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.

  • $\begingroup$ I can't argue with your last line, but I wonder would this be considered a mathSE post, rather than physics. $\endgroup$ – user108787 Aug 7 '16 at 19:43
  • $\begingroup$ Related: physics.stackexchange.com/q/43069/2451 , physics.stackexchange.com/q/45227/2451 , physics.stackexchange.com/q/216846/2451 and links therein. $\endgroup$ – Qmechanic Aug 7 '16 at 19:49
  • $\begingroup$ What structure does the inner product of a Hilbert space used in QM have? Proceed from there ... $\endgroup$ – Sanya Aug 7 '16 at 21:07
  • $\begingroup$ $<a,b>=<b,a>^{*}$ $\endgroup$ – 7919 Aug 7 '16 at 22:28
  • 1
    $\begingroup$ You just answered your question yourself ;) The inner product gives you the antilinearity ... $\endgroup$ – Sanya Aug 8 '16 at 8:06

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 $\lambda^*\left\langle a\right |$.


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