Wave function units

Here’s a question I started thinking about this morning. What are the units on the wave function. Its not as simple as you might think.

If we start with the Schrodinger Equation i\hbar \frac{d}{dt}\mid\psi\rangle = \hat{H}\mid\psi\rangle then we can see that \mid\psi\rangle has no units since the product \hbar\frac{d}{dt} has units of energy as does the Hamiltonian. You may say, “yes but that does not preclude \mid\psi\rangle from having units since it is on both sides of the equation.” That’s true certainly. So the Schrodinger equation does not determine the units of the wave function. But we also know that \langle\psi\mid\psi\rangle=1. So it looks as though \mid\psi\rangle is unitless.

But if we look at the wave function in position space, \psi (\vec{x},t) we know that the integral of the wave function squared over all space is one, so

\int \mid \psi(\vec{x},t)\mid^2 dV=1

where dV is the volume element for the configuration space under consideration. Then it must be the case that \psi(\vec{x},t) has units of 1/V^{1/2}

So the question is, how did \mid\psi\rangle get units? We know that \psi(\vec{x},t)=\langle\vec{x}\mid\psi\rangle so presumably the position basis carries units? This might be obvious, but its not to me. Our usual basis elements in physics are unitless arrows along axes that have a scale with units. Certainly \hat{i},\hat{j}, \hat{k} don’t have units. Considering the fact that \langle x\mid x'\rangle =\delta(x-x') and that the delta function have units of reciprocal volume (so it integrates to 1). It does seem to make sense that while the “wave-ket” is unitless, the wave function \psi(x,t) has units, given to it by projection onto the position basis. The product of these components, \psi(x,t) and the basis elements \mid x\rangle give back the wave-ket,

\mid\psi\rangle=\int\psi(x,t)\mid x\rangle dx

and the units all work.

NOTES: I’ve neglected to write \mid\psi(t)\rangle though I should have since the ket is time dependent regardless of its basis representation. I also think its a profitable discussion since many lists of ‘misconceptions’ in quantum include the units of a wave function, but perhaps we should be careful of what precisely we mean but wave function? This has no doubt been ‘made careful’ by many individuals before me though.


About because0fbeauty

Fascinated by the way mathematics and physics interact, captivated by visual and tactile mathematics and hoping to become a better expositor of these things is why I blog...occasionally...when I remember.
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5 Responses to Wave function units

  1. As a follow up question I’ll post on later, what would the units be on the momentum representation of momentum?

  2. Why did you assume that \langle\psi\rvert and \lvert\psi\rangle have the same units?

    • That’s a valid question/criticism. Because I couldn’t see any way in which it would change. Although as I think about it I suppose the map from kets to bras could, itself, carry units. That would alter the conclusion that the wave-ket or wave-bra doesn’t carry units. Though I think it still follows that the wave function and the wave-ket have different units. What are your thoughts? Thanks for calling me on that.

  3. hawzhin taha says:

    but we can not find by the probability density so by this way we can find it esely which equal to 1/m2

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