# Motivation for transition probabilities in quantum mechanics

In the area of the time-dependent perturbation theory, if we have $$\hat{H}(t) = \hat{H}_0 + \hat{V}(t).$$ The Schrodinger equation is $$i \hbar \frac{d | \Psi(t) \rangle}{dt} = (\hat{H}_0 + \hat{V}(t))| \Psi(t) \rangle$$ where in the Shrodinger picture we have $$|\Psi(t) \rangle = \sum_n c_n(t)e^{\frac{-iE_n}{\hbar}}| \psi_n \rangle$$ In the interaction picture we have $$i \hbar \frac{d | \Psi(t) \rangle_{I}}{dt} = \hat{V}_{I}(t) | \Psi(t) \rangle_{I}$$ where $$| \Psi(t) \rangle_{I} = \hat{U}(t,t_i)| \Psi(t_i) \rangle_I$$ we therefore get $$i \hbar \frac{d \hat{U}(t, t_i)}{dt} = \hat{V}_i(t)\hat{U}_{I}(t,t_i).$$

Question: Short question, why does it follow that the transition probability corresponding to a transition from an initial unperturbed state $| \psi_i \rangle$ to another unperturbed state $| \psi_f \rangle$ is $$P_{if}(t) = |\langle \psi_f | \hat{U}_{I}(t,t_i)| \psi_i \rangle |^2$$ also why is the transition probability in terms of the expansion coefficients given by $$P_{if}(t) = |c_{f}^{0}+c_{f}^{1} +...|^2?$$ Are these postulates?

Thanks.

Question 1 :

First of all let us remember the link between the Schrödinger and Interaction Picture : $$|\psi(t)\rangle_I = e^{i\hat{H_0}t/\hbar} |\psi(t)\rangle_S \quad \text{and} \quad \hat{O}_I(t) = e^{i\hat{H_0}t/\hbar} \hat{O}_S(t) e^{-i\hat{H_0}t/\hbar}$$

If I understand well your notation $| \psi_f \rangle$ and $| \psi_i \rangle$ are given unperturbed states (that is eigenvectors of $\hat{H}_0$ with eigenvalues $E_f$ and $E_i$) where $f$ stands for final and $i$ stands for initial.

Let me explain : The operator $\hat{U}_I(t,t_i)$ is the evolution operator in the interaction picture. By definition when you apply it to the initial state $|\psi(t_i)\rangle_I = | \psi_i \rangle$, it makes it evolve to the state at time $t$ in the interaction picture, that is $| \psi(t) \rangle_I$ : $$| \psi(t) \rangle_I = \hat{U}_I(t,t_i) | \psi(t_i) \rangle_ I = \hat{U}_I(t,t_i) | \psi_i \rangle$$

Now, the probability to measure the state $| \psi(t) \rangle_I$ as $| \psi_f \rangle$ is also by definition the scalar product squared : $$P_{if}(t) = |\langle \psi_f | \psi(t)\rangle_I|^2$$

Hence : $$P_{if}(t) = |\langle \psi_f| \hat{U}_I(t,t_i) | \psi_i \rangle|^2$$

As a remark this is equal to : $$P_{if}(t) = |\langle \psi_f| \psi(t) \rangle_S|^2 = |\langle \psi_f|e^{-i(\hat{H}_0+\hat{V}_i(t))t/\hbar}| \psi_i \rangle|^2$$

Question 2

Using the remark above : $$P_{if}(t) = |\langle \psi_f| \psi(t) \rangle_S|^2$$ $$P_{if}(t) = |\langle \psi_f | \sum_n c_n(t) e^{-iE_nt/\hbar} | \psi_n \rangle|^2$$ $$P_{if}(t) = |c_f(t) e^{-iE_ft/\hbar}|^2 = |c_f(t)|^2$$

• Thanks for your answer. Two questions: Why do you have $| \psi(t_i) \rangle_{I} = | \psi_i \rangle$, is $|\psi_i \rangle$ not a stationary state in the Shrodinger picture? Should the equivalent state in the interaction picture not be $| \psi(t_i) \rangle_{I} = e^{\frac{i \hat{H}_{0} t_i}{\hbar}}| \psi_i \rangle$? – Alex Mar 7 '17 at 9:49
• Answering your first question : You are trying to compute the transition probability between $|\psi_i\rangle$ and $|\psi_f\rangle$. Hence the initial state that you are starting from is $|\psi_i\rangle$. On the other hand, the notation $|\psi(t_i)\rangle$ corresponds to the state at initial time $t_i$. Hence the equality in the Schrödinger picture. As a remark, $|\psi_i\rangle$ is a stationary state of the unperturbed hamiltonian $\hat{H}_0$ and not $\hat{H}(t)$ a priori. – mhham Mar 7 '17 at 10:57
• Now, you are right when you say that in the interaction picture the equivalent of $|\psi_i\rangle$ is $e^{i\hat{H}_0 t/\hbar} |\psi_i\rangle$. Hence if you consider everything in the interaction picture : $$P_{if}(t) = |_I\langle \psi_f | \hat{U}_I(t,t_i) | \psi_i \rangle_I|^2 = |\langle \psi_f | e^{-i\hat{H}_0 t/\hbar} \hat{U}_I(t,t_i) e^{i\hat{H}_0 t_i/\hbar} | \psi_i \rangle|^2$$ $$P_{if}(t) = | e^{i(E_f t - E_i t_i)/\hbar} \langle \psi_f | \hat{U}_I(t,t_i) | \psi_i \rangle|^2 = | \langle \psi_f | \hat{U}_I(t,t_i) | \psi_i \rangle|^2$$ This gives the wanted result. – mhham Mar 7 '17 at 10:59
• This, in a way also answers your second question. In terms of probabilities, thanks to the presence of $| \quad |^2$, an eigenstate of $\hat{H}_0$ gives the same result in the Schrödinger and Interaction picture. And finally, all the different pictures (Schrödinger, Heisenberg, Interaction) give the same results when it comes to deriving probabilities, expectation values, and so on. This is important because these are the physical values that we have access to in experiments ! They should not be dependent on the picture that we use to make the calculations. – mhham Mar 7 '17 at 11:07
• The first equality written by JDoe is unfortunately wrong. Indeed $|\psi(t)\rangle_I = e^{i\hat{H}_0 t/\hbar}|\psi(t)\rangle_S$ and not $|\psi(t)\rangle_I = e^{i\hat{H}_0 t/\hbar}|\psi_i(t)\rangle_S$. The Schrödinger and Interaction states relation is valid at same time $t$ for both, not $t$ for the interaction term and $t_i$ for the Schrödinger term. I hope my explanation is OK – mhham Mar 7 '17 at 11:13