At time t=0 a particle is represented by the wave function ?(x, 0) = { A, 0<x<b/2; 0, otherwise. Normalize the wave function (find A). ?b; ?(2/b); ?(b/2); ?(1/b)
Added by Holly J.
Close
Step 1
In mathematical terms, this is expressed as: ∫ |Ψ(x,0)|² dx = 1 In this case, the wave function is given as Ψ(x,0) = A for 0 < x < b/2 and Ψ(x,0) = 0 otherwise. Show more…
Show all steps
Your feedback will help us improve your experience
Suman K and 63 other Physics 101 Mechanics educators are ready to help you.
Ask a new question
Labs
Want to see this concept in action?
Explore this concept interactively to see how it behaves as you change inputs.
Key Concepts
Recommended Videos
In the region $0 \leq x \leq a$, a particle is described by the wave function $\psi_{1}(x)=-b\left(x^{2}-a^{2}\right)$. In the region $a \leq x \leq w,$ its wave function is $\psi_{2}(x)=(x-d)^{2}-c .$ For $x \geq w, \psi_{3}(x)=0 .(a)$ By applying the continuity conditions at $x=a,$ find $c$ and $d$ in terms of $a$ and $b .(b)$ Find $w$ in terms of $a$ and $b$.
The wave function for a particle is given by $\psi(x)=A e^{-|x| / a}$ , where $A$ and $a$ are constants. (a) Sketch this function for values of $x$ in the interval $-3 a < x < 3 a$ . (b) Determine the value of $A$ . (c) Find the probability that the particle will be found in the interval $-a < x < a$ .
Cal W.
The wave function for a quantum particle confined to moving in a one-dimensional box is $$ \psi(x)=A \sin \left(\frac{n \pi x}{L}\right) $$ Use the normalization condition on $\psi$ to show that $$ A=\sqrt{\frac{2}{L}} $$ Suggestion: Because the length of the box is $L,$ the wave function is zero for $x<0$ and for $x>L$ , so the normalization condition (Eq. 41.7$)$ reduces to $$ \int_{0}^{L}|\psi|^{2} d x=1 $$
Recommended Textbooks
University Physics with Modern Physics
Physics: Principles with Applications
Fundamentals of Physics
Transcript
18,000,000+
Students on Numerade
Trusted by students at 8,000+ universities
Watch the video solution with this free unlock.
EMAIL
PASSWORD
