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Dimensional Analysis Dimensional analysis Calvin W. Johnson September 9, 2014 1 Dimensional analysis Dimensional analysis is a powerful and important tool in calculations. In fact, you should make it a habit to automatically use dimensional analysis to check not only your answers but also your equations. Later on we will see we can even use it to check integrals! Let me give an example: the Schr¨odingerequation for free particles (with no potential energy). In one dimensional systems, the time-dependent Schr¨odinger equation is ¯h2 @2 @ − = i¯h (1) 2m @x2 @t while the time-independent equation is ¯h2 @2 − = E : (2) 2m @x2 In the above, x is position, t is time, m is mass, and E energy.h ¯ is Planck's constant (actually it is h=2π) which has dimensions of action = momentum × length = energy × time. Let's dimensionally analyze the second equation first; we can do that even without fully understanding it. Because both equations are linear in the wave- function (meaning appears at most first-order, no 2 or 3) , the dimension of is not relevant here; in other words, the dimensions of cancel out. (Below I will discuss how we can find the dimension of .) The right-hand side of the equation has (leaving off the dimension of ) the dimension of energy. The left hand side has (again leaving off ) dimensions [¯h]2 [action]2 = (3) [m][x]2 mass · length2 where the square brackets [:::] signify \dimension of." The next step will likely take some trial and error; don't feel bad if it takes you a few, or many, tries. I'm simply not writing down for you all the wrong steps I took. The fundamental dimensions (units, see below for more on the distinction) are mass, length, and time, and others such as energy, momentum, and action, 1 are composites of these. But we have a hint: we have an equation. We do not expect something like time = length; mass that would be illogical. So we take action = energy × time and substitute it in the left-hand side: [action]2 energy2time2 = mass · length2 mass · length2 (Again, if you first try action = momentum × length there is nothing wrong or bad about that, it may or may not lead you to the correct answer as quickly.) At first this doesn't look helpful, but what we do is take out one of the energies [action]2 energytime2 = energy × mass · length2 mass · length2 and then we remember, I hope, that mass × length2 energy = time2 (which I remember from kinetic energy = (1=2)mv2 and velocity v = length=time). We then see that [action]2 energytime2 mass · length2 time2 = energy× = energy× × = energy; mass · length2 mass · length2 time2 mass · length2 (4) in other words, we get dimensions of energy on both the left-hand and right-hand sides, so both sides of the equation agree in dimensions. Hurray! What about the time-dependent Schr¨odingerequation? The left-hand side remains the same. The right-hand side, ignoring as usual, is [¯h] (5) [t] (the unit imaginary, i, is dimensionless of course). This is easier, as [¯h] = action = energy × time and we get [¯h] energy × time = = energy (6) [t] time so both side match once again. If we take the Schr¨odinger equation including potential energy, we have (for the time-independent equation) ¯h2 @2 − + V (x) = E : (7) 2m @x2 2 so the potential energy, V , must have dimensions of energy, as it does. Let's explore a little further with potential energy. I'll take a couple of cases. The harmonic oscillator. This is a fundamental problem in quantum me- chanics as well as in classical mechanics and during this course we will study it in detail. Here V (x) = (1=2)m!2x2, where ! is the frequency of the oscillator and hence has dimensions of (time)−1. Although we expect the dimensions to work out to be energy, we need to check! So: 1 1 m!2x2 = mass · · length2 (8) 2 time2 which are the dimensions for energy (again, remember kinetic energy). So far so good. We can do something more, however. Ultimately we want to solve ¯h2 @2 1 − + m!2x2 = E : 2m @x2 2 Now the energy E, which will be a number (in fact, an `eigenvalue' as we will see), so it cannot depend upon x, but it will, of course, have to have dimensions of energy. When we solve this particular equation, we will find that the solutions with have E proportional to ¯h!. Does this make sense? Yes: 1 [¯h!] = action · frequency = energy · time · = energy: (9) time Good! In fact, the energy has come out in terms if ¯h, !, and m because those are the only dimensionful constants in the problem. It just so happens here that the energy does not depend on mass m. The linear potential. Suppose V (x) = kx. What must the dimensions of k be? Well, we want energy = [k] × length; (10) so k has the ungainly dimensions of energy/length. (It's okay if you break it down to, say, mass-length/time2 but I don't find this additionally helpful). When we solve the equation ¯h2 @2 − + kx = E : 2m @x2 the only dimensionful constants are ¯h, m, and k, and E must be proportional to some combination of them. It takes some trial and error to figure it out. It helps to see that energy/mass = (length/time)2. But [¯h] = energy · time and [k] = energy=length, then time [¯hk] = energy2 × (11) length 3 and squaring this (yes, I know it seems crazy, but you have to try crazy things once in a while) time2 [(¯hk)2] = energy4 × : length2 Hmm. If we divide by mass m, energy time2 [(¯hk)2=m] = energy3 × × (12) mass length2 and use energy/mass = (length/time)2 then we have [(¯hk)2=m] = energy3 (13) So finally 1=3 ¯h2k2 (14) m has dimensions of energy. I realize this kind of manipulation is far from trivial; you have to be willing to try different combinations and be persistent. The most important messages you should get from these notes is that you have to pay attention to dimensions (and in fact, dimension are more important than actual numbers, because if the dimensions are wrong, the numbers don't matter). As a matter of reflex you should try to always confirm the dimensionality of an answer. We will get some practice in this when we solve the harmonic oscillator and the hydrogen atom. You try: confirm directly that expression (14) has dimensions of energy. You try (1): If we take the potential energy V (x) = λx4, what are the dimensions of λ? (See below for solution) You try (2): For the potential energy V (x) = λx4, which of the following combinations of constants have dimensions of energy? (it is possible that some, all, or none are). h¯ 3=2 (a) λ hm¯ 2=3 (b) λ2 2=3 h¯4λ2 (c) m 1=3 h¯4λ (d) m2 (answer will be given in class) Note Although sometimes units and dimensions are used interchangeably (and even I do sometimes as well, which is lazy of me), technically units refer to a specific value, such as seconds, meters, kilograms, which dimensions refers to time, length, mass, respectively. I will not nitpick on this distinction but you should be aware of it. 4 2 Answers to selected problems (1) [λ] = energy/length4. 5.
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