Note 10 Back to Ch. 05 · The Hamiltonian and the Schrödinger Equation
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The calculation The commutator formulas

In the card we stated four formulas about commutators and used them to determine the Hamiltonian matrix. Here we prove them.

Suppose the function f(X)f(X) can be expanded in a power series

f(X)=∑n=−∞+∞fnXnf(X)=\sum_{n=-\infty}^{+\infty}f_nX^n

The derivative of this function is

df(X)dX=∑−∞+∞fnnXn−1\frac{df(X)}{dX}=\sum_{-\infty}^{+\infty}f_nnX^{n-1}

So to prove 1a we must verify the following identity

[K,∑n=−∞+∞fnXn]=i∑n=−∞+∞nfnXn−1\left[K,\sum_{n=-\infty}^{+\infty}f_nX^n\right]=i\sum_{n=-\infty}^{+\infty}nf_nX^{n-1}

Analogously, to prove 2a we must verify the following identity

[X,∑−∞∞fnKn]=−i∑−∞∞nfnKn−1\left[X,\sum_{-\infty}^\infty f_nK^n\right]=-i\sum_{-\infty}^\infty nf_nK^{n-1}

We will prove that the terms of the sums on the left-hand sides are equal, one by one, to the terms of the sums on the right-hand sides:

[K,fnXn]=infnXn−1⇔[K,Xn]=inXn−1[X,fnKn]=−infnKn−1⇔[X,Kn]=−inKn−1\begin{aligned} [K,f_nX^n] & =inf_nX^{n-1}\Leftrightarrow \\ [K,X^n] & =inX^{n-1} \\ [X,f_nK^n] & =-inf_nK^{n-1}\Leftrightarrow \\ [X,K^n] & =-inK^{n-1} \end{aligned}∀n∈1…∞\forall n\in1\dots\infty

We carry out a proof by induction.

For n=1n=1 we must verify that

[K,X]=iI  [X,K]=−iI\begin{aligned} [K,X] & =iI\;[X,K] \\ & =-iI \end{aligned}

Let us begin with the first. Consider a generic function ψ(x)\psi(x)

[K,X]ψ(x)=KXψ(x)−XKψ(x)=iddx(xψ(x))−xddxψ(x)=iψ(x)+xddxψ(x)−xddxψ(x)=iψ(x)\begin{aligned} [K,X]\psi(x) & =KX\psi(x)-XK\psi(x) \\ & =i\frac{d}{dx}(x\psi(x))-x\frac{d}{dx}\psi(x) \\ & =i\psi(x)+x\frac{d}{dx}\psi(x)-x\frac{d}{dx}\psi(x) \\ & =i\psi(x) \end{aligned}

so [K,X]ψ(x)=iψ(x)[K,X]\psi(x)=i\psi(x). Being true for every ψ(x)\psi(x), we can deduce [K,X]=iI[K,X]=iI.

The second is now obvious, indeed [X,K]=−[K,X]=−iI[X,K]=-[K,X]=-iI.

Now we prove that if the formulas

[K,Xn]=inXn−1  [X,Kn]=−inKn−1\begin{aligned} [K,X^n] & =inX^{n-1}\;[X,K^n] \\ & =-inK^{n-1} \end{aligned}

are true for n, then they are true also for n+1n+1 and for n−1n-1.

Let us begin with the first and prove that if it is true for n, then it is also true for n+1n+1

[K,Xn+1]=KXn+1−Xn+1K=KXn+1−XnXK=\begin{aligned} \left[K,X^{n+1}\right] & =KX^{n+1}-X^{n+1}K \\ & =KX^{n+1}-X^nXK= \end{aligned}

applying the formula valid for n=1n=1

=KXn+1+Xn(iI−KX)=KXn+1+iXn−XnKX=iXn+(KXn−XnK)X=\begin{aligned} & =KX^{n+1}+X^n(iI-KX) \\ & =KX^{n+1}+iX^n-X^nKX \\ & =iX^n+(KX^n-X^nK)X= \end{aligned}

applying the formula valid for n

=iXn+inXn−1X=iXn+inXn=i(n+1)XnAs we wanted to prove.\begin{aligned} & =iX^n+inX^{n-1}X \\ & =iX^n+inX^n \\ & =i(n+1)X^n\quad\text{As we wanted to prove.} \end{aligned}

Now we prove that if it is true for n, then it is also true for n−1n-1

[K,Xn−1]=KXn−1−Xn−1K=X−1XKXn−1−Xn−1K=\begin{aligned} \left[K,X^{n-1}\right] & =KX^{n-1}-X^{n-1}K \\ & =X^{-1}XKX^{n-1}-X^{n-1}K= \end{aligned}

applying the formula valid for n=1n=1

=X−1(−iI+KX)Xn−1−Xn−1K=−iXn−2+X−1KXn−X−1XnK=−iXn−2+X−1(KXn−XnK)\begin{aligned} & =X^{-1}(-iI+KX)X^{n-1}-X^{n-1}K \\ & =-iX^{n-2}+X^{-1}KX^n-X^{-1}X^nK \\ & =-iX^{n-2}+X^{-1}(KX^n-X^nK) \end{aligned}

applying the formula valid for n

−iXn−2+X−1inXn−1=−iXn−2+inXn−2=i(n−1)Xn−2As we wanted to prove.\begin{aligned} -iX^{n-2}+X^{-1}inX^{n-1} & =-iX^{n-2}+inX^{n-2} \\ & =i(n-1)X^{n-2}\quad\text{As we wanted to prove.} \end{aligned}

For 2a the steps are identical.

The 3rd and the 4th are practically obvious, and hold for any operator A, indeed

[A,An]=AAn−AnA=An+1−An+1=0\begin{aligned} [A,A^n] & =AA^n-A^nA \\ & =A^{n+1}-A^{n+1} \\ & =0 \end{aligned}
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La Quantistica · Note No. 10 · Rev. 2026 F. Palma