On the Explicit Expression of a Time-dependent Hamiltonian in Heisenberg Picture
Published:
The Heisenberg Hamiltonian
\[\hat{H}_H\left(t,t_0\right)=U^\dagger \left(t,t_0\right) \hat{H}_S\left(t\right) U(t,t_0)\]We’ll prove the following Theorem.
Statement of the Theorem
\[\hat{H}_H\left(t,t_0\right)=\hat{H}_S\left(t\right) + \sum_{n=1}^{\infty} \left(-\frac{i}{\hbar}\right)^n \int_{t_0}^{t}dt_1 \int_{t_0}^{t_1}dt_2 \cdots \int_{t_0}^{t_{n-1}}dt_n \ \hat{C}_n\left(t,t_1,t_2,\cdots,t_n\right)\]where
\[\hat{C}_n\left(t,t_1,t_2,\cdots,t_n\right) = \left[\cdots \left[\left[\hat{H}_S\left(t\right),\hat{H}_S\left(t_1\right) \right],\hat{H}_S\left(t_2\right)\right] \cdots, \hat{H}_S\left(t_n\right) \right]\]and it can be easily seen that
\[\hat{C}_n\left(t,t_1,t_2,\cdots,t_n\right) = \left[\hat{C}_{n-1}\left(t,t_1,t_2,\cdots,t_{n-1}\right),\hat{H}_S\left(t_n\right) \right]\]To deal with the problem, we first sketch the proofs of several lemmas.
Lemma 1
Statement of Lemma 1
\[\int_{t_0}^{t}dt_1 \int_{t_0}^{t_1}dt_2 \cdots \int_{t_0}^{t_{n-1}}dt_n = \int_{t_0}^{t}dt_n \int_{t_{n}}^{t}dt_{n-1} \cdots \int_{t_2}^{t}dt_1\]Proof of Lemma 1
Proof: By induction.
- Starting point:
where LHS and RHS both correspond to the integration over
\[D=\{(t_1,t_2)\ | \ t_0\leq t_1 \leq t,\ t_0\leq t_2 \leq t, \ t_2 \leq t_1\}\]- Induction: suppose we already have
Then
\[\begin{align*} & \int_{t_0}^{t}dt_1 \int_{t_0}^{t_1}dt_2 \cdots \int_{t_0}^{t_{n-1}}dt_n \int_{t_0}^{t_{n}}dt_{n+1}\\ =& \boxed {\int_{t_0}^{t}dt_1 \int_{t_0}^{t_1}dt_{n+1}} \int_{t_{n+1}}^{t_1}dt_n \cdots \int_{t_3}^{t}dt_2 \\ =& \int_{t_0}^{t}dt_{n+1} \boxed{ \int_{t_{n+1}}^{t}dt_{1} \int_{t_{n+1}}^{t_1}dt_n } \cdots \int_{t_3}^{t}dt_2 \\ =& \cdots \\ =& \int_{t_0}^{t}dt_{n+1} \int_{t_{n+1}}^{t}dt_{n} \cdots \int_{t_3}^{t}dt_2 \int_{t_2}^{t}dt_1 \quad \quad \quad \Box \end{align*}\]Lemma 2
Statement of Lemma 2
\[\frac{\partial}{\partial t_0} U\left(t, t_0\right) = \frac{i}{\hbar} U\left(t, t_0\right) H_S (t_0)\]Proof of Lemma 2
Proof: From
\[U\left(t, t_0\right) = \mathbb{1} + \sum_{n=1}^{\infty} \left(-\frac{i}{\hbar}\right)^n \int_{t_0}^{t}dt_1 H_S(t_1) \int_{t_0}^{t_1}dt_2 H_S(t_2) \cdots \int_{t_0}^{t_{n-1}}dt_n H_S(t_n)\]This is not hard to prove. (details omitted) \(\Box\)
Lemma 3
Statement of Lemma 3
\[\frac{\partial}{\partial t_0} H_H\left(t, t_0\right) = \frac{i}{\hbar} \left[H_H\left(t, t_0\right), H_S (t_0)\right]\]Proof of Lemma 3
