How exactly does squaring operators (e.g. <p^2> work?

In summary, squaring an operator is not as simple as just squaring its components. The differential part of the operator changed, too.
  • #1
Zacarias Nason
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I'm having to essentially piece together a framework of background knowledge to understand parts of a QM class in which I'm lacking prerequisites; one of the things that I noticed that really confused me was the square of the momentum operator <p>, and how that translated into the integrand of an expectation value; Squaring an operator as far as I'm as aware isn't as simple as just squaring its components, because, for the momentum operator defined as [tex] \langle p^2 \rangle = \int_{-\infty}^{\infty}\psi^*(x)p^2\psi(x) = \int_{-\infty}^{\infty}\psi^*(x)\bigg(-\hbar^2 \frac{\partial^2}{\partial x^2}\psi(x)\bigg)[/tex]
The differential part of the operator changed, too! This isn't just squaring, what is it? [tex] \frac{d}{dx}\frac{d}{dx} \neq \frac{d^2}{dx^2} [/tex] If I'm given some arbitrary operator A and told to calculate [tex] \langle A^2\rangle [/tex] How do I know what the operator "squared" takes the form of?
 
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  • #2
The differential part of the operator changed, too! This isn't just squaring, what is it?

operators are supposed to operate in a space - so a A^2 must be equivalent to A(op). A(op) , a product say operating on the right...
 
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  • #3
I don't understand what that meant; I don't have any working knowledge of spaces, regardless of what they are.
drvrm said:
a product say operating on the right...
I also have no idea of what A(op) is supposed to indicate. A is the operator, and I'm only guessing "op" is shorthand for "operator", so unless A(op) is a really unnecessarily roundabout way of saying (A)(A) or something, I don't get it.
 
  • #4
Alright then, thanks for the help(?).
 
  • #5
##p^2\psi=pp\psi=p(p\psi)##, that is, you apply ##p## to ##\psi## to get a new function ##\psi_1##, then you apply ##p## to ##\psi_1##.

Likewise for any other operator.
 
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  • #6
Thank you very much!
 

1. How does the squaring operator work?

The squaring operator is a mathematical operation that involves multiplying a number by itself. For example, if we have the number 3, squaring it would give us 3 x 3 = 9. This is represented by the symbol "^2" or "²", and is commonly used in algebra and other mathematical equations.

2. What is the purpose of using a squaring operator?

The main purpose of using a squaring operator is to find the square of a number, which is useful in various mathematical calculations. It can also be used to simplify equations and express them in a more concise form.

3. How does the squaring operator affect negative numbers?

The squaring operator can be applied to negative numbers as well. When we square a negative number, the result is always positive. For example, (-3)^2 = 9. This is because multiplying a negative number by itself results in a positive number.

4. Can the squaring operator be used for other types of values besides numbers?

Yes, the squaring operator can be applied to other types of values such as variables, matrices, and vectors. In these cases, the squaring operator is used to represent the operation of multiplying the value by itself, similar to how it works with numbers.

5. What is the difference between the squaring operator and the square root operator?

The squaring operator and the square root operator are inverse operations of each other. While the squaring operator gives the square of a number, the square root operator gives the number that when squared, gives the original number. For example, the square root of 9 is 3, because 3^2 = 9.

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