I've encountered a Hamiltonian of the following form while studying quantum mechanical systems:

$$ \hat H = \left[\left( \frac{\partial}{\partial x} + c \right)^2 + \cos x \right] $$

where c is a constant. I'm wondering how I can expand this expression, specifically the binomial expansion of an operator and a number. If we imagine the differential operator as a matrix like this

$$ \hat A = \begin{pmatrix} 0 & 1 & 0& \cdots & \\ 0 & 0 & 1 & \cdots & \\ \vdots & \vdots & \vdots & \ddots \\ \end{pmatrix} $$

then we seem to get a contradiction because a matrix plus a scalar is undefined.

Usually we would approach the Hamiltonian in the context of the Schr$\mathrm{\ddot o}$dinger equation, $$ \hat H \psi(x)=E\psi(x) $$ where $ \hat H$ to operates on $\psi(x)$, so the $\frac{\partial}{\partial x}$ operator can act on the function $before$ we add the operator and the scalar together, avoiding the problem above. But I'm also uncomfortable with this. Starting with this,

$$ \left( \frac{\partial}{\partial x} + c \right)^2 \psi (x)$$

is it possible to expand like this?

$$ \left(\frac{\partial^2}{\partial x^2} + 2 b\frac{\partial}{\partial x} + b^2 \right) \psi (x) = \frac{\partial^2\psi (x)}{\partial x^2} + 2b\frac{\partial \psi (x)}{\partial x} + b^2 \psi (x)$$

I've never learned any rules about how operators behave all by themselves during algebra and arithmetic. I usually like to think in terms of infinite matrices, but I'm worried about the contradiction above.

Please help. How can I be sure that this binary expansion is in accordance with the axioms of mathematics.

  • $\begingroup$ Remember that representations of operators make sense only when applied to vectors: therefore in order to expand the above Hamiltonian you should apply it to something first (in particular the derivative will then follow the Leibniz rule). $\endgroup$ – gented Sep 15 '17 at 12:59
  • $\begingroup$ What do you mean when you say "make sense"? I represented the derivative operator as an infinite matrix with ones along the superdiagonal, right? Why doesn't that make sense? $\endgroup$ – psitae Sep 15 '17 at 13:10
  • $\begingroup$ Well, you first use some matrix representation (I am not sure where that comes from), then you use the derivative symbol again, then you don't specify what domain of functions you're taking so that those representations hold. In order for all to make sense you should define the representation first and then use that one only. Also notice that in general you cannot expand the square of a derivative the way you did (in this case it works fine because $c$ is a number, but in general you should follow the Leibniz rule). $\endgroup$ – gented Sep 15 '17 at 14:10
  • $\begingroup$ Yeah operators on their own don't make much sense. Without going to a matrix representation of an operator, writing out $\partial/\partial x$ on its own shouldn't make any sense. Neither should $\partial/\partial x + c$. Consider the operator $\hat{a} = 2 +$. $\hat{a}^2$ means $\hat{a}\hat{a} = (2+)(2+)$. This operator is equivalent to $\hat{a} = 4 +$. Now consider $\hat{b} = 3\cdot$. The operation $\hat{a}\hat{b}f(x) = 2 + 3f(x)$ is not in general $\hat{b}\hat{a}f(x) = 3(2+f(x))$. This is why we have to define an order of operations when we first start learning math as you know $\endgroup$ – DWade64 Sep 15 '17 at 14:25

The meaning of something like $A^2\psi(x)$ is that you apply $A$ twice in a row to $\psi(x)$. In your specific case, with $c$ a number: \begin{align} \left(\frac{\partial}{\partial x}+c\right)^2\psi(x)&= \left(\frac{\partial}{\partial x}+c\right)\left(\psi'(x)+c\psi(x)\right)\\ &=\psi''(x)+ 2c\psi'(x)+c^2\psi(x) \end{align} because $c$ commutes with $\partial_x$. More generally, if you had an expression like $(A+B)^2\psi(x)$ you'd need to keep track of the ordering: \begin{align} (A+B)^2\psi(x)&=(A+B)(A\psi(x)+B\psi(x))\, ,\\ &=A^2\psi(x)+AB\psi(x)+BA\psi(x)+B^2\psi(x) \end{align} with no a priori reason to suppose that $AB=BA$.

Edit: I'm not sure why you would think that a differential operator is a matrix of the type you suggest. You'd have to specify a basis for this.

As an easy counterexample, given that $\partial_x\sim p_x$ up to some unimportant factors, one easily verifies that $\langle \psi_n\vert p_x\vert\psi_m\rangle$ for harmonic oscillator eigenstates $\vert\psi_m\rangle$ does not produce a matrix of the form your suggest. Rather, $p_x$ is a hermitian matrix with non-zero entries when $m=n\pm 1$.


Your scalar is an operator, namely this scalar times the identity, normally dropped to save superfluous symbols.

Using the obvious operator identity (you may imagine an arbitrary function f on the right, if you are not used to operator calculus), $$ \partial_x ~ e^{-cx}= e^{-cx} (-c+\partial_x), $$ it follows that $$ e^{cx} \hat{H} e^{-cx}= \partial_x^2 + \cos x ~, $$ evidently easier to address.

Do you have experience with the Mathieu equation?

Edit in response to question : Well, clear and present danger indeed, $$ \hat{H} \psi = E\psi \qquad \iff \qquad (\partial_x^2 + \cos x) ~(e^{cx} \psi) = E ~ (e^{cx} \psi)~. $$

  • $\begingroup$ Thanks for your answer.Yes. I know about the Mathieu equation $\endgroup$ – psitae Sep 20 '17 at 13:43
  • $\begingroup$ ... My system is just dangerously close to the Mathieu equation, and I was wondering if I can use it to solve for the energy eigenvalues of the system. Perhaps that's for another post, though $\endgroup$ – psitae Sep 20 '17 at 13:44
  • $\begingroup$ What you added to your response a few days ago is indeed the Mathieu equation. I'm wondering if my original Hamiltonian (all the way at the top) applies for c != 0. Do you have any idea how to approach this? If not, I'll link to the new question soon. $\endgroup$ – psitae Sep 23 '17 at 6:52

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