0
$\begingroup$

Consider a Hamiltonian $H$ with discrete eigenvalues $\{E_n\}_{n=1}^\infty$ and eigenstates $\{\psi_n\}_{n=1}^\infty$.

Suppose I prepare a state $\psi=c_1\psi_1+c_2\psi_2$ (normalized) and make a measurement of the "energy". What will I find? The Born rule tells me that I will measure the energy to be one of two values: \begin{align} \mathcal E=\cases{E_1\quad \text{with probability} |c_1|^2\\ E_2\quad \text{with probability} |c_2|^2 \tag{1} } \end{align} But there is another possibility that seems natural to me: the measured energy could be the expectation value: $\langle \psi|H |\psi\rangle=E_1 |c_1|^2+E_2|c_2|^2$.

So my question is: which of these possibilities is realised? Will I measure the energy to be $\langle H \rangle$ with certainty, or will I measure the energy according to (1)?

$\endgroup$
1
  • $\begingroup$ Technically, you often measure neither. For instance, in spectroscopic measurements (almost all quantum measurements taken in a lab), you will actually measure the difference between the energy levels. Now, since we can arbitrarily define the zero of energy, this is basically just measuring the energy levels, but I find it interesting that mathematically Fermi’s Golden Rule doesn’t give the results as the pure energies, but their differences. $\endgroup$ Oct 23, 2023 at 11:04

3 Answers 3

3
$\begingroup$

Born's rule is correct. Your measurement result will be $E_1$ or $E_2$, but not $E_\text{average}$.

It is like rolling a die. You get a $1$, $2$, $3$, $4$, $5$ or $6$. But you never get a $3\frac 12$.

$\endgroup$
2
$\begingroup$

The measured energy follows the Born rule. It may help to consider the following simplified setup. Consider a spin-1/2 system in the state $\frac{\mid\uparrow\rangle+\mid\downarrow\rangle}{\sqrt{2}}$. You measure the spin $\sigma_z$ of the system. Will you ever measure the expectation value of the spin, $\langle\sigma_z\rangle=0$? (No.)

$\endgroup$
2
  • 1
    $\begingroup$ So I have to find the spectrum of the operator in question to deduce the outcomes and then use Born’s rule to deduce the probabilities? $\endgroup$
    – dennis
    Oct 23, 2023 at 11:12
  • $\begingroup$ @dennis Yes, that is correct. $\endgroup$
    – DanDan0101
    Oct 23, 2023 at 21:53
2
$\begingroup$

If you have set out to measure the energy of your system, you will measure the energy in accordance with the Born Rule. The energy expectation value is in essence the average result of measuring $\hat{E}$ on the state $\psi$ over a series of measurements.

To understand this better, suppose you prepare a very large number of identical systems each in the same state $\psi$. The energy $E$ is then measured on all of these identical systems; the results of these measurements are $E_1,E_2,E_3,...., E_n,...;$ the corresponding probabilities of occurrence are $P_1,P_2,P_3,...,P_n,...$ The average value of all these repeated measurements is called expectation value of $\hat{E}$ with respect to the state $\psi$.

So, your single measurement of energy will not yield the result $\langle H \rangle$.

$\endgroup$

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.