Consider a system whose state is initially $\left|\psi(t_i)\right\rangle$. At a later time $t_f$, its state will be $$\left|\psi(t)\right\rangle=\mathcal{U}(t_i,t_f)\left|\psi(t_i)\right\rangle$$ where $\mathcal{U}(t_i,t_f)=\exp(-i\hat{H}(t_f-t_i)/\hbar)$ is the time evolution operator (in the case of a time-independent Hamiltonian).

Now consider the matrix elements of $\mathcal{U}(t_i,t_f)$ in the position eigenbasis $\{\left|x\right\rangle\}$. We define the propagator as $$\left\langle x_f\middle|\mathcal{U}(t_i,t_f)\middle|x_i\right\rangle$$

How can I show that the propagator as defined above can also be written as the transition amplitude $$\left\langle x_f,t_f\middle|x_i,t_i\right\rangle$$ ?


After thinking about it for a bit, I came up with this: since



$$\left|x_f,t_f\right\rangle=\exp(-i\hat{H}t_f/\hbar)\left|x_f\right\rangle\quad\Rightarrow\quad\left\langle x_f,t_f\right|=\left\langle x_f\right|\exp(i\hat{H}t_f/\hbar)$$

then it follows that $$\left\langle x_f,t_f\middle|x_i,t_i\right\rangle=\left\langle x_f\middle| \exp(-i\hat{H}(t_i-t_f)/\hbar) \middle| x_i\right\rangle$$

however this produces the wrong sign in the exponent, and I also believe I might be mixing the Heisenberg and Schrodinger pictures.


1 Answer 1


Unless you have an alternative definition of the transition amplitude given in class I think this question is just a matter of notation / definition. When you write $\langle x_f, t_f | x_i, t_i\rangle$ you mean precisely the matrix element of the evolution operator (the name for this matrix element is the kernel, or propagator) in the position space representation.

Note in particular that since you're working in the Schrödinger picture the basis states $\{ |x\rangle \} $ are time independent.

Edit: Given your comment, if the states $|x_{i}, t_{i}\rangle$ are supposed to be taken in the Heisenberg picture then $ |x_{i}, t_{i}\rangle = e^{\frac{i}{\hbar} \hat{H} t_{i}} |x_{i}\rangle$ where $|x_{i}\rangle$ is the Schrödinger picture operator at time zero. Likewise for $|x_{f}, t_{f}\rangle$. Conjugating and taking the inner product we get $$\langle x_{f}, t_{f} | x_{i}, t_{i}\rangle = \langle x_{f} | e^{-\frac{i}{\hbar} \hat{H}t_{f}} e^{\frac{i}{\hbar} \hat{H}t_{i}} | x_{i}\rangle = \langle x_{f} | e^{-\frac{i}{\hbar} \hat{H}(t_{f} - t_{i})} | x_{i}\rangle \langle x_{f} | \hat{U}(t_{f}, t_{i}) | x_{i} \rangle$$ as required.

  • $\begingroup$ Hmm Sakurai on page 111 notes that either expression "can elegantly be written as" the other. Nakahara (page 19) also sketches a proof that I do not understand. $\endgroup$
    – martin
    Commented Jan 20, 2020 at 3:13
  • $\begingroup$ I've updated my answer $\endgroup$
    – nox
    Commented Jan 20, 2020 at 3:24

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service and acknowledge you have read our privacy policy.

Not the answer you're looking for? Browse other questions tagged or ask your own question.