Question

I am familiar with the momentum operator in quantum mechanics, and have used it in problems, in at least one dimension. However, I am reading a new text in effort to absorb any details about the formalization that I might not yet be aware of.

Sakurai and Napolitano in Modern Quantum Mechanics 3rd edition, are giving a derivation of the momentum operator effect on a vector with known position. They are starting with an expression that they derived in preceding sections, using one minus the imaginary unit times fixed momentum times displacement (over hbar)

I am having extreme difficulty trying to understand formulas 1.247, 1.248, 1.249 and 1.250 -- if this question is answered successfully I should be able to correctly interpret the rest of the page.

The authors are using the same symbol (x′)
in three different ways in the same formula. This seems to go against all rules of notation I've ever encountered, and in this case, it creates real ambiguity.

a free variable on the left hand side, which should be interepreted as constant, just the name of the vector component we are considering (in 1.249 and 1.250)

the variable of integration in the middle (in 1.247 and 1.248); as everybody knows, these are not really variables in the greater expression, but just determine what the integral is doing to the expression integrated, i.e. how the expression is being turned into a function so that it can be integrated w.r.t. that specific variable as opposed to some other. Here x′
is also the symbol they chose for that

the variable of differentiation on the right (in 1.247, 1.248, 1.249 and 1.250). Same problem as with the integral, not really a variable outside the scope of the formula being differentiated, as in all derivatives, but due to having the same symbol, its meaning is ambiguous here. Now there is the question as to how parentheses would close the expression:

In addition there is a question as to where parentheses are to be inserted. The authors have taken up the habit of putting the dx
right next to the integral sign, which I've gotten used to, but in this case it seems ambiguous in regards to whether

either the derivative is its own expression and is turned into a function then integrated
the derivative is not part of the integral and the derivative, once taken, is just multiplied on the right of the integral expression after it is also taken, and they are separate, just two factors
the derivative is somehow re-taken for every point in the region integrated, and then the sum of infinitesimals of these are taken as the integral
I'm attaching the page in question from the above text

I was hoping someone would provide the same formula but using notation with three different names for variables in these three different contexts of the formulas.

Of course I know the simple one-dimensional momentum operator expression that you can get on Wikipedia. But I was hoping someone would clarify this in a way that I can understand what point the authors are trying to make. Clearly there is something important to be learned concerning deriving it in this specific way, that is supposed to be more general.

Answer 1

Distinct symbols can be used to clarify the use of the momentum operator in quantum mechanics, ensuring clear differentiation between position, integration variable, and differentiation variable.

Step-by-step explanation:

The confusion arising from the notation in the mentioned formulas can be clarified by differentiating the usage of the symbol x' by substituting it with distinct symbols for clarity. In quantum mechanics, it is important to distinguish between the action of an operator and its result on a function. For instance, we can consider the momentum operator in the x-direction, given by the expression -i\(\hbar\) d/dx. This operator acts on a wave function \(\psi(x), and when calculating the expectation value, the operator is applied to the wave function, which is then multiplied by the complex conjugate of another wave function before integration.

To avoid confusion, we can use x for the position vector, x' for the variable of integration, and x'' for the variable of differentiation. For example, the expectation value of momentum can be expressed as:

\(\int_{-\infty}^{\infty} \psi^{*}(x) \left(-i\hbar \frac{d}{dx''}\right) \psi(x') dx'\)

Where:

• x represents the specific position component being considered.
• x' is the variable of integration.
• x'' is the variable in the derivative operator acting on the wave function.