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I have a notational problem, I know when you define bra and ket you are defining an inner product, but you can see it as an linear operation where the linear operators (bras) act on vectors (kets), but of the same form you can think of the kets as operator over bras but in this case this operators are antilinear, I would think this operator would have to be linear as well.

Therefore I have a doubt there, why do i need to use in this case an antilinear operator? Probably the answer is the inner product is always positive or equal zero given the object (probability) is always positive or equal zero and I use the inner product to calculate this magnitude, which it's really the only thing important here.

peterh
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7919
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  • I can't argue with your last line, but I wonder would this be considered a mathSE post, rather than physics. –  Aug 07 '16 at 19:43
  • Related: http://physics.stackexchange.com/q/43069/2451 , http://physics.stackexchange.com/q/45227/2451 , http://physics.stackexchange.com/q/216846/2451 and links therein. – Qmechanic Aug 07 '16 at 19:49
  • What structure does the inner product of a Hilbert space used in QM have? Proceed from there ... – Sanya Aug 07 '16 at 21:07
  • $<a,b>=<b,a>^{*}$ – 7919 Aug 07 '16 at 22:28
  • Well, $<a,\alpha b>=\alpha <a,b>$, then bras act as linear operator. – 7919 Aug 07 '16 at 22:30
  • Whereas $<\alpha a,b>=\alpha^{*}<a,b>$, then kets act as antilinear operators – 7919 Aug 07 '16 at 22:31
  • $<\alpha a,\alpha b>=norm(\alpha)^{2}<a,b>$ if $b=a$ then $<a,a>$ is greater than or equal to 0. – 7919 Aug 07 '16 at 22:38
  • My doubt exactly is if i can interpret the kets as linear operators over bras, i ask it because my first option would be to interpret them so. But the inner product imposes some rule over the operation in order to get something. – 7919 Aug 07 '16 at 22:50
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    You just answered your question yourself ;) The inner product gives you the antilinearity ... – Sanya Aug 08 '16 at 08:06
  • @7919 - You can interpret kets as linear operators over bras. Linear operators on a vector space $V$ are themselves a vector space known as the dual vector space $V^$. The dual of a dual vector space $V^{}$ is then the space of linear operators on $V^$. An elementary lemma of linear algebra states that $V^{*} \cong V$. Thus, the original vectors (kets) of $V$ can themselves be interpreted as linear operators acting on $V^$ (bras). See http://math.stackexchange.com/questions/170481/motivating-to-understand-double-dual-space – Prahar Aug 08 '16 at 19:14
  • Exact, I would want somebody says that. My doubt it's exactly there, I mean I can't interpret kets as linear operator because the inner product gives me the antilinearity and If I suppose kets are elements of $V^{**}$ then kets are linear operators and therefore I reach a contradiction, I mean by a side I claim kets are antilinear and by the other side i say they are linears , Is that possible or am I mixing concepts? Thanks a lot!! – 7919 Aug 08 '16 at 21:31

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Bras are linear operators from kets to scalars, since the bra $\left\langle a\right |$ sends the linear combination $\lambda\left|b\right\rangle+\mu\left|c\right\rangle$ to $\lambda\left\langle a,b\right\rangle+\mu\left\langle a,c\right\rangle$.

Kets also act linearly on bras: the ket $\left| a\right\rangle$ sends the linear combination $\lambda\left\langle b\right|+\mu\left\langle c\right|$ to $\lambda\left\langle b,a\right\rangle+\mu\left\langle c,a\right\rangle$.

But the transformation that turns a ket into its corresponding bra ($\left| a\right\rangle\mapsto\left\langle a\right|$) is antilinear: it sends $\lambda\left| a\right\rangle$ to $\lambda^*\left\langle a\right |$.

Oscar Cunningham
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