As far as my knowledge goes, in quantum field theory, energy-momentum relation is imposed. Is my understanding correct, and just in case my underlying understanding is incorrect, can anyone explain how energy-momentum relation is imposed?
In Feynman diagram picture, virtual particles are known not to obey energy-momentum relation, but as far as I know, they only arise because Feynman diagram is essentially perturbation picture. Is this correct understanding?
I assume that by "energy-momentum relation" you mean $$ p^2=m^2 $$ where $ab\equiv a^0 b^0-\boldsymbol a\cdot\boldsymbol b$. Here, $p=(E,\boldsymbol p)$.
Note that this relation is kinematical, not dynamical. It's origin is the same as in classical mechanics; there is nothing "quantum-mechanical" about that. In essence, the relation above is a definition: mass is always defined as $$ m^2\equiv p^2 $$
That this is a sensible definition is based on the fact that $p^2$ is a Lorentz scalar. This is independent of the dynamics of the theory; it is a consequence of the properties of Lorentz transformations, which are the same in the classical and quantum mechanical cases. So no, this relation is not really "imposed", but is nothing more than a convenient definition.
As you already know, the fact that $p^2$ is a Lorentz scalar means that it is independent of the reference frame. This means that you can measure $m^2$ by placing yourself in the rest frame of the particle, and measuring its energy. As $p^2=E^2-\boldsymbol p^2$, and $\boldsymbol p=\boldsymbol 0$ in the rest frame, we have $m=E_\mathrm{rest}$. We use this to actually measure the mass of a particle (though in practice there are other experimental methods that are much more convenient). In short: mass is, by definition, the energy of a particle in its rest frame.
In a Feynman diagram, you can always set $p^2=\mu^2$, but $\mu$ is not the mass of anything, because $p$ is an integration variable (i.e., a Fourier variable). It is not the momentum of anything. In this sense, the definition $$ \mu^2\equiv p^2 $$ is valid but useless; $\mu$ is a meaningless parameter and it cannot be measured.
In regards to virtual particles, I really think that you should read the PSE posts What actually are virtual particles? and Do virtual particles actually physically exist?. Long story short, a "virtual particle" has nothing to do with actual particles. They are Wick contractions of interacting fields; they are a rather technical concept that is very often misused by non-experts. My recommendation is that you shouldn't try to understand this concept until you have studied the basics of QFT. Otherwise, chances are that you will use the concept incorrectly, as many people do.
Finally, I'm not sure what you mean by "perturbation picture". Perhaps you mean "interaction picture". If this is actually what you mean, then let me point out that you can introduce Feynman diagrams without any mention to the interaction picture, for example using Feynman's path integral. In this sense, the fact that Feynman diagrams can also be introduced using the interaction picture is rather tangential to your question. It has nothing to do with $p^2=m^2$.
