- #1
kvl
- 10
- 0
In standard quantization procedure we should apply commutation rules [tex][p,q]=i[/tex]. But let's do a simple calculation:
[tex] i \langle q_0 | q_0 \rangle = \langle q_0 | [p,q] | q_0 \rangle = \langle q_0 | pq | q_0 \rangle - \langle q_0 | qp | q_0 \rangle = q_0 \langle q_0 | p | q_0 \rangle - q_0 \langle q_0 | p | q_0 \rangle) = 0 [/tex]
which means that in Hilbert space q can have no eigenvectors other then zero (the same holds for p). But then, what does completeness of q means ? How can q be an observable ?
[tex] i \langle q_0 | q_0 \rangle = \langle q_0 | [p,q] | q_0 \rangle = \langle q_0 | pq | q_0 \rangle - \langle q_0 | qp | q_0 \rangle = q_0 \langle q_0 | p | q_0 \rangle - q_0 \langle q_0 | p | q_0 \rangle) = 0 [/tex]
which means that in Hilbert space q can have no eigenvectors other then zero (the same holds for p). But then, what does completeness of q means ? How can q be an observable ?