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Oliver321
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- Why is it that |-x> ≠ -|x>, even though kets are vectors and thus obey linearity?
I am a bit confused about how kets in dirac notation are working.
I read on wikipedia, that kets are linear, so |a*Φ>=a*|Φ>.
Also I read (https://ocw.mit.edu/courses/physics...all-2013/lecture-notes/MIT8_05F13_Chap_04.pdf) that this is not true for the position state ket ( or non denumerable bases in general): |-x> ≠ -|x>
My question is now: Why is this the case? If I am understanding it right, the position state ket is representing the position of a particle in that way, that a particle with ket |2> is at the point 2 on the x axis. Or expressed otherwise: |2> =(0,0,...,1,0,0,...), where the second representation is a vector in an infinite space where every component (or base vector) is representing a real number (in this example the number 2). So it would be valid to say
|-2> =(0,0,...,-1,0,0,...) = -(0,0,...,1,0,0,...)= -|2>
Probabliy I have a wrong understanding of what kets really are. I would appreciate every help!
I read on wikipedia, that kets are linear, so |a*Φ>=a*|Φ>.
Also I read (https://ocw.mit.edu/courses/physics...all-2013/lecture-notes/MIT8_05F13_Chap_04.pdf) that this is not true for the position state ket ( or non denumerable bases in general): |-x> ≠ -|x>
My question is now: Why is this the case? If I am understanding it right, the position state ket is representing the position of a particle in that way, that a particle with ket |2> is at the point 2 on the x axis. Or expressed otherwise: |2> =(0,0,...,1,0,0,...), where the second representation is a vector in an infinite space where every component (or base vector) is representing a real number (in this example the number 2). So it would be valid to say
|-2> =(0,0,...,-1,0,0,...) = -(0,0,...,1,0,0,...)= -|2>
Probabliy I have a wrong understanding of what kets really are. I would appreciate every help!