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fxdung
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Why can't we interprete /x> in relativistic QFT as position eigenstate?And by the way what is the difference between /x> and /1x>=Phi(field operator)(x)/0>?
In non-relativistic QFT it's the same. In relativistic QFT one changes the normalization of ##\phi## so that it becomes Lorentz invariant, the consequence of which is that ##|1x\rangle## differs from ##|x\rangle##. See e.g. myfxdung said:And by the way what is the difference between /x> and /1x>=Phi(field operator)(x)/0>?
In relativistic quantum field theory (QFT), the concept of position becomes more complicated due to the uncertainty principle. Unlike in non-relativistic quantum mechanics, where position operators can have well-defined eigenvalues, in relativistic QFT, the position operator does not have a well-defined eigenvalue. This is because position and momentum are conjugate variables, and the uncertainty principle states that they cannot both be precisely measured at the same time. Therefore, the state vector /x> cannot be interpreted as a position eigenfunction in relativistic QFT.
The uncertainty principle in relativistic QFT states that the position and momentum of a particle cannot be precisely measured at the same time. This means that the state vector /x> cannot be interpreted as a position eigenfunction, as the position operator does not have a well-defined eigenvalue in this context. Instead, the state vector represents a superposition of all possible positions of the particle, with each position having a certain probability of being measured.
While the state vector /x> cannot be interpreted as a position eigenfunction in relativistic QFT, it can still be used to describe the position of a particle in a probabilistic sense. The state vector represents a superposition of all possible positions of the particle, and the probability of measuring a specific position is given by the squared magnitude of the corresponding coefficient in the state vector.
In non-relativistic quantum mechanics, position is a well-defined observable that can be precisely measured. In relativistic QFT, however, the uncertainty principle affects the interpretation of position, as it does not have a well-defined eigenvalue. Instead, the concept of position becomes more probabilistic, with the state vector representing a superposition of all possible positions of the particle.
Aside from the uncertainty principle, there are other factors that prevent the state vector /x> from being interpreted as a position eigenfunction in relativistic QFT. These include the fact that the position operator does not commute with the Hamiltonian, and the fact that the state vector must be Lorentz invariant, which restricts its form and makes it difficult to interpret in terms of position. Additionally, the concept of position becomes more complicated in relativistic QFT due to the presence of multiple particles and interactions between them.