# Time evolution of state in first and second quantization

I was trying to work out the time evolution of a single particle state in second quantization and got something apparently contradicting with the first quantized picture.

For a system with energy eigenvalues $$\{\varepsilon_i\}$$, the second quantized Hamiltonian is $$\hat{H}=\sum_i \varepsilon_i \hat{c}^\dagger_i \hat{c}_i$$. In Heisenberg picture the creation operator at time $$t$$ is obtained by $$\frac{d}{dt}\hat{c}^\dagger_i = \frac{1}{i\hbar}[\hat{c}^\dagger_i, \hat{H}]$$ which gives $$\hat{c}^\dagger_i(t) = e^{i\varepsilon_it/\hbar}\hat{c}^\dagger_i(0)$$.

If we now consider the time evolution of an eigenstate $$|j\rangle$$, at $$t=0$$ we have $$|\psi(t=0)\rangle=\hat{c}_j^\dagger(0)|vac\rangle=|j\rangle$$ where $$|vac\rangle$$ is the vacuum state. At time $$t$$ we have $$|\psi(t)\rangle = \hat{c}_j^\dagger(t)|vac\rangle=e^{i\varepsilon_jt/\hbar}\hat{c}^\dagger_j(0)|vac\rangle=e^{i\varepsilon_jt/\hbar}|\psi(0)\rangle = e^{i\varepsilon_jt/\hbar}|j\rangle$$.

However, if we simply consider the time evolution operator in first quantization, $$U(t)=\exp{(-iHt/\hbar)}$$, then we expect an eigenstate $$|j\rangle$$ has time evolution $$|\psi(t)\rangle=e^{-i\varepsilon_jt/\hbar}|j\rangle$$.

The two results are off by a minus sign in front of the $$i$$ in the exponential. I don't think the two method should give a different answer but where is the problem in my calculation?

For a system with energy eigenvalues $$\{\varepsilon_i\}$$, the second quantized Hamiltonian is $$\hat{H}=\sum_i \varepsilon_i \hat{c}^\dagger_i \hat{c}_i$$. In Heisenberg picture the creation operator at time $$t$$ is obtained by $$\frac{d}{dt}\hat{c}^\dagger_i = \frac{1}{i\hbar}[\hat{c}^\dagger_i, \hat{H}]$$...

This is fine.

If we now consider the time evolution of an eigenstate $$|j\rangle$$, at $$t=0$$ we have $$|\psi(t=0)\rangle=\hat{c}_j^\dagger(0)|vac\rangle=|j\rangle$$ where $$|vac\rangle$$ is the vacuum state.

Ok. This definition for a $$t=0$$ state is fine.

At time $$t$$ we have $$|\psi(t)\rangle = \hat{c}_j^\dagger(t)|vac\rangle=\ldots$$.

No. This is wrong. At time t you have: $$|\psi(t)\rangle = e^{-iHt}|\psi(0)\rangle\;.$$

You don't just get to redefine time evolution however you would like. Remember, the purpose of the definition $$\mathcal{O}(t) = e^{iHt}\mathcal{O}(0)e^{-iHt}$$ is to preserve expectation values between the Heisenberg and Schrodinger pictures, such that $$\langle \psi(0)|\mathcal{O}(t)|\psi (0)\rangle = \langle \psi(t)|\mathcal{O}(0)|\psi (t)\rangle\;$$

However, if we simply consider the time evolution operator in first quantization, $$U(t)=\exp{(-iHt/\hbar)}$$,

This the time evolution operator that acts on the state regardless of if it is "first quantization" or "second quantization."

The two results are off by a minus sign in front of the $$i$$ in the exponential. I don't think the two method should give a different answer but where is the problem in my calculation?

The problem is the first calculation you present is wrong.