Solving for $\hat{A}\psi(x)=\psi(x+b)$

  • Thread starter CalcYouLater
  • Start date
In summary, the conversation discusses the operator \hat{A}=exp({b}\frac{d}{dx}), where b is a constant, and how it can be used to show that \hat{A}\psi(x)=\psi(x+b). The exponential notation is explained and it is noted that learning operator basics and notation can be challenging.
  • #1
CalcYouLater
54
0

Homework Statement



Given operator [tex]\hat{A}=exp({b}\frac{d}{dx})[/tex] where b is a constant, show that:

[tex]\hat{A}\psi(x)=\psi(x+b)[/tex]

Homework Equations


The Attempt at a Solution



I have no idea with this one. If I understand correctly the operator, A, says take the derivative of (something) with respect to x, multiply the result by b, and then exponentiate it. A pointer would be much appreciated here.
 
Physics news on Phys.org
  • #2
It's usually a good idea to consider an infinitesimal change first and then deduce the result for a finite change.
 
  • #3
You haven't quite gotten the meaning of the exponential notation correct, which is why you're confused :wink: A function of an operator is defined by its series expansion,
[tex]f(\hat{D}) = f_0 + f_1 \hat{D} + f_2 \hat{D}^2 + \cdots[/tex]
Typically the function f is one that can be expanded in a Maclaurin series (Taylor series around 0), so in that case the coefficients [itex]f_n[/itex] are just the same coefficients that appear in the Maclaurin series:
[tex]f_n = \frac{1}{n!}f^{(n)}(0)[/tex]
For the exponential function, the series is
[tex]e^x = 1 + x + \frac{x^2}{2} + \cdots[/tex]
so the operator [tex]e^{\hat{D}}[/tex] is defined as
[tex]e^{\hat{D}} = 1 + \hat{D} + \frac{\hat{D}^2}{2} + \cdots[/tex]
In your case, [itex]\hat{D}[/itex] is multiplication by a constant composed with the derivative operator [itex]\mathrm{d}/\mathrm{d}x[/itex], so the notation [itex]\hat{D}^n[/itex] is effectively telling you to take the n'th derivative and then multiply by [itex]b^n[/itex]. This would of course be applied to whatever the operator is acting on - for example, in
[tex]e^{b\mathrm{d}/\mathrm{d}x}\psi(x)[/tex]
each term would be produced by differentiating ψ(x) some number of times and then multiplying by b that many times.
 
  • #4
Oh I see! Thank you both for responding. Interesting that it yields the series expansion for psi of (x+b). Learning operator basics and notation is turning out harder than I expected.
 
  • #5


I can help you understand this problem. The operator \hat{A} represents an exponential function with a variable inside, where the variable is the derivative of something with respect to x multiplied by a constant b. This operator can be applied to any function \psi(x) to give a new function, which is \psi(x) shifted by b units to the right. In other words, \hat{A}\psi(x) is equivalent to \psi(x+b). This can be seen by substituting x+b into the original function \psi(x) and then applying the operator \hat{A}. Therefore, the solution to this problem is simply \hat{A}\psi(x) = \psi(x+b). I hope this helps!
 

Related to Solving for $\hat{A}\psi(x)=\psi(x+b)$

1. What does the equation $\hat{A}\psi(x)=\psi(x+b)$ represent?

The equation represents a translation operator, where $\hat{A}$ is the operator that translates the wave function $\psi(x)$ by a distance of $b$ units in the x-direction.

2. How is $\hat{A}$ defined?

$\hat{A}$ is defined as the operator that performs the translation operation on the wave function, which is given by $x\rightarrow x+b$.

3. What is the significance of solving for $\hat{A}\psi(x)=\psi(x+b)$?

Solving for this equation allows us to understand the behavior of wave functions under translation operations and how it affects the overall wave function.

4. How can $\hat{A}$ be applied in real-world scenarios?

The translation operator $\hat{A}$ has various applications in quantum mechanics, such as in the study of quantum tunneling and scattering problems.

5. Can the translation operator $\hat{A}$ be combined with other operators?

Yes, $\hat{A}$ can be combined with other operators, such as the momentum operator, to study more complex quantum systems.

Similar threads

  • Advanced Physics Homework Help
Replies
10
Views
838
  • Advanced Physics Homework Help
Replies
1
Views
560
Replies
5
Views
2K
  • Advanced Physics Homework Help
Replies
16
Views
526
  • Advanced Physics Homework Help
Replies
4
Views
1K
  • Advanced Physics Homework Help
Replies
1
Views
2K
  • Advanced Physics Homework Help
Replies
1
Views
1K
  • Quantum Physics
Replies
9
Views
1K
  • Advanced Physics Homework Help
2
Replies
44
Views
3K
  • Advanced Physics Homework Help
Replies
2
Views
1K
Back
Top