After a while, you just get tired. An honest science blogger can only handle so much science jargon thrown around without meaning, only a limited amount of Choprawoo and quantum flapdoodle. How long can anyone with integrity, curiosity and a dose of genuine knowledge endure the trumpeting that, say, the brain’s limited ability to recover after injury is evidence for some quantum spirit? The brain is living flesh, made of living cells: by the same so-called logic, the scabbed knees of childhood are all evidence of quantum skin.

After a while, a deep reserve of psyche cries out, “Enough! If my freedom means aught, I must *stop* responding to these charlatans and move beyond. I must send a message of my own, a message which is not a reaction but an expression unto itself. I must sing the quantum genuine!”

In my case, this means another post on supersymmetric quantum mechanics.

**RECAP**

Last time, we deduced some interesting properties of Hamiltonians which can be factored into operators and adjoints:

[tex]H_1 = A^\dag A,\ H_2 = AA^\dag.[/tex]

We observed that [tex]H_1[/tex] and [tex]H_2[/tex] are *isospectral.* That is, while the forms of their eigenfunctions may be different, the eigenvalues associated with those functions are the same; or, in physical terms, the wavefunctions have different shapes, but the energies match. The only exception is the ground state: if [tex]H_1[/tex] has a zero-energy ground state, then [tex]H_2[/tex] will not. Furthermore, the operator [tex]A[/tex] maps eigenstates of [tex]H_1[/tex] into those of [tex]H_2[/tex], and the operator [tex]A^\dag[/tex] maps eigenstates in the reverse direction:

We can sketch the full set of [tex]H_1[/tex] eigenstates as a ladder with, potentially, an infinite number of rungs. The eigenstates of [tex]H_2[/tex] sit right alongside them; the bottom “rung” is missing, as we mentioned earlier.

That “missing rung” has an important implication: instead of mapping the ground state of [tex]H_1[/tex] into a state of [tex]H_2[/tex], the operator [tex]A[/tex] annihilates it.

[tex]A\ket{\psi_0^{(1)}} = 0.[/tex]

We could, therefore, solve for the wavefunction [tex]\psi_0^{(1)}(x)[/tex] by solving a first-order, linear differential equation — a simpler task than tackling the full, second-order Schrödinger Equation.

Now, we’ll make a qualitative statement which we’ll firm up in a moment. If the Hamiltonian [tex]H_2[/tex] were “roughly similar” to [tex]H_1[/tex], then we should be able to “pull the same trick again,” writing it as

[tex]H_2 = A_2^\dag A_2,[/tex]

and inventing a third Hamiltonian,

[tex]H_3 = A_2 A_2^\dag,[/tex]

which will be isospectral with [tex]H_2[/tex] except for a ground state satisfying

[tex]A_2\ket{\psi_0^{(2)}} = 0.[/tex]

Then, why not, we do the same thing a whole bunch of times over again, building a whole hierarchy of Hamiltonia stretching off beyond the margin of our page.

Note that the “bottom rung” in each “ladder” satisfies a first-order differential equation, and that we can build any state of [tex]H_1[/tex] by stacking up a sequence of [tex]A^\dag[/tex] operators. Start with a state annihilated by some [tex]A_n[/tex] and then apply

[tex]A^\dag_{n-1}A^\dag_{n-2}\cdots A^\dag_{1}[/tex]

to get an eigenstate of [tex]H_1[/tex] there on the left edge. This is reminiscent of the way we solve the harmonic oscillator using creation and annihilation operators, with the difference that instead of building *up,* we’re working in from the *side.*

**DEFINITION OF SHAPE INVARIANCE**

To see this in action, we need to solidify the notion of rough resemblance we invoked above. Suppose we have a potential [tex]V_1[/tex] associated with some system of interest. If we compute the superpotential [tex]W(x)[/tex] and find that the partner potential [tex]V_2[/tex] has a similar shape to [tex]V_1[/tex], we can calculate the energy states of [tex]V_1[/tex] in a remarkably straightforward way. In this context, “similar shape” means that if [tex]V_1[/tex] depends upon a set of parameters, we can reach [tex]V_2[/tex] by substituting different values of those parameters into [tex]V_1[/tex]. More mathematically, two partner potentials [tex]V_1[/tex] and [tex]V_2[/tex] are said to be *shape invariant* if they satisfy

