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I was looking up expectation value of energy for a free particle on the following webpage:


It says that $E=\frac{p^2}{2m}$ and therefore $\langle E\rangle=\frac{\langle p^2\rangle}{2m}$

This leads to $$\langle E\rangle=\int\limits_{-\infty}^{+\infty}\Psi^*\frac{-\hbar^2}{2m}\frac{\partial^2}{\partial x^2}\Psi\,dx$$

However, it also has on the bottom of the page: "In general, the expectation value for any observable quantity is found by putting the quantum mechanical operator for that observable in the integral of the wavefunction over space".

Now, I know that the operator for $E$ is $i\hbar\frac{\partial}{\partial t}$. So shouldn't $\langle E\rangle$ be: $$\langle E\rangle=\int\limits_{-\infty}^{+\infty}\Psi^*(i\hbar)\frac{\partial}{\partial t}\Psi\,dx$$

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2 Answers 2

up vote 2 down vote accepted

First things first: the operator which corresponds to the energy is the Hamiltonian, typically written as $H$. So when you want to get the expectation value of the energy, you evaluate $\langle H\rangle$.

Now, there are multiple ways to do this. One way is to use the Schroedinger equation to get

$$\langle H\rangle = \left\langle i\hbar\frac{\partial}{\partial t}\right\rangle = \int\Psi^*(x,t) i\hbar\frac{\partial}{\partial t}\Psi(x,t)\,\mathrm{d}x\tag{1}$$

That calculation is completely general, i.e. it is valid in any situation for which the Schroedinger equation applies.

Another way to get $\langle H\rangle$ is to use the definition of the Hamiltonian operator, which in nonrelativistic QM is

$$H = -\frac{p^2}{2m} + V(x,t) = -\frac{\hbar^2}{2m}\frac{\partial^2}{\partial x^2} + V(x,t)$$

That gives you

$$\begin{align}\langle H\rangle &= \biggl\langle -\frac{\hbar^2}{2m}\frac{\partial^2}{\partial x^2} + V(x,t)\biggr\rangle \\ &= \int\Psi^*(x,t)\biggl(-\frac{\hbar^2}{2m}\frac{\partial^2}{\partial x^2} + V(x,t)\biggr)\Psi(x,t)\,\mathrm{d}x\tag{2}\end{align}$$

Either (1) or (2) works in general.

For a free particle only, the potential $V(x,t)$ is zero, and you get

$$\langle H\rangle = \int\Psi^*(x,t)\biggl(-\frac{\hbar^2}{2m}\frac{\partial^2}{\partial x^2}\biggr)\Psi(x,t)\,\mathrm{d}x$$

which is the expression you saw on that web page.

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As David's answer notes, these two expressions are both correct. They should be equal to each other in all cases. For a particular case, you can verify that they are equal to each other by actually working both out. As a supplement to Davids answer, I thought I would go through this particular case.

I am going to cheat, because I know that the wave function for the free particle is $\Psi(x,t) = A\exp(-iEt/\hbar)\exp(i\sqrt{2mE}x/\hbar)$. Take the relevant derivatives: $$\frac{\partial}{\partial t}\Psi(x,t) = (-iE/\hbar) \Psi(x,t) $$ $$\frac{\partial^2}{\partial x^2} \Psi(x,t) = (i\sqrt{2mE}/\hbar)^2\Psi(x,t) = (-2mE/\hbar)\Psi(x,t)$$ Plug those into your two expressions for $\langle E \rangle$; the constants will all cancel out, giving you $\langle E \rangle = E\int\limits_{-\infty}^{+\infty}\Psi^*\Psi\,dx = E$ for both.

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