Proof in heisenbergs uncertainty relation involving bra-ket

  • Thread starter lavster
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  • #1
lavster
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hey, can someone show me the step between these two lines of equations please:

[tex](\Delta A)^2=<\psi|A^2|\psi>-<\psi|A|\psi>^2[/tex]
[tex]=<\psi|(A-<A>)^2|\psi>[/tex]

where A is an operator and [tex]\psi[/tex] is the wavefunction and [tex]<A>[/tex] is the expectation value of A
 

Answers and Replies

  • #2
Doc Al
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[tex]=<\psi|(A-<A>)^2|\psi>[/tex]
Just expand this expression, realizing that <A> is just a number. ([tex]<A> = <\psi|A|\psi>[/tex])
 
Last edited:
  • #3
lavster
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thanks for quick reply! however im still not getting it. could you write it out expliitly please?
 
  • #4
Doc Al
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thanks for quick reply! however im still not getting it. could you write it out expliitly please?
[tex]<\psi|(A-<A>)^2|\psi> = <\psi|A^2 -2A<A> + <A>^2 |\psi>[/tex]

I'll let you do the rest.
 
  • #5
Fredrik
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Just a little LaTeX tip: Use \langle and \rangle instead of < and >. (Doc AI's answer is good, so I have nothing to add, except the complete solution, but you should try it yourself first. Note: "just a number" really means "just a number times the identity operator". OK, I guess I did have something to add :smile:).
 
  • #6
Doc Al
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I thought those bras and kets looked a bit off. :uhh: (Thanks, Fredrik!)
 
  • #7
Frame Dragger
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I thought those bras and kets looked a bit off. :uhh: (Thanks, Fredrik!)

Must... not... say... I... prefer... bras... off... must... not say... to staff...ARGGGH.. Too late. :smile: Be gentle!
 

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