[tex]V_2(x;a_1) = V_1(x;a_2) + R(a_1)[/tex]

where [tex]a_1[/tex] and [tex]a_2[/tex] are sets of parameters, and [tex]a_2[/tex] is uniquely determined by [tex]a_1[/tex] (that is, [tex]a_2 = f(a_1)[/tex] for some function [tex]f[/tex]). The residual term [tex]R(a_1)[/tex] does not depend upon the coordinate [tex]x[/tex]. We shall see that if [tex]H_1[/tex] has a shape-invariant partner potential in [tex]H_2[/tex], the entire spectrum of [tex]H_1[/tex] can be computed through operator manipulations.

In the material which follows, we’ll work in “natural units,” to save ourselves from carrying around excess notational baggage. Our operators will therefore take the form

[tex]A = \partial_x + W(x)[/tex]

and

[tex]A^\dag = -\partial_x + W(x).[/tex]

**HIERARCHY OF HAMILTONIANS**

We are usually interested in examining the physics of a particular system, for which we have written an [tex]H_1[/tex]. (We are restricted to work in one dimension; however, many higher-dimensional problems can be attacked through separation of variables. When studying the hydrogen atom, the radial dependence may be factored out of the Coulomb Hamiltonian to give a pseudo-one-dimensional potential; when studying the Landau effect, a convenient choice of gauge shows that along one axis, electrons behave as if in a harmonic oscillator potential.) As mentioned above, we can slide the energy scale up and down, so for convenience we fix the [tex]n = 0[/tex] eigenstate of [tex]H_1[/tex] to have zero energy:

[tex]E_0^{(1)}(a_1) = 0.[/tex]

We can solve for the ground state knowing the superpotential to show that

[tex]\psi_0^{(1)}(x;a_1) = C\exp\left(-\int^x W(y;a_1)dy\right)[/tex]

with [tex]C[/tex] being a constant of normalization. (Having the freedom to choose this constant makes the lower limit of the integral arbitrary.)

To investigate the spectrum of [tex]H_1[/tex], we construct the series of Hamiltonia [tex]\{H_s\}[/tex], where *s* ranges over the positive integers:

[tex]H_s = -\partial_x^2 + V_1(x;a_s) + \sum_{k=1}^{s-1}R(a_k),[/tex]

in which we have defined

[tex]a_s = f^{s-1}(a_1).[/tex]

Here, [tex]f^{s-1}(a_1)[/tex] denotes applying the transformation [tex]f[/tex] to the parameter set [tex]a_1[/tex] a total of [tex]s-1[/tex] times. Each time, we get a new, uniquely determined set of parameters. With this definition, it is easy to see that [tex]H_s[/tex] and [tex]H_{s+1}[/tex] are SUSY partners:

[tex]H_{s+1} = -\partial_x^2 + V_1(x;a_{s+1}) + \sum_{k=1}^{s}R(a_k),[/tex]

which is also equal to

[tex] -\partial_x^2 + V_2(x;a_s) + \sum_{k=1}^{s-1}R(a_k).[/tex]

Because [tex]H_s[/tex] and [tex]H_{s+1}[/tex] are SUSY partners, they share the same eigenenergies, save for the ground state, whose energy is just given by the sum of the residuals,

[tex]E_0^{(s)} = \sum_{k=1}^{s-1}R(a_k).[/tex]

