The matrix representation of the Hamiltonian is given in the position basis: $\{|+a\rangle,|-a\rangle\}$. $$H = \begin{pmatrix} E_{0} & E_{-A} \\ E_{-A} & E_{0} \end{pmatrix}$$

The initial state of a system is also given in terms of the position basis:

$\mid \psi \rangle = \frac{1}{5}[3|+a\rangle + 4|-a\rangle]$.

To find the probability of the a particular energy eigenvalue, why is it incorrect to choose the coefficient of $|-a\rangle$ so that it would be $\left(\frac{4}{5}\right)^2$?

Instead, we have to transform to the Hamiltonian basis, using the eigenvectors of the Hamiltonian. Why can't we use the position basis?


1 Answer 1


In quantum mechanics observables are represented by operators and the outcome of a given measurement can only be an eigenvalue of this operator. After the measurement, the state of the system will be projected to the eigenstate of the operator corresponding to the measured eigenvalue (consider for simplicity the case where you don't have more than one eigenstate corresponding to one eigenvalue).

Now, in you case, the probability $|4/5|^2$ corresponds to the following: $$ P(\text{-}a) = |\langle \text{-}a|\psi\rangle|^2. $$ This is the probability of projecting (by measurement) the state of the system into $|-a\rangle$. This corresponds to computing the probability of obtaining the eigenvalue $\text{-}a$ as a result of your measurement. If you want to know the probability of being in a give energy state, you have to compute the following probability $$ P(E) = |\langle E|\psi\rangle|^2. $$ where $|E\rangle$ is the state with energy $E$ (eigenstate of the Hamiltonian). Therefore, in order to compute the braket $\langle E|\psi\rangle$ you need to write $|\psi\rangle $ in the basis of Hamiltonian's eigenstates: in this basis, the solution is trivial.


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