# Is it possible to calculate the momentum of an electron precisely?

Which of the following is not correct with regard to the uncertainty principle?

1. An electron in an atom cannot be defined by a well-defined orbit.
2. The momentum of an electron cannot be measured exactly (correct answer)
3. A harmonic oscillator possesses a zero-point energy
4. Measurement of one variable in an atomic system can affect subsequent measurements of certain other variables

To calculate an electron's momentum precisely, it must have a totally undeterminable position! Since our apparatus have some definite size, you have to confine it in some definite space.

$$\Delta x \Delta p \geq \frac{\hbar}{2}$$

Since we want to make $$\Delta p$$ to approach zero, $$\Delta x$$ must approach infinity, which is not possible. So we can't measure the electron's precise momentum. So option (2) shouldn't be the answer.

Moreover, as far as I know, Heisenberg's uncertainty principle restricts us from measuring two quantities(which doesn't commute each other) simultaneously. We can measure them subsequently (one after another). So option (4) should be the answer.

P.S. I know that it is not necessary that we will get the same measurement on repeating the experiment, we can get any of the eigenvalues of that observable. But this is not related to Heisenberg's uncertainty principle. It's related to observer's effect.

• As orthocresol's answer suggests in his remarks on option 2, the uncertainty in position can approach infinity. We can see that this must be true by looking at the case of the free particle, which has a momentum eigenfunction as its wavefunction. – Tyberius Feb 6 '18 at 19:33
• @Tyberius Just nitpicking: I'd say "can have" instead of "has" - after all, the wavefunction of the free particle can well be a superposition of momentum eigenstates. The momentum eigenstates $\psi_i(x)$ are the stationary states which satisfy the time-independent Schrodinger equation Hφ = Eφ, but any linear combination $$\Psi(x,t) = \sum_i c_i\psi_i(x)\exp(-\mathrm i E_it/\hbar)$$ will satisfy the time-dependent Schrodinger equation $\mathrm i\hbar (\partial \Psi/\partial t) = -(\hbar^2/2m)(\partial^2\Psi/\partial x^2)$. Of course, this does not affect your actual argument in any way. – orthocresol Feb 6 '18 at 19:50

To preface this answer I suggest reading a proper QM text which should address many of your issues. Griffiths' Introduction to Quantum Mechanics served me well but you may prefer any of the other excellent texts out there.

### Option 4

This depends on your interpretation of quantum mechanics, but in the usual Copenhagen interpretation, the measurement of an observable causes the wavefunction of the system to collapse into the corresponding eigenstate. To give a more concrete example, if we have energy eigenstates $|1\rangle, |2\rangle, |3\rangle, \cdots$ with corresponding eigenvalues $E_1, E_2, E_3, \cdots$ then the wavefunction may be written as a linear combination of energy eigenstates. [This is guaranteed to be true because the Hamiltonian is a hermitian operator, and hermitian operators have complete sets of eigenstates.]

$$|\psi\rangle = c_1|1\rangle + c_2|2\rangle + c_3|3\rangle + \cdots$$

When you measure the energy of this system, the chance of getting the result $E_i$ is equal to $|c_i|^2$. Let's say you measure the energy to be $E_1$. Immediately after the measurement, the wavefunction of the system collapses to $|1\rangle$:

$$|\psi\rangle \overset{\text{collapse}}{\longrightarrow} |1\rangle$$

Now let's say you want to measure some other observable $A$. Again, you can express the original wavefunction $|\psi\rangle$ as a linear combination of eigenstates of $A$.

$$|\psi\rangle = a_1|A_1\rangle + a_2|A_2\rangle + a_3|A_3\rangle + \cdots$$

and the probability of measuring the eigenvalue $A_i$ would be $|a_i|^2$. However, the new wavefunction after collapse will have different coefficients in this expansion:

$$|1\rangle = a_1'|A_1\rangle + a_2'|A_2\rangle + a_3'|A_3\rangle + \cdots; \qquad a_i \neq a_i' \text{ (in general)}$$

and hence the probability of measuring the eigenvalue $A_i$ is now $|a_i'|^2$, which is in general, not the same as $|a_i|^2$.

In the absolute simplest case, we can consider measuring the energy again immediately after measuring the energy and obtaining the value $E_1$. Since the measurement caused the wavefunction of the system to collapse into $|1\rangle$, a second measurement done immediately after this will inevitably return the same value $E_1$ again.

Hence option 4 is true, which means it is not the answer you are looking for. Measurement of an observable causes wavefunction collapse, which in turn causes the probabilities of measuring other eigenvalues to change.

### Option 2

First, we note that for a measurement of an observable to have no uncertainty, the system must be in an eigenstate of the relevant operator. This follows from the definition of uncertainty

$$\sigma_A^2 = \left\langle (A - \langle A\rangle)^2 \right\rangle$$

If the wavefunction of an electron is an eigenstate of the momentum operator $\hat{p}$, then a measurement of the momentum will always return the same value. In real life, if we carry out this measurement, will there be some uncertainty due to the precision of the machine? Yes, of course. Is it practical to set up an electron in a momentum eigenstate? Probably not. (By the way, your electron cannot really be constrained within your apparatus, because of quantum tunnelling. That is, unless the walls of your apparatus are an infinitely thick infinite potential energy barrier.)

But, that question is not asking you how you carry something out in the lab. The question is asking purely about theoretical considerations - and there is no fundamental theoretical problem with having an electron in a momentum eigenstate and measuring its momentum to arbitrary precision without changing its state. You are correct that its position will be indeterminate, but that is not a problem in the context of this question.

For interest, an "ideal measurement" of momentum is described in Shankar's Principles of Quantum Mechanics, 2nd ed. (p 123):

Consider a particle in a momentum eigenstate $|p\rangle$. The postulate tells us that if the momentum in this state is measured we are assured a result $p$, and that the state will be the same after the measurement (since $|\psi\rangle = |p\rangle$ is already an eigenstate of the operator $P$ in question). One way to measure the momentum of the particle is by Compton scattering, in which a photon of definite momentum bounces off the particle.

Let us assume the particle is forced to move along the $x$-axis and that we send in a right-moving photon of energy $\hbar\omega$ that bounces off the particle and returns as a left-moving photon of energy $\hbar\omega'$. (How do we know what the photon energies are? We assume we have atoms that are known to emit and absorb photons of any given energy.) Using momentum and energy conservation:

\begin{align} cp' &= cp + \hbar(\omega + \omega') \\ E' &= E + \hbar(\omega - \omega') \end{align}

it is now possible from this data to reconstruct the initial and final momenta of the particle:

\begin{align} cp &= -\frac{(\hbar\omega + \hbar\omega')}{2} + \sqrt{1 + \frac{m^2c^4}{\hbar^2\omega\omega'}}\frac{\hbar\omega - \hbar\omega'}{2} \\ cp' &= \frac{(\hbar\omega + \hbar\omega')}{2} + \sqrt{1 + \frac{m^2c^4}{\hbar^2\omega\omega'}}\frac{\hbar\omega - \hbar\omega'}{2} \\ \end{align}

Since the photon always loses energy to the particle (as is clear in the particle rest frame) $\omega' < \omega$ and by sending $\omega \to 0$, we can make the change in momentum $p' - p$ arbitrarily small.

• I have a little doubt regarding option 4 ! Question is which of the following is not correct regarding the "Heisenberg's Uncertainty principle" , how is it connected to uncertainty principle. – Aditya Shrivastava Feb 7 '18 at 9:50
• I would agree that it is not directly related to the uncertainty principle. – orthocresol Feb 7 '18 at 21:29