The SUSY between [tex]H_1[/tex] and [tex]H_2[/tex] means that they are isospectral; the [tex]n=1,2,3,\ldots[/tex] eigenstates of [tex]H_1[/tex] have the same energies as the [tex]n = 0,1,2\ldots[/tex] states of [tex]H_2[/tex]. The same holds true for [tex]H_2[/tex] and [tex]H_3[/tex], [tex]H_3[/tex] and [tex]H_4[/tex], [tex]H_{17}[/tex] with [tex]H_{18}[/tex] and so on until we run out of bound states. (For a problem like the hydrogen atom, we never do: [tex]n[/tex] ranges upward to infinity.) We “lose” one state, the lowest-energy ground state, every time we go from [tex]H_s[/tex] to [tex]H_{s+1}[/tex]. Following the degeneracies through and liberally applying the transitive law, we see that the *n*-th eigenstate of [tex]H_1[/tex] is degenerate with the ground state of [tex]H_n[/tex]. Therefore, *we can recover the entire spectrum of [tex]H_1[/tex] from the equation above.*

We can get the eigenfunctions as well as the eigenvalues by this approach. Up to a normalization constant,

[tex]\psi_n^{(s+1)} \propto A_s \psi_{n+1}^{(s)},[/tex]

so knowing the eigenfunctions of [tex]H_1[/tex] would allow us to compute those of [tex]H_2[/tex]. The proportionality works in reverse, too:

[tex]\psi_n^{(s)} \propto A^\dag_{s+1}\psi_{n-1}^{(s+1)}.[/tex]

Therefore, if we knew the ground-state wavefunction of [tex]H_n[/tex], we could apply a sequence of operators [tex]A^\dag_n,A^\dag_{n-1},\ldots[/tex] until we reached a state in the spectrum of [tex]H_1[/tex]. But, we *do* know the ground state of [tex]H_n[/tex]: it is that state which is annihilated by the operator [tex]A_n[/tex]. Remember that [tex]A_n[/tex] involves a single first-order derivative [tex]\partial_x[/tex], so solving [tex]A_n \psi = 0[/tex] only involves solving a first-order differential equation. In this way, for any [tex]H_1[/tex] which has a shape-invariant potential (SIP), we can *exactly* solve for its eigenvalues and eigenfunctions.

As we noted earlier, this is reminiscent of the operator-based solution of the harmonic oscillator. With that in mind, the interested reader can for an exercise see what happens when the superpotential takes the simple form

[tex]W(x) = x.[/tex]

**READING**

- Fred Cooper, Avinash Khare, Uday Sukhatme, “Supersymmetry and Quantum Mechanics”
*Phys.Rept.***251**(1995): 267–385. Available as arXiv:hep-th/9405029.

**SUSY QM SERIES**

- Part 1: Superalgebra
- Part 2: Shape Invariance
- Part 3: Two-Body Problem
- Part 4: Separation of Variables
- Part 5: The Hydrogen Atom
- Part 6: Intermezzo: The Dirac Equation
- Part 7: The 1D Dirac Hamiltonian

Great post. Thanks for the arXiv reference

“Hamiltonia?” Really? That sounds like a pluralization of the pseudo-Latin “Hamiltonium.”

As crappy as it sounds, I’d go with Hamiltonians. Though I could certainly be mistaken on the proper conjugation.

Something in me had been wanting to say that for

years,and I finally succumbed. With luck, it shall not happen again.You know, I actually followed that just fine without flinching until the first integral. The graphics help, what can I say? Anyway, I don’t know that I would ever need the math for a high school physics class–which I haven’t taught in five years, anyway: enrollment is down, sadly–but I enjoy the challenge of trying to follow it all.

I saw this material in the first term of my junior year “at university,” so I doubt that it would be needed in high-school physics; we’d have to fix a whole host of broken things first, before we had to care about this! However, given how much time was wasted during my public-school education (pre-calculus? what in blazes was up with

that?) it’s not impossibly hard to imagine speeding things up by a couple years.I’m sorry to hear that enrollment in physics classes is down, which I believe you mentioned before. We seem to be so adept at giving the next generation a poor start in life; if only there were prizes for it.