# Consistent calculation of the total momentum in quantum physics

When we are considering a one-dimensional case in quantum physics, we can calculate the momentum by solving the eigenvalue equation:

$$-i \hbar \frac{d}{d x} \psi(x, t) = p_x \psi(x, t).$$

If we would like to extend our problem to 2D or 3D we would just introduce momentum operators for the new dimensions, which at the end would give us a set of operators:

$$\hat{p}_x = -i \hbar \frac{\partial}{\partial x},$$ $$\hat{p}_y = -i \hbar \frac{\partial}{\partial y},$$ $$\hat{p}_z = -i \hbar \frac{\partial}{\partial z},$$

which can be written in a compact form:

$$\hat{p} = -i \hbar \nabla.$$

I can't grasp the connection between action of such an operator of total momentum on the wavefunction and the eigenvalue equation. I suppose we can write down a general formula for a wavefunction in the 3D case as $$\psi(\bar{r}, t) = A e^{i \bar{k} \cdot \bar{r} - \omega t}$$, which gives a scalar (because of the dot product between $$\bar{k}$$ and $$\bar{r}$$). It makes sense for me, because for a given parameters $$\bar{r}$$ and $$t$$ it should give a probability amplitude.

But the action of $$\nabla$$ on a scalar field gives a vector field (or a gradient). So for an equation

$$-i \hbar \nabla \psi(\bar{r}, t) = p \psi(\bar{r}, t),$$

on the left hand side we get a vector, but on the right hand side we get a scalar?

It would make sense for me to apply $$\hat{p}_x$$, $$\hat{p}_y$$ and $$\hat{p}_z$$ to the wavefunction separately and put the results into a vector getting $$\bar{p} = (p_x, p_y, p_z)$$, but I don't quite like it.

At the same time momentum is represented by a vector and energy by a scalar. So e.g. in a time dependent Schrödinger equation

$$i\hbar \frac{\partial}{\partial t} \psi(\bar{r}, t) = \bigg [ -\frac{\hbar^2}{2m} \nabla^2 + V(\bar{r}, t) \bigg ] \psi(\bar{r}, t)$$

we have $$\nabla^2 = \frac{\partial^2}{\partial x^2} + \frac{\partial^2}{\partial y^2} + \frac{\partial^2}{\partial z^2}$$, which is a scalar operator. So there is much sense in getting vector while calculating momentum and scalar while calculating energy, but I can't make it strict.

I might be wrong in many places, so some clarification would be fantastic.

• If you put all those $\hat{p}_i$ together to $\nabla$, your eigenvalue also has to be a vector of the separate eigenvalues. – Jan2103 Aug 22 at 11:33
• Is there some more strict way of writing this down, than saying that $\nabla$ just enforces $p$ on the right hand side to be a vector (that's the only way, in which I can explain this right now)? – brzepkowski Aug 22 at 12:02
• You say that you can combine the three momentum operators in a compact form, so what bothers you about saying the same for the eigenvalues? I mean, those two combinations go hand in hand. – Jan2103 Aug 22 at 12:18
• Ok, that makes sense. Thank you! – brzepkowski Aug 22 at 12:19
• Is this $$-i\hbar\nabla\psi(\mathbf{r},t)=-i\hbar\left(\hat{\mathbf{x}}\frac{\partial}{\partial x}+\hat{\mathbf{y}}\frac{\partial}{\partial y}+\hat{\mathbf{z}}\frac{\partial}{\partial z}\right)\psi(\mathbf{r},t)=\left(\hat{\mathbf{x}}p_x+\hat{\mathbf{y}}p_y+\hat{\mathbf{z}}p_z\right)\psi(\mathbf{r},t)=\mathbf{p}\psi(\mathbf{r},t)$$ not what you're looking for? – Alfred Centauri Aug 22 at 12:20

$$-i\hbar\nabla\psi(\mathbf{r},t)=-i\hbar\left(\hat{\mathbf{x}}\frac{\partial}{\partial x}+\hat{\mathbf{y}}\frac{\partial}{\partial y}+\hat{\mathbf{z}}\frac{\partial}{\partial z}\right)\psi(\mathbf{r},t)=\left(\hat{\mathbf{x}}p_x+\hat{\mathbf{y}}p_y+\hat{\mathbf{z}}p_z\right)\psi(\mathbf{r},t)=\mathbf{p}\psi(\mathbf{r},t)$$