# Heisenberg's derivation of Schrödinger's Equation [duplicate]

In Heisenberg's book "The Physical Principles of the Quantum Theory", he presents the following derivation of the Schrödinger Equation from his own, Matrix-based, Quantum Mechanics.

A matrix $x$ has components $x(a'a'')$ (the indices indicate that in this coordinate system the matrix $a$ is diagonal, that is, the basis vectors are eigenvectors of $a$). He starts by proposing the existence of matrices of continuous indices, such that matrix multiplication $$xy(a'a'')=\sum_{a'''}x(a'a''')y(a'''a'')$$ is replaced by $$xy(a'a'')=\int_{-\infty}^{\infty}x(a'a''')y(a'''a'')da'''$$ He then proposes that the continuous correspondent of the Identity Matrix $I$ is the Dirac-Delta function, such that $$I(a'a'')=\delta(a'-a'')$$ His derivation of the Schrödinger Equation is as follows. To transform the Hamiltonian Matrix $H$ from a coordinate system in which the coordinate matrix $q$ is diagonal into a system in which the Hamiltonian is diagonal, in which the Hamiltonian will be denoted $W$, the components of which are the system's allowed energy levels. The equations for this transformation are: $$S^{-1}HS=W$$ or $$HS=SW$$ such that $$\int H(q'q'')S(q''W')dq''=S(q'W')W'$$ with $W'$ being $W$'s diagonal components He then proposes (in one dimension) the following forms for the $q$ and $p$ matrices in the position basis" $$q(q'q'')=q'\delta (q'-q'')$$ $$p(q'q'')=\frac{\hbar}{i}\delta'(q'-q'')$$ He then shows they obey the Commutation Relation $pq-qp=\frac{\hbar}{i}$ but his proof of the commutation relation uses integration by parts incorrectly, having the integral of the Dirac Delta's derivative be $$\int f(\xi)\delta(a-\xi)d\xi=f'(a)$$ instead of $$\int f(\xi)\delta(a-\xi)d\xi=-f'(a)$$ Usingintegration by parts incorrectly once again, he argues that $$qS(q'W')=\int q'\delta(q'-q'')S(q''W')dq''=q'S(q'W')$$ and $$pS(q'W')=\frac{\hbar}{i}\int \delta'(q'-q'')S(q''W')=\frac{\hbar}{i}\dfrac{d}{dq'}S(q'W')$$ So we replace in $H(q,p)S=SW$ $$H\left(q',\frac{\hbar}{i}\dfrac{d}{dq'}\right)S=W'S$$ Which is evidently the Time-Independent Schrödinger Equation, which can be solved for the energy Eigenvalues $W'$

But this is derived with a wrong value for $p$ because the derivation uses a wrong process of integration by parts. The immediately obvious choice would be to have $p(q'q'')=-\frac{\hbar}{i}\delta'(q'-q'')$, so that integrated with the transformation matrix it becomes$\frac{\hbar}{i}\dfrac{d}{dq'}$, but I just can't seem to make it fit the Commutation Relation.

Can anyone help solve this?

• Possible duplicate by OP: physics.stackexchange.com/q/419668/2451 – Qmechanic Aug 1 '18 at 8:57
• It is not a duplicate. Last time I asked the question I misspelled the second equation, I meant to have the $a$ and $\xi$ in the same order. I got a whole bunch of replies about the derivative of the delta function being odd because of that misspelling and now I can't delete my question. – user140323 Aug 1 '18 at 12:59

You posted this already last week. Heisenberg is not incorrect. The right answer is $$\int d\xi \ f(\xi) \delta’(a-\xi)=\int dx \ f(x+a) \delta’(-x) = - \int dx \ f(x+a) \delta(x) = - [-f’(a)] =f’(a)$$ I have shifted the variable $x=\xi -a$ and used oddness property $\delta’(-x)=-\delta(x)$. If you’re confused by an answer, ask clarifying questions