Confused in David Bohm's *Quantum Theory*

Question

In discussing matrix mechanics Bohm says in chapter 16 of Quantum Theory that the $rm$ element of operator $A$, $a_{rm}$, is given by $$a_{rm} = \int dx\ \psi^*_r(x)A\psi_m(x) \hspace{1in} (Eqn. 2)$$ He then goes on to discuss the 1D harmonic oscillator specifically. And he says at the beginning of ch. 16 sec. 4: $$\left(x+\frac{ip}{\hbar}\right)_{mn}=\int dx\ \psi^*_n(x)\left(x+\frac{ip}{\hbar}\right)\psi_m(x) \;\;\;$$ Note that the indices in the integrand here are swapped compared to the indices in his Eqn. 2. AFAICT, this is not just a notation change, but actually has consequences, since two equations down he derives $$\left(x+\frac{ip}{\hbar}\right)_{mn} = \sqrt{2m}\ \ \delta_{m-1,n}$$ So $x+ip/\hbar$ becomes a raising operator. But AFAICT if we calculate $(x+ip/\hbar)_{mn}$ consistently with Eqn. 2, we have $$\left(x+\frac{ip}{\hbar}\right)_{mn}=\int dx\ \psi^*_m(x)\left(x+\frac{ip}{\hbar}\right)\psi_n(x)$$ which evaluates to $$\left(x+\frac{ip}{\hbar}\right)_{mn}=\sqrt{2n}\ \delta_{m,n-1}$$ and so $x+ip/\hbar$ becomes a lowering operator.

I presume that David Bohm is smarter than I am. So can someone tell me what I am missing?

Answer 1

In discussing matrix mechanics Bohm says in chapter 16 of Quantum Theory that the $rm$ element of operator $A$, $a_{rm}$, is given by $$a_{rm} = \int dx\ \psi^*_r(x)A\psi_m(x) \hspace{1in} (Eqn. 2)$$

OK. This is fine.

And he says at the beginning of ch. 16 sec. 4: $$\left(x+\frac{ip}{\hbar}\right)_{mn}=\int dx\ \psi^*_n(x)\left(x+\frac{ip}{\hbar}\right)\psi_m(x) \;\;\;$$ Note that the indices in the integrand here are swapped compared to the indices in his Eqn. 2. AFAICT, this is not just a notation change, but actually has consequences...

This index swap appears to be a typo, since it is inconsistent with your "Eqn. 2" above. Either it is a typo or else you are not correctly telling us the analog of $\hat A$ in this context.

So can someone tell me what I am missing?

The matrix elements for the raising and lowering operators are worked out in detail in many references, including online references.

For simplicity I will set $\hbar = m\omega = 1$, and so we have: $$ \hat a \equiv \frac{1}{\sqrt{2}}\left(\hat x + i\hat p\right)\;, $$ which is the usual lowering operator, and for which the matrix elements are well known: $$ a_{nm}\equiv \langle \psi_n|\hat a|\psi_m\rangle = \sqrt{m}\langle n|m-1\rangle = \sqrt{m}\delta_{n,m-1}\;. $$

It is similarly well known that for the raising operator, we have $$ a^\dagger_{nm}\equiv \langle \psi_n|\hat a^\dagger|\psi_m\rangle = \sqrt{m+1}\langle \psi_n|m+1\rangle = \sqrt{m+1}\delta_{n,m+1} $$

so $x+ip/\hbar$ becomes a lowering operator

Not "becomes," but is. Or, more precisely, is proportional to the conventional lowering operator.