowed that, say, <x|M|x> + <x|N|x> = <x|(M+N)|x>, but that didn't work, and I tested this with a real numerical example so that definitely doesn't work. I need to somehow end up with
<f|g> + <g|f> = i<x|(AB - BA)|x>
but I haven't seen how objects like these add together before.
How is this done?
 
Last edited:

Answers and Replies

  • #2
CompuChip
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Err, aren't bras and kets linear?
I.e. (A + B)|x> = A|x> + B|x> and <x|(m |y> + n |z>) = m <x|y> + n <x|z>?

So then what you claim not to be true (<x|M + N|y> = <x|M|y> + <x|N|y>) is actually true, and the proof consists of adding up the first two expressions in your post.
 
  • #3
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Err, aren't bras and kets linear?
I.e. (A + B)|x> = A|x> + B|x> and <x|(m |y> + n |z>) = m <x|y> + n <x|z>?

So then what you claim not to be true (<x|M + N|y> = <x|M|y> + <x|N|y>) is actually true, and the proof consists of adding up the first two expressions in your post.
ahh I think I know where I went wrong. I worked out M|x>, then <x|M|x>, did the same for N, then added M+N and then calculated (M+N)|x> and finally <x|(M+N)|x>. That definitely didn't give <x|(M+N)|x> = <x|M|x> + <x|N|x>.
 
  • #4
CompuChip
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I still think it shouldn't matter.
Basically, <x|M|x> means nothing other than M(x) in function notation, and the sum of two functions M and N is defined as (M + N)(x) = M(x) + N(x).
If you first add two linear operators, evaluate them on an element and choose a basis, or first choose a basis (the same for both, obviously), evaluate them on an element and add the results, should not matter.

Or, if you prefer to look at it as an (uncountably-dimensionally) matrix operation, <x|M|y> is just the element in position (x, y) of the matrix M, and you know how adding matrices works.
 
  • #5
jambaugh
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...That definitely didn't give <x|(M+N)|x> = <x|M|x> + <x|N|x>.
It should have. How are you parsing this?

In general:
i.) [tex]\langle x | (|y\rangle + |z\rangle) = \langle x|y\rangle +\langle x |z\rangle[/tex]
ii.) [tex](\langle x| + \langle y|)|z\rangle = \langle x|z\rangle +\langle y |z\rangle[/tex]
iii.) [tex]\langle x | (M + N) = \langle x|M +\langle x |N[/tex]
iv.) [tex](M+N)|x\rangle = \langle M|x\rangle +\langle N |x\rangle[/tex]
By what you said you tried, combining the i. and iii. or combining ii. and iv. should give what you didn't.

BTW, typesetting math Example:
{tex} \langle x | (M+N) | x\rangle = \langle x | M | x\rangle + \langle x | N | x \rangle {/ tex}
with {}--> [] gives
gives
[tex] \langle x | (M+N) | x\rangle = \langle x | M | x\rangle + \langle x | N | x \rangle [/tex]
 
  • #6
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[tex] | x\rangle = \left(\begin{array}{c}1&2\end{array}\right)[/tex], [tex] M = \left(\begin{array}{cc}1&2\\2&1\end{array}\right)[/tex], [tex] N = \left(\begin{array}{cc}1&0\\1&0\end{array}\right)[/tex]

[tex] M| x\rangle = \left(\begin{array}{c}5\\4\end{array}\right)[/tex]
[tex] N| x\rangle = \left(\begin{array}{c}1\\1\end{array}\right)[/tex]

[tex] \langle x |M| x\rangle =13[/tex],
[tex] \langle x |N| x\rangle =3[/tex],
[tex] \langle x |N| x\rangle + \langle x |M| x\rangle = 16[/tex]

[tex] M+N = \left(\begin{array}{cc}2&2\\3&1\end{array}\right)[/tex]
[tex] (M+N)| x\rangle = \left(\begin{array}{c}6\\5\end{array}\right)[/tex]
[tex] \langle x |(M+N)| x\rangle =16[/tex]

heh, looks like it does work after all... what the hell is wrong with me...
 
Last edited:
  • #7
CompuChip
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Phew, what a relief... linear algebra still works! :D
 

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