Why Is a Given State an Eigenstate of Field Operators in Quantum Field Theory?

chientewu
Messages
7
Reaction score
0
Hi,

I am studying Peskin's An Introduction To Quantum Field Theory. On the beginning of page 284, the authors say We can turn the field \phi_S(x_1)|\phi_1\rangle=\phi_1(x_1)|\phi_1\rangle. I tried hard to prove this relation but still can't get it right. Could anyone give me some hints? Thanks.
 
Last edited:
Physics news on Phys.org
It's a bit strangely formulated. The generalized kets |\varphi \rangle are defined as generalized eigenstates of the field operators (here in the Schrödinger picture of time evolution), i.e.,
\hat{\phi}_S(\vec{x}_1) |\varphi \rangle = \varphi(x) |\varphi \rangle.
Note that \hat{\phi}_S is a field operator in the Schrödinger picture while \varphi is a (complex or real-valued, depending on whether you describe charged or strictly neutral scalar bosons) c-number field.
 
Last edited:
  • Like
Likes Holophagus
vanhees71 said:
Note that \hat{\phi}_S[/itex] is a field operator in the Schrödinger picture while \varphi is a (complex or real-valued, depending on whether you describe charged or strictly neutral scalar bosons) c-number field.
<br /> This is difficult to read. Please format your LaTeX correctly, or use UTF. <img src="https://cdn.jsdelivr.net/joypixels/assets/8.0/png/unicode/64/1f609.png" class="smilie smilie--emoji" loading="lazy" width="64" height="64" alt=":wink:" title="Wink :wink:" data-smilie="2"data-shortname=":wink:" />
 
Bill_K said:
This is difficult to read. Please format your LaTeX correctly, or use UTF. :wink:

Done (butt no UTF, which is hard to read either ;-)).
 
vanhees71 said:
It's a bit strangely formulated. The generalized kets |\varphi \rangle are defined as generalized eigenstates of the field operators (here in the Schrödinger picture of time evolution), i.e.,
\hat{\phi}_S(\vec{x}_1) |\varphi \rangle = \varphi(x) |\varphi \rangle.
Note that \hat{\phi}_S is a field operator in the Schrödinger picture while \varphi is a (complex or real-valued, depending on whether you describe charged or strictly neutral scalar bosons) c-number field.

Thanks! That makes sense but I still don't understand why this given state is an eigenstate of field operators with eigenvalue being the field amplitudes at some specific position.
 
chientewu said:
Thanks! That makes sense but I still don't understand why this given state is an eigenstate of field operators with eigenvalue being the field amplitudes at some specific position.

Because that's how we're defining the state |\phi_1\rangle. We're trying to pick out the state in the Hilbert space that, when you hit it with the field operator at any position x, will give you an eigenvalue equal to the c-number \phi(x). It's a way of going from states in a Hilbert space to simple c-number functions, so that you can perform the functional integral over them.
 
Last edited:
  • Like
Likes Holophagus
Insights auto threads is broken atm, so I'm manually creating these for new Insight articles. Towards the end of the first lecture for the Qiskit Global Summer School 2025, Foundations of Quantum Mechanics, Olivia Lanes (Global Lead, Content and Education IBM) stated... Source: https://www.physicsforums.com/insights/quantum-entanglement-is-a-kinematic-fact-not-a-dynamical-effect/ by @RUTA
If we release an electron around a positively charged sphere, the initial state of electron is a linear combination of Hydrogen-like states. According to quantum mechanics, evolution of time would not change this initial state because the potential is time independent. However, classically we expect the electron to collide with the sphere. So, it seems that the quantum and classics predict different behaviours!
Back
Top