Find expectation value for combination of states - Physics Stack Exchange most recent 30 from physics.stackexchange.com 2019-09-17T15:29:39Z https://physics.stackexchange.com/feeds/question/355241 https://creativecommons.org/licenses/by-sa/4.0/rdf https://physics.stackexchange.com/q/355241 0 Find expectation value for combination of states Luan Nico https://physics.stackexchange.com/users/22659 2017-09-03T13:38:46Z 2017-09-03T19:38:23Z <p>I know how to find the expectation value for certain operators when the state is a simple state; for instance, if I want the expectation value of the energy of a state $\Psi = \Phi_{n, l, m}$, I know $\langle E_{n, l, m} \rangle = E_0/n^2$.</p> <p>But what if my state is a combination of simple states? If $\Psi = \sum_i q_i \cdot \Phi_{n_i, l_i, m_i}$?</p> <p>What I tried to do follows. Let's call $\Phi_{n_i, l_i, m_i} = \Phi_i$ for simplicity.</p> <p>The expectancy value of $\langle E \rangle = \int \text{d}^3r \Psi^* E \Psi = \int dr^3 \left[\sum_i q_i^* \Phi_i\right] E \left[\sum_i q_i \Phi_i \right]$.</p> <p>Here, when I distribute, I will get $\langle E \rangle = \int \text{d}^3r \sum_i\left[ q_i^* q_i \Phi_i E \Phi_i^*\right] + R$, where R are the mixed terms. Now I know the states $\Phi_i$ are orthogonal, so inside the integral $\Phi_i^* \Phi_j = 0$, but can I say $\Phi_i^* E \Phi_j = 0$ for $i \neq j$?</p> <p>Because then, if I can, I get $R = 0$ and $\langle E \rangle = \int \text{d}^3r \sum_i\left[q_i^* q_i \Phi_i E \Phi_i^*\right] = \sum_i\left[q_i^* q_i \int \text{d}^3r \Phi_i E \Phi_i^*\right] = \sum_i\left[q_i^* q_i \langle E_i\rangle\right]$, which reduces the problem to one I can solve.</p> <p>Is this logic correct?</p> https://physics.stackexchange.com/questions/355241/-/355246#355246 1 Answer by ZeroTheHero for Find expectation value for combination of states ZeroTheHero https://physics.stackexchange.com/users/36194 2017-09-03T13:53:02Z 2017-09-03T14:05:28Z <p>This is simplest to illustrate with two terms. Suppose $$\Psi=q_1\Phi_{1}+q_2\Phi_2$$ then \begin{align} \langle E\rangle &amp;= \int dr^3 \left(q_1\Phi_{1}+q_2\Phi_2\right)^* \hat H \left(q_1\Phi_{1}+q_2\Phi_2\right)\, ,\\ &amp;= \int dr^3 \left(q_1\Phi_{1}+q_2\Phi_2)\right)^* \left(E_1 q_1\Phi_{1}+E_2 q_2\Phi_2)\right)\, \\ &amp;=q_1^*q_1 E_1 \int dr^3 \Phi^*_{1}\Phi_{1} + q_2^*q_2 E_2 \int dr^3 \Phi^*_{2}\Phi_{2}\\ &amp;=+q_1^*q_2 E_2 \int dr^3 \Phi^*_{1}\Phi_{2}+ q_2^*q_1 E_1 \int dr^3 \Phi^*_{2}\Phi_{1}\, . \tag{1} \end{align} using $\hat H\Phi_k=E_k\Phi_k$ throughout.</p> <p>Note that my steps differ from yours by clearly identifying the Hamiltonian operator $\hat H$ on the right of the equality sign, and by properly labelling each energy $E_k$ in using $\hat H\Phi_k=E_k\Phi_k$. Since $E_k$ is just a real constant, then indeed $$\int dr^3 \Phi^*_i E_j \Phi_j= E_j \int dr^3 \Phi^*_i \Phi_j =\delta_{ij}$$ since $\int dr^3 \Phi^*_i \Phi_j=\delta_{ij}$.</p> <p>Thus the terms of the last line of (1) disappear by orthogonality as you correctly guessed and you are left with $$\langle E\rangle = \vert q_1\vert^2 E_1 +\vert q_2\vert^2 E_2$$ since the $\Phi$'s are normalized. </p> <p>Since $\vert q_k\vert^2$ is the probability of finding your system in the state $\Phi_k$, the average value is the weighted average energy of the system, with each energy value weighted by the probability for the system of being in state $\Phi_k$.</p>