- #1
mkarydas
- 8
- 0
I have the following question for anyone who can help:
Suppose in Dirak notation i have the following state:
/ψ> = 2 /u1> + /u2> where /u1>,/u2> are the two first eigen kets of energy of the infinite square well.
That means <x/u1>= Asin(π x /L) and <x/u1>= Asin(2 π x /L) which gives :
<x/ψ>= 2Asin(π x /L) + Asin(2 π x /L)
Everything ok so far. But what would happen if i chose <x/u1>= - Asin(π x /L) instead of
<x/u1>= Asin(π x /L) which is perfectly legitimate. Then <x/ψ> would become:
<x/ψ>= -2Asin(π x /L) + Asin(2 π x /L) which is not the same as
<x/ψ>= 2Asin(π x /L) + Asin(2 π x /L) ..( you can not get to one another by multiplying a constant phase so the are different)
So the question is does /ψ> = 2 /u1> + /u2> determine the state of a system or not because it seems that it depends on my free choice of base kets /u1>, /u2> ?
Suppose in Dirak notation i have the following state:
/ψ> = 2 /u1> + /u2> where /u1>,/u2> are the two first eigen kets of energy of the infinite square well.
That means <x/u1>= Asin(π x /L) and <x/u1>= Asin(2 π x /L) which gives :
<x/ψ>= 2Asin(π x /L) + Asin(2 π x /L)
Everything ok so far. But what would happen if i chose <x/u1>= - Asin(π x /L) instead of
<x/u1>= Asin(π x /L) which is perfectly legitimate. Then <x/ψ> would become:
<x/ψ>= -2Asin(π x /L) + Asin(2 π x /L) which is not the same as
<x/ψ>= 2Asin(π x /L) + Asin(2 π x /L) ..( you can not get to one another by multiplying a constant phase so the are different)
So the question is does /ψ> = 2 /u1> + /u2> determine the state of a system or not because it seems that it depends on my free choice of base kets /u1>, /u2> ?