- #1
Ahmes
- 78
- 1
Hi,
I try to understand the proof for the uncertainty principle for two Hermitian operators A and B in a Hilbert space. My questions are rather general so you don't need to know the specific proof.
The first thing I couldn't get into my head was the definition of uncertainty
[tex](\Delta A)^2=\langle\psi|(A- \langle A \rangle)^2|\psi\rangle[/tex]
Using the properties of the inner product I can be convinced this is equivalent to the ordinary definition of [tex](\Delta A)^2=\langle A^2 \rangle-{\langle A \rangle}^2[/tex] BUT [tex]A[/tex] is an operator and [tex]\langle A \rangle[/tex] is a number! How can [tex](A- \langle A \rangle)^2[/tex] even be defined?
Second - somewhere else in the proof a new vector is defined [tex]|\varphi \rangle=C| \psi \rangle[/tex] OK, they can do it, but then they say that if so, then also [tex]\langle \varphi |= \langle \psi | C^\dagger[/tex] and I just can't see why... Why the dagger I mean
Can anyone help?
Thanks in advance.
I try to understand the proof for the uncertainty principle for two Hermitian operators A and B in a Hilbert space. My questions are rather general so you don't need to know the specific proof.
The first thing I couldn't get into my head was the definition of uncertainty
[tex](\Delta A)^2=\langle\psi|(A- \langle A \rangle)^2|\psi\rangle[/tex]
Using the properties of the inner product I can be convinced this is equivalent to the ordinary definition of [tex](\Delta A)^2=\langle A^2 \rangle-{\langle A \rangle}^2[/tex] BUT [tex]A[/tex] is an operator and [tex]\langle A \rangle[/tex] is a number! How can [tex](A- \langle A \rangle)^2[/tex] even be defined?
Second - somewhere else in the proof a new vector is defined [tex]|\varphi \rangle=C| \psi \rangle[/tex] OK, they can do it, but then they say that if so, then also [tex]\langle \varphi |= \langle \psi | C^\dagger[/tex] and I just can't see why... Why the dagger I mean
Can anyone help?
Thanks in advance.
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