\[\frac{\partial}{\partial t_0} H_H\left(t, t_0\right) = \left(\frac{\partial}{\partial t_0}U^\dagger \left(t,t_0\right) \right) \hat{H}_S\left(t\right) U(t,t_0) + U^\dagger(t,t_0) \hat{H}_S\left(t\right) \left(\frac{\partial}{\partial t_0}U \left(t,t_0\right)\right)\]From Lemma 2, \(\displaystyle \frac{\partial}{\partial t_0} U^\dagger\left(t, t_0\right) = -\frac{i}{\hbar} H_S (t_0) U^\dagger\left(t, t_0\right)\). Substitute them back to the previous equation, we finish the proof of Lemma 3. \(\Box\)
Lemma 4
Statement of Lemma 4
Let
\[f(t,t_0) = \hat{H}_S\left(t\right) + \sum_{n=1}^{\infty} \left(-\frac{i}{\hbar}\right)^n \int_{t_0}^{t}dt_1 \int_{t_0}^{t_1}dt_2 \cdots \int_{t_0}^{t_{n-1}}dt_n \ \hat{C}_n\left(t,t_1,t_2,\cdots,t_n\right)\]We have
\[\frac{\partial}{\partial t_0} f\left(t, t_0\right) = \frac{i}{\hbar} \left[f\left(t, t_0\right), H_S (t_0)\right]\]Proof of Lemma 4
Proof: From Lemma 1,
\[f(t,t_0) = \hat{H}_S\left(t\right) + \sum_{n=1}^{\infty} \left(-\frac{i}{\hbar}\right)^n \int_{t_0}^{t}dt_n \int_{t_{n}}^{t}dt_{n-1} \cdots \int_{t_2}^{t}dt_1 \ \hat{C}_n\left(t,t_1,t_2,\cdots,t_n\right)\]Therefore,
\[\frac{\partial}{\partial t_0} f(t,t_0) = - \sum_{n=1}^{\infty} \left(-\frac{i}{\hbar}\right)^n \int_{t_{n}}^{t}dt_{n-1} \cdots \int_{t_2}^{t}dt_1 \ \hat{C}_n\left(t,t_1,t_2,\cdots,t_{n-1},t_0\right)\]Rewrite
\[\hat{C}_n\left(t,t_1,t_2,\cdots,t_{n-1}, t_0\right) = \left[\hat{C}_{n-1}\left(t,t_1,t_2,\cdots,t_{n-1}\right),\hat{H}_S\left(t_0\right) \right]\]and notice
\[\hat{C}_0\left(t\right) = \hat{H}_S\left(t\right)\]we get
\[\begin{align*} \frac{\partial}{\partial t_0} f(t,t_0) &= \frac{i}{\hbar} \sum_{n=1}^{\infty} \left(-\frac{i}{\hbar}\right)^{n-1} \int_{t_{n}}^{t}dt_{n-1} \cdots \int_{t_2}^{t}dt_1 \ \left[\hat{C}_{n-1}\left(t,t_1,t_2,\cdots,t_{n-1}\right),\hat{H}_S\left(t_0\right) \right] \\ &=\frac{i}{\hbar} \left[f\left(t, t_0\right), H_S (t_0)\right] \end{align*}\]This finishes our proof of Lemma 4. \(\Box\)
Proof of the Theorem
Now we get back to the desired theorem.
Proof of the Theorem:
From Lemma 3 and Lemma 4,
\[\frac{\partial}{\partial t_0} H_H\left(t, t_0\right) = \frac{i}{\hbar} \left[H_H\left(t, t_0\right), H_S (t_0)\right]\] \[\frac{\partial}{\partial t_0} f\left(t, t_0\right) = \frac{i}{\hbar} \left[f\left(t, t_0\right), H_S (t_0)\right]\]which means \(H_H\left(t, t_0\right)\) and \(f\left(t, t_0\right)\) satisfies the same differential equation.
On the other hand,
\[H_H\left(t, t_0=t\right) = f\left(t, t_0=t\right) = H_S(t)\]which means they share the same initial condition.
Thus, for any \(t_0<t\), we have
\[H_H\left(t, t_0\right) = f\left(t, t_0\right)\]This marks the end of our proof. \(\Box\)