# Two expressions for expectation value of energy

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

http://hyperphysics.phy-astr.gsu.edu/hbase/quantum/expect.html

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: $$\langle E\rangle=\int\limits_{-\infty}^{+\infty}\Psi^*(i\hbar)\frac{\partial}{\partial t}\Psi\,dx$$

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.

First off, there are some very misleading answers given above. Introductory quantum courses fail to properly discuss "time." It is a parameter, not an observable. E(operator)=ih(bar) d/dt has no meaning. That operator simply discribes the time evolution of a wavefuntion that is complex. So it does not describe a physical observable. I know this might be hard to swallow because of the uncertainty relation between time and energy. But make no mistake, the "time" discussed in the uncertainty principal is more of an interval as opposed to an observable. Maybe it is easier to realize what I mean by thinking about a question: when is the zero point for time? We can not establish one like we can establish a spatial origin. So to calculate the expectation value of the energy required the use of the Hamiltonian.

Second, the notation. = . Either is acceptable.

Third off. The free particle is really tricky. A single eigenstates is not a physically realizable state. Only a linear combination is physically realizable. This might seem inaccurate sense V=0 in the infinite well too, but those states are physically realizable. What you should note is that the well is comprised of bound states where the index for the eigenstates is discrete (ie: n=1,2,3,...), but the free particle is comprised of scattering states where the index for the eigenstates is continuos (ie: n= any and ALL real numbers). As a result, involves a Fourier transformed wavefunction.

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.