- #1
TheCanadian
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I am reading through a quantum optics book where they are deriving the equations for a quantized EM field and one of the paragraphs state:
"In Section 6.1, the problem has been set in the Hamiltonian form by expressing the total energy (6.55) of the system comprising charges and electromagnetic field in terms of the pairs of conjugate canonical variables ##(r_\mu, p_\mu)## and ##(Q_l, P_l)##. Canonical quantization consists in replacing these pairs of canonical variables by pairs of Hermitian operators with commutators set equal to ##i\hbar##."
Now I understand using the operators ## \hat{x} = x## and ##\hat{p_x} = -i\hbar \frac{\partial}{\partial x} ## that the commutator is ##i\hbar## but what necessitates all canonical variables requiring this as the result of their commutator? I guess I'm just missing something fairly fundamental here; namely on how these operators were established in the first place (i.e. why does momentum involve a derivative and position simply a scalar multiplication) and the intuition/meaning behind a commutator value of precisely ##i\hbar##.
"In Section 6.1, the problem has been set in the Hamiltonian form by expressing the total energy (6.55) of the system comprising charges and electromagnetic field in terms of the pairs of conjugate canonical variables ##(r_\mu, p_\mu)## and ##(Q_l, P_l)##. Canonical quantization consists in replacing these pairs of canonical variables by pairs of Hermitian operators with commutators set equal to ##i\hbar##."
Now I understand using the operators ## \hat{x} = x## and ##\hat{p_x} = -i\hbar \frac{\partial}{\partial x} ## that the commutator is ##i\hbar## but what necessitates all canonical variables requiring this as the result of their commutator? I guess I'm just missing something fairly fundamental here; namely on how these operators were established in the first place (i.e. why does momentum involve a derivative and position simply a scalar multiplication) and the intuition/meaning behind a commutator value of precisely ##i\hbar##.