# What is the impact of the time-independent Hamiltonian operator on the observation probabilities?

If you assume that $$H$$ is a time-independent Hamiltonian, by the Schrodinger equation, the state evolution $$|\Psi(t)\rangle$$ is given by $$\left( \sum e^{\frac{-i \lambda_j}{\hslash} t} |{ v_j }\rangle \langle v_j | \right)(|\Psi(t_0) \rangle)$$, where $$\lambda_j$$ is an eigenvalue of the operator $$H$$, and $$v_j$$ is the corresponding eigenvector.

In such a case, it is easy to show that $$| \langle v_j | \Psi(t) \rangle |^2 = | \langle v_j | \Psi(t_0) \rangle |^2$$. It makes sense also. Measuring the energy at time $$t_0$$ or time $$t$$ should be the same, if one assumes that the energy is time-independent.

Reading from some lecture notes on the web (related to a physics course), if any other Hermitian operator $$L$$ is chosen to do some measurement on any other property of the system, $$| \langle v^* | \Psi(t) \rangle |^2 = | \langle v^* | \Psi(t_0) \rangle |^2$$, where $$v^*$$ is an eigenvector of this new operator. Thus, the probability of observing $$\lambda^*$$ (the corresponding eigenvalue associated to eigenvector $$v^*$$ of $$L$$) is

at time $$t_0$$ : $$|\langle v^* | \Psi(t_0) \rangle|^2 = \left| \sum \langle v^* | a_i(t_0) v_i \rangle \right|^2$$

at time $$t > t_0$$ : $$|\langle v^* | \Psi(t) \rangle|^2 = \left| \sum \langle v^* | e^{-i \lambda_i t / \hslash} a_i(t_0) v_i \rangle \right|^2$$

At this point, I cannot figure out if my interpretation of the lecture notes (source seems very much reliable) is wrong and those two probabilities are different, or if these two probabilities are really equal.

• In general (i.e., when $L$ and $H$ do not commute), the probabilities are different. Jun 11, 2021 at 11:04
• But in this case, I just want to see if there is a difference to the measure $L$ at time $t_0$ or later on at time $t$. I do not plan to measure with $H$ meanwhile.
– Jean
Jun 11, 2021 at 11:39
• I'm not sure I understand your comment. The expressions at $t_0$ and $t$ are not the same as there is interference of probability amplitudes at $t$, i.e. each term in the sum is multiplied by $e^{-i\lambda_i t/\hbar}$ so that $\sum a_i \langle v^*\vert v_i\rangle \ne \sum a_i e^{-i\lambda_i t/\hbar} \langle v^*\vert v_i\rangle$. Then of course the probabilities will be different. Jun 11, 2021 at 12:03
• Suppose that $\langle v^*| = \langle v_k|$ is a eigenvector of $H$, $\sum a_i \langle v^*|v_i\rangle = a_k \langle v_k|v_k\rangle$. Its squared magnitude is simply $|a_k|^2$. The squared magnitude of the second term is simply $|e^{-i \lambda_k t / \hslash} a_k |^2 = |a_k|^2$. The time related factor has a magnitude of 1. It simply disappears. Back to Fabian comment, if $L$ and $H$ commute, i.e., if they have the same eigenvectors, the same result can be derived. So, in some cases, the factor $e^{-i \lambda_k t / \hslash}$ may disappear, explaining my previous comment.
– Jean
Jun 11, 2021 at 17:59

This is best understood with a specific example rather than a general mess of indices. Take $$H=\hbar\omega \sigma_z=\hbar\omega \left(\begin{array}{cc}1&0 \\ 0 &-1\end{array}\right)$$ and $$\Psi(0)=\frac{1}{\sqrt{2}}\left(\begin{array}{c} 1\\ 1\end{array}\right)=\frac{1}{\sqrt{2}}\vert +\rangle +\frac{1}{\sqrt{2}}\vert -\rangle$$, where $$\sigma_z\vert\pm\rangle =\pm \vert\pm\rangle$$.
Take $$\hat L=\sigma_x=\left(\begin{array}{cc} 0 & 1 \\ 1&0 \end{array}\right)$$, with eigenvectors $$\vert \pm 1\rangle = \frac{1}{\sqrt{2}}\left(\vert +\rangle \pm \vert -\rangle\right)$$ . At $$t=0$$, the probability of getting the eigenvalue $$+1$$ for $$\sigma_x$$ is $$\vert\langle +1\vert \Psi(0)\rangle\vert^2=1\, .$$ After a time $$t$$, this probability is \begin{align} \vert\langle +1\vert \Psi(t)\rangle\vert^2&= \vert\frac{1}{\sqrt{2}}\langle +1\vert +\rangle e^{-i \omega t}+\frac{1}{\sqrt{2}} \langle +1\vert -\rangle e^{i \omega t}\vert^2\, ,\\ &=\vert \frac{1}{2} e^{-i \omega t}+\frac{1}{2} e^{i \omega t}\vert^2= \cos^2(\omega t)\, . \end{align} Thus the effect of the different factors $$e^{\pm i\omega t}$$ produce time-dependent interference between the eigenstates of $$\hat H$$. In particular, at $$t=0$$ the probability is $$1$$ but at $$\omega t=\pi/2$$ the probability is $$0$$.