I'm a beginner in quantum mechanics, and I'm pretty much confused with the momentum of electron in hydrogen atom.

The wave function of the electron in 1s orbital of hydrogen atom is $\Psi_{1s} = \frac{1}{{a_0}^{\frac{3}{2}}\sqrt\pi}.e^{\frac{-r}{a_0}}$, $a_0$ is the Bohr radius. The momentum operator is $\hat p = -i \hbar \nabla $. The momentum of the electron can be found using the equation, $\hat p \Psi = \bar p \Psi$.

Which yields $$\bar p = \frac{i \hbar}{ra_0} (x \hat i + y \hat j + z \hat k)$$which is imaginary, which doesn't make any sense to me.

My questions are:

Is the momentum really imaginary? If yes, is it because of the imaginary momentum operator? If no, what is the correct way to find the momentum?

  • 6
    $\begingroup$ [The expectation value of the momentum (and the position) is zero - not what you have calculated, which is sort of like half the integral you need but not in spherical polar coordinates. It also has to be an observable, hence not imaginary.] Why the operator is imaginary is a result of (at least) Bohr's quantization constraint and the uncertainly relation in quantum mechanics - you can accept this as faith or work through the math. I highly recommend Part II, Chapter 9 of David Bohm's ${\it Quantum\ Theory}$, Dover (1989). $\endgroup$
    – Todd Minehardt
    Jan 3 at 15:57
  • 1
    $\begingroup$ The Wikipedia page also gives a simple rationale in terms of a generalized plane wave. $\endgroup$
    – Paul
    Jan 3 at 17:45

2 Answers 2


Why is the momentum operator imaginary?

The simplest explanation hinges on the fact that observables are represented by Hermitian operators in quantum mechanics. Once we accept this, then we can show that the momentum operator $\hat{p} = -\mathrm{i}\hbar\nabla$ is Hermitian precisely because of the factor of $\mathrm{i}$. We need to show that for any two states $f$ and $g$, $\langle f | \hat{p} | g\rangle = \langle g | \hat{p} | f\rangle^*$. For simplicity we work in one dimension, such that $\hat{p} = -\mathrm{i}\hbar(\mathrm{d}/\mathrm{d}x)$. The proof is taken from Binney and Skinner's The Physics of Quantum Mechanics (p 36), although truthfully it is quite straightforward, the main idea being integration by parts:

$$\begin{align} \langle f | \hat{p} | g\rangle &= -\mathrm{i}\hbar \int_{-\infty}^{\infty} f^* \frac{\mathrm{d}g}{\mathrm{d}x} \,\mathrm{d}x \\[8pt] &= -\mathrm{i}\hbar \left\{ \left[f^*g\right]_{-\infty}^{\infty} - \int_{-\infty}^{\infty} g \frac{\mathrm{d}f^*}{\mathrm{d}x} \,\mathrm{d}x \right\} \\[8pt] &= \mathrm{i}\hbar \int_{-\infty}^{\infty} g \frac{\mathrm{d}f^*}{\mathrm{d}x} \,\mathrm{d}x \\[8pt] &= \left( -\mathrm{i}\hbar \int_{-\infty}^{\infty} g^* \frac{\mathrm{d}f}{\mathrm{d}x} \,\mathrm{d}x \right)^* \\[8pt] &= \langle g | \hat{p} | f \rangle ^* \end{align}$$

(Well-behaved functions vanish as $x \to \pm \infty$, hence the term $[f^*g]_{-\infty}^{\infty}$ goes to zero.) In a similar way, it can be shown that without the the operator $\mathrm{d}/\mathrm{d}x$ itself is not Hermitian, and thus cannot be an operator corresponding to an observable quantity in QM.

The momentum of the electron can be found using the equation, $\hat p \Psi = \bar p \Psi$

This is an eigenvalue equation for the momentum, but it does not actually give you the value of the momentum. When you apply an operator to a state, you get out another state, and there's nothing that forbids this second state from being complex / imaginary. It's only the final expectation value at the end that has to be real, and this expectation value is given by $\langle \psi | \hat{p} | \psi \rangle$.

Is the momentum really imaginary?

It turns out that the factor of $\mathrm{i}$ doesn't matter, because if you calculate this correctly, then the momentum turns out to be zero. This makes sense, because all directions in the hydrogen atom are equal, and it can't be that the electron has a net momentum in some direction.

