1
$\begingroup$

The expectation value of the momentum of a one-dimensional wave function $\Psi$ is

$$\langle p \rangle = \int_{-\infty}^{\infty} \Psi^* \hat{p}\Psi \, dx = \int_{-\infty}^{\infty} \Psi^* \left(-i \hbar \frac{\partial}{\partial x}\right)\Psi \, dx.$$

The square of the uncertainty of the momentum is

$$\sigma_p^2 = \langle p^2 \rangle - \langle p \rangle ^2.$$

From the reading I have done, it seems that $\langle p^2 \rangle$ is calculated as

$$\langle p^2 \rangle = \int_{-\infty}^{\infty} \Psi^* \hat{p}^2 \Psi \, dx = \int_{-\infty}^{\infty} \Psi^* \left(-\hbar^2 \frac{\partial^2}{\partial x^2}\right) \Psi \, dx.$$

This is what confuses me. Why do we square just the operator? To me, that doesn't seem like we are squaring the momentum. I don't understand how squaring the momentum operator is equivalent to squaring the momentum. I believe my confusion stems from a misunderstanding of what momentum is in quantum mechanics and the difference between momentum and the momentum operator.

$\endgroup$
2
  • 1
    $\begingroup$ Have you seen physics.stackexchange.com/a/57740/291677? $\endgroup$ May 5, 2021 at 17:06
  • 1
    $\begingroup$ I would advise you to read the first few chapters of Landau & Lifshitz (Quantum Mechanics : Non-relativistic theory). They explain very well why and how observables are represented by operators $\endgroup$ May 5, 2021 at 20:09

2 Answers 2

1
$\begingroup$

When you do:

$$ \langle p \rangle = \int_{-\infty}^{\infty} \Psi^* \hat{p}\Psi \, dx = \int_{-\infty}^{\infty} \Psi^* \left(-i \hbar \frac{\partial}{\partial x}\right)\Psi \, dx \; , $$

you are actually doing:

$$ \langle p \rangle = \langle \Psi | \ \hat{p} \ | \Psi \rangle \; , $$

which is the expectation value of the momentum, as you know it.

For the actual measurement of the momentum of a system in a given state, the operation is:

$$ \hat{p} \ |\Psi\rangle \; , $$

here, $\hat{p}$ is the momentum operator and $| \Psi \rangle$ is a vector (in the Hilbert space of the problem) representing the state of the system. For recovering the wave function related to the state one must perform a projection of $| \Psi \rangle $ in the real space (coordinate representation), that goes like this:

$$ \langle x | \Psi \rangle \equiv \Psi(x) \; , $$

this is Quantum Mechanics in Dirac Notation.

In this perspective is more clear to see why we square just the operator. It is because the measurement of a physical quantity (an observable) is given by the application of an operator to a state of the system. So, the first $\langle p \rangle$ is not the true measurement of momentum, but the average of the possible values of momentum.

That said, when you want to calculate the expectation value $\langle p^2 \rangle$, you do:

$$ \langle p^2 \rangle = \langle pp \rangle = \langle \Psi | \ \hat{p} \hat{p} \ | \Psi \rangle = \langle \Psi | \ \hat{p}^2 | \Psi \rangle = \int_{-\infty}^{\infty} \Psi^* \hat{p}^2 \Psi \, dx \; . $$

$\endgroup$
0
$\begingroup$

By definition \begin{align} \langle \hat A\rangle := \int_{-\infty}^{\infty} dx \Psi^* \hat A \Psi \end{align} for any operator so apply this to $\hat A=\hat p$ and $\hat A=\hat p^2$. In practice, the square of an operator means the operator acts twice: \begin{align} \hat A^2\Psi=\hat A\hat A\Psi=\hat A\left(\hat A\Psi\right)\, . \end{align} Of course from the definition \begin{align} \langle \hat A\rangle^2=\left( \int_{-\infty}^{\infty} dx \Psi^* \hat A \Psi\right) \left( \int_{-\infty}^{\infty} dx \Psi^* \hat A \Psi\right) \ne \int_{-\infty}^{\infty} dx \Psi^* \hat A^2 \Psi =\langle \hat A^2\rangle \end{align}

$\endgroup$

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.