Is energy of a quantum mechanical moving particle conserved?

From the Schroedinger equation $$H\psi=E\psi,$$ if we want to measure the total energy of a quantum mechanical moving particle, then we have to apply the Hamiltonian operator to the wave function and as a result we get the eigenvalue of the Hamiltonian.

Now let a particle move under a central force field where the total energy is constant. Now I want to measure the energy of the particle so I have to apply Hamiltonian and I get one of eigenvalues. If I repeat the process several times, for every time I will get a different eigenvalue or the same, which are not predictable.

So I am confused with this result. If the total energy of the particle is conserved, then how I get the different values?

• Not if we want to measure, but if we want to predict. We don't measure by applying operators, but by using measurement apparatuses. Feb 27 '15 at 3:28

If the particle is in an eigenstate of the Hamiltonian, you will get the same energy eigenvalue every time.

We know that energy is conserved because the Hamiltonian obviously commutes with itself. The only time it is not conserved is if the Hamiltonian depends explicitly on time.

The equation you wrote $$H|\psi\rangle=E|\psi\rangle$$ is the time-independent Schrödinger equation for an energy eigenstate. I.e., the state you are considering is already an eigenstate of the Hamiltonian with energy $E$. Therefore, as mentioned in the other answer, its time evolution is a simple phase factor, and you will always measure $E$ if you keep on just measuring the energy.

I think what you want to know about is a general wavefunction (call it $\Psi$) that is not necessarily an eigenstate of the Hamiltonian. In this case we still have the time dependent Schrödinger equation: $$H|\Psi\rangle=i\frac{\partial |\Psi\rangle}{\partial t}\;.$$ In this case the expectation value of the energy is $$\langle\Psi|H|\Psi\rangle\;,$$ and the probability that you measure some energy $E_\alpha$ is given by $$\left|\langle\alpha|\Psi\rangle\right|^2\;,$$ where $$H|\alpha\rangle=E_\alpha|\alpha\rangle$$

Once you actually do make a measurement and find the value $E_\alpha$, the state collapses from $|\Psi\rangle$ to $|\alpha\rangle$, and any subsequent measurement of the energy is guaranteed to be $E_\alpha$.

Once you measure an eigenvalue, you collapse into a particular eigenstate. Let's call that state $| \alpha \rangle$, such that $\hat{H} | \alpha \rangle = E_\alpha | \alpha \rangle$. Now we use the time-dependent Schrodinger equation: $$i \hbar \frac{\mathrm{d}}{\mathrm{d} t} | \alpha (t) \rangle = H | \alpha \rangle \\ = E_\alpha | \alpha \rangle \\ \implies \frac{\mathrm{d}}{\mathrm{d} t} | \alpha (t) \rangle = -i \frac{E_\alpha}{\hbar} | \alpha \rangle$$

The solution to this differential equation is the time-dependent ket: $$| \alpha(t) \rangle = e^{-i \frac{E_\alpha}{\hbar} t} | \alpha \rangle$$

So you can see that while this ket has a time-dependent phase on it, it's always in the same energy eigenstate as you first measured, and will stay there forever.

You say "Now let a particle move under a central force field where the total energy is constant. Now I want to measure the energy of the particle so I have to apply Hamiltonian and I get one of eigenvalues."

By applying the Hamiltonian to the state of the particle you can predict its energy, i.e. on the paper, not measure. We measure in the lab, with suitable apparatuses.

The state of a particle is not necessarily an energy eigenstate. If you say that "If I repeat the process several times, for every time I will get a different eigenvalue or the same," that means that the particle state, i.e. the wave-function, is not an eigenstate of the Hamiltonian.

In short, the fact that the particle is in a central force field, doesn't guarantee that the particle is an eigenstate of the Hamiltonian. What is conserved in this field, in your case, is the average value of the energy. Indeed, denoting the Hamiltonian by $\hat H$, and the wave-function by $|\psi\rangle = \sum_i C_i |\psi_i\rangle$ where $C_i$ are constants, we get

$$\langle E \rangle = \sum_{j,i} C^*_j C_i \langle \psi_j|\hat H|\psi_i \rangle$$

$$= \sum_{j,i} C^*_j C_i E_i\langle \psi_j|\psi_i \rangle$$

$$= \sum_{j,i} C^*_j C_i E_i \delta_{i,j} = \sum_i |C_i|^2 E_i . \tag{i}$$

So, since the average energy is a constant,

$$\frac {d}{dt} \langle E \rangle = 0. \tag{ii}$$

About the fact that at each measurement you obtained one of the values $E_i$ was explained by the user "hft", it is the effect of the so-called "collapse".

• perhaps you could've written the time evolved state $U(t)|\psi\rangle$ instead, thus to provide a self-contained proof of the fact that energy is indeed conserved, in accordance with Ehrenfest's theorem. Feb 27 '15 at 10:46
• @Phoenix87 1) I chose the simplest way to explain. 2) Ehrenfest has a couple of theorems. But I recall that they have to do with analogy with classical situations. I don't recall smth. in the direction you mention. So, would you tell me to which theorem you refer? Or, alternatively, would you post an answer of your own? It's a good idea to give the OP different variants of explanation. Feb 27 '15 at 10:55
• @Phoenix87 I think that the best answer were to tell the user that the operator of average commutes with the Hamiltonian. But I wasn't sufficiently sure. Is this commutation correct? Feb 27 '15 at 11:00
• what's the operator of average? Feb 27 '15 at 11:05
• @Phoenix87 $\sum V_i \hat P_i$ where $V_i$ is the eigenvalue of the operator $\hat V$, in our case the Hamiltonian, and $\hat P_i$ is the projector on the eigenstate nr. $i$ of $\hat V$. Feb 27 '15 at 11:12