Calculating it correctly, however, is a slightly tricker topic than expected. In this case, the momentum operator $\hat{p}$ is a vector operator ($\equiv (\hat{p}_x, \hat{p}_y, \hat{p}_z)$), and consequently when you integrate it you need to be very careful to not mix up vector and scalar quantities. If you wish to do this in spherical coordinates, I first suggest going to Physics Stack Exchange and searching for questions about "momentum of hydrogen atom". You will find lots of similar questions asking about the factor of $\mathrm{i}$ and how it doesn't make sense.

Personally, I think a more direct, and slightly easier approach, is to simply integrate one (Cartesian) component of the vector at a time. Thus, we can try to evaluate $\langle \psi | \hat{p}_x | \psi \rangle$. For simplicity, I will drop the normalisation constant and use $\psi = \exp{(-r/a_0)}$. Also, note that

$$\begin{align} \frac{\mathrm{d}r}{\mathrm{d}x} &= \frac{\mathrm{d}}{\mathrm{d}x} (x^2 + y^2 + z^2)^{1/2} \\ &= \frac{1}{2}(x^2 + y^2 + z^2)^{-1/2} \cdot (2x) \\ &= \frac{x}{r} \end{align}$$

Thus, we have that

$$\begin{align} \langle \psi | \hat{p}_x | \psi \rangle &= -\mathrm{i}\hbar \int_{-\infty}^{\infty} \int_{-\infty}^{\infty} \int_{-\infty}^{\infty} \exp\left(-\frac{r}{a_0}\right) \frac{\mathrm{d}}{\mathrm{d}x}\left[\exp\left(-\frac{r}{a_0}\right)\right] \,\mathrm{d}x\,\mathrm{d}y\,\mathrm{d}z \\[8pt] &= -\mathrm{i}\hbar \iiint \exp\left(-\frac{r}{a_0}\right) \frac{\mathrm{d}}{\mathrm{d}r}\left[\exp\left(-\frac{r}{a_0}\right)\right]\frac{\mathrm{d}r}{\mathrm{d}x} \,\mathrm{d}x\,\mathrm{d}y\,\mathrm{d}z \\[8pt] &= -\mathrm{i}\hbar \iiint \exp\left(-\frac{r}{a_0}\right) \left(-\frac{1}{a_0}\right)\exp\left(-\frac{r}{a_0}\right)\frac{x}{r} \,\mathrm{d}x\,\mathrm{d}y\,\mathrm{d}z \\[8pt] &= \frac{\mathrm{i}\hbar}{a_0} \iiint \exp\left(-\frac{2r}{a_0}\right) \frac{x}{r} \,\mathrm{d}x\,\mathrm{d}y\,\mathrm{d}z \\[8pt] \end{align}$$

At this point, it's easiest to bring in symmetry arguments. Notice that $r = \sqrt{x^2 + y^2 + z^2}$ is an even function of $x$, $y$, and $z$. Thus, the integrand is odd with respect to $x$ (because of the extra factor of $x$), but even with respect to $y$ and $z$.* The entire integral therefore goes to zero.

The integrals for $\hat{p}_y$ and $\hat{p}_z$ proceed in exactly the same way.

* Although it also diverges at the origin. I'd be very glad if someone could tell me how to rigorously resolve this.

  • $\begingroup$ When you say "simplest" I'm not sure I agree! $\endgroup$
    – matt_black
    Jan 4 at 12:22
  • $\begingroup$ @matt_black, the least complicated? :-) $\endgroup$
    – orthocresol
    Jan 4 at 13:31
  • $\begingroup$ It doesn't give any intuition about the "why" and requires heavy math (which might be the best rigorous answer but doesn't help those without the math knowledge). I might try a simpler answer with an analogy. $\endgroup$
    – matt_black
    Jan 4 at 13:42
  • $\begingroup$ @matt_black, I don't really think integration by parts is all that heavy: I expect that most people studying QM should have seen it before. However, I agree it's not intuitive. I've never seen this explained without the maths before, though, so I'd be pretty keen to see what you have. $\endgroup$
    – orthocresol
    Jan 4 at 13:44
  • $\begingroup$ "I'd be very glad if someone could tell me how to rigorously resolve this." - Spherical coordinates? $\endgroup$
    – Antimon
    Jan 5 at 2:04

Because periodic waves can be represented using exponential functions using imaginary arguments

I'm going to attempt a non-rigorous explanation that might give some intuition as to why imaginary numbers appear in the mathematics of quantum stuff. Most of this doesn't require a knowledge of what a Hermitian operator is or about vector spaces or eigenvalues. You need to understand these if you want to really want to be a quantum mechanic, but the basic intuition can come from a single very famous math equation and the much simpler math used to study periodic signals in electronic circuits.

The first thing you need to understand is that electron orbitals are described by wave functions. This is an essential foundation of QM derived from de Broglie's insight that particles can be described as waves. This implies that solutions for possible orbitals of an electron orbiting a hydrogen atom must be quantised (like the possible vibrations of a violin string): waves that don't meet certain strict requirements will cancel out just as only certain waves on a vibrating string are possible. I'm simplifying a lot (not least because orbitals are 3D unlike violin strings and the "waves" are spherical harmonics (don't ask, but you will be familiar with the pictures of possible orbital shapes)).

If you accept this highly simplified picture (and given how fundamental it is to everything quantum related you really have to) the question arises what sort of math can be used to describe the wave functions?

And here we have to look at one of the most famous equations in mathematics: Euler's function:

$e^{i {\theta}} = \cos(\theta) + i \sin(\theta) $

The point here is that a periodic function can be described by an exponential function with an imaginary exponent (one containing i or the $ \sqrt {-1} $ ). And using this identity makes manipulating the math of periodic functions a great deal easier. This insight is not confined to quantum chemistry, it is essential to electrical engineering where it provides a compact way to describe the behaviour of electrical circuits involving anything more complicated than pure resistance when AC is involved. In most circuits the voltage and current vary over time in a periodic way. But the voltage and current are usually not in phase because of the inductance and capacitance of the circuit. The exponential captures the nuances of this periodic behaviour in a simple mathematical way.

Ultimately the exponential form of the math of electrical circuits captures the periodic solutions to differential equations describing the components of the circuit and the result is periodic functions describing voltage and current.

Chemically relevant wavefunctions are far more complex (not least they are 3D as I mentioned before). But they represent valid solutions to the differential equations describing the electron wave function and those solutions are periodic waves in 3D space.

So the basic leap is that, once you have accepted electrons are waves, the math describing them will involve periodic functions and the best way to describe those is exponential functions with imaginary arguments because, Euler's formula.

After that leap, QM gets complicated and you start having to grasp concepts like Hilbert spaces, Hermitian operators and Eigenvalues. But the essential leap is to realise the implications of Euler's formula: wave behaviour can be captured by exponential functions with imaginary inputs.

  • $\begingroup$ I'm afraid I don't really follow. You've suggested why QM as an entire field requires complex numbers; but that doesn't necessarily imply that the momentum operator $\hat{p}$ must have a factor of i in it, which is what OP's specific question is about (or at least, that's the way I understood it). $\endgroup$
    – orthocresol
    Jan 4 at 17:48
  • $\begingroup$ @orthocresol I'm not trying to be rigorous, just to give an intuition as to why anything with wave equations is going to involve complex numbers. Your argument is that the math doesn't work if you don't use i. My argument is a very general illustration illustrating the intuition that i is required to give wavefunction like solutions when solving differential equations. QM is a complex example of this but it all comes back to Euler: if you want waves, you need factors of i in the math. $\endgroup$
    – matt_black
    Jan 4 at 18:28
  • $\begingroup$ Hmm, but you can get wavefunctions even if the operator itself is real: the Hamiltonian, for example, is real and you can solve $\hat{H}\psi = E\psi$ to get a set of wavefunctions (e.g. the orbitals). Imho, the solution to this has to somehow focus on momentum specifically, not wavefunctions in general. Perhaps, the link I'm looking for is that the eigenfunctions of momentum are plane waves, which (as your answer explains) are necessarily complex (or formally $\exp(-\mathrm{i}\vec{k}\cdot\vec{x})$), explaining the factor of i in the momentum operator. Either way, thanks for posting! $\endgroup$
    – orthocresol
    Jan 4 at 18:45
  • $\begingroup$ @orthocresol I'm not trying to get into the deeper quantum explanation of why those specific momentum equations have to have an i. I'm just trying to create a very simple illustration (eg the comparison with much simpler electrical signals) where the math is much simpler than QM but where i makes wave solutions work. $\endgroup$
    – matt_black
    Jan 4 at 18:50
  • $\begingroup$ Yeah, I do get what you're trying to say. I just don't think that was the question OP was asking. But it's OK; I've posted what I think is an answer, so that's enough from me! I don't want to be, or come off as, confrontational here. $\endgroup$
    – orthocresol
    Jan 4 at 18:58

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