Homework Help is Here – Start Your Trial Now!
Science
Advanced Physics
Going from -(h_bar^2/2m) (d^2/dx^2) ψ to the momentum operator squared (1/2m) p_hat^2, how is the negative sign lost?  I must be missing something fundamental since it looks to me like momentum operator ->.  p_hat^2 = (-ih_bar d/dx)(-ih_bar d/dx)=+i^2 h_bar^2 (d^2/dx^2)= -h_bar^2 (d^2/dx^2) ? Thank you!

Going from -(h_bar^2/2m) (d^2/dx^2) ψ to the momentum operator squared (1/2m) p_hat^2, how is the negative sign lost?  I must be missing something fundamental since it looks to me like momentum operator ->.  p_hat^2 = (-ih_bar d/dx)(-ih_bar d/dx)=+i^2 h_bar^2 (d^2/dx^2)= -h_bar^2 (d^2/dx^2) ? Thank you!

icon
Related questions
Question
100%

Going from -(h_bar^2/2m) (d^2/dx^2) ψ to the momentum operator squared

(1/2m) p_hat^2, how is the negative sign lost?  I must be missing something fundamental since it looks to me

like momentum operator ->.  p_hat^2 = (-ih_bar d/dx)(-ih_bar d/dx)=+i^2 h_bar^2 (d^2/dx^2)= -h_bar^2 (d^2/dx^2) ?

Thank you!

which describes simple harmonic oscillation (about the point xo), with (it is customary to eliminate the spring constant in favor of the classical constant k = V"(xo). That's why the simple harmonic oscillator is so impon oscillatory motion is approximately simple harmonic, as long as the amplitu The quantum problem is to solve the Schrödinger equation for the potent generalizes to a broad class of potentials in supersymmetric quantum (Probea Equation 2.42). As we have seen, it suffices to solve the time-indepente Note that V" (x)) 0, since by assumption xo is a minimum. Only in the rare case Vw To begin with, let's rewrite Equation 2.45 in a more suggestive form: msi W. Robinett, Quantum Mechanics (Oxford University Press, New York, 1997), Section 14 23 We'll encounter some of the same strategies in the theory of angular momentum ( arbitrary potential, in the local minimum. V"(xo) (x- xo)2, V (x) 2 well, wha From 2.41 V(x) = -mo²x? As anticipat * and p: it operators A 2.11 P 27 equation: あ* V In this notat 2m dx? + mw*x*v = EV. 2um In the literature you will find two entirely different approaches to this mi a straightforward “brute force" solution to the differential equation, using method; it has the virtue that the same strategy can be applied to many abe fact, we'll use it in Chapter 4 to treat the hydrogen atom). The second is adi algebraic technique, using so-called ladder operators. I'll show you te first, because it is quicker and simpler (and a lot more fun);- if you war series method for now, that's fine, but you should certainly plan to study lra We need slippery to them a "tes you'll be le [x. F .23 Dropping th 2.3.1 Algebraic Method This lovely + (max)²| = Ev, 2m trode 24 Put a hat C 25 In a deep 22 do not com not even approximately simple harmonic. it to derive (Chapter
Transcribed Image Text:which describes simple harmonic oscillation (about the point xo), with (it is customary to eliminate the spring constant in favor of the classical constant k = V"(xo). That's why the simple harmonic oscillator is so impon oscillatory motion is approximately simple harmonic, as long as the amplitu The quantum problem is to solve the Schrödinger equation for the potent generalizes to a broad class of potentials in supersymmetric quantum (Probea Equation 2.42). As we have seen, it suffices to solve the time-indepente Note that V" (x)) 0, since by assumption xo is a minimum. Only in the rare case Vw To begin with, let's rewrite Equation 2.45 in a more suggestive form: msi W. Robinett, Quantum Mechanics (Oxford University Press, New York, 1997), Section 14 23 We'll encounter some of the same strategies in the theory of angular momentum ( arbitrary potential, in the local minimum. V"(xo) (x- xo)2, V (x) 2 well, wha From 2.41 V(x) = -mo²x? As anticipat * and p: it operators A 2.11 P 27 equation: あ* V In this notat 2m dx? + mw*x*v = EV. 2um In the literature you will find two entirely different approaches to this mi a straightforward “brute force" solution to the differential equation, using method; it has the virtue that the same strategy can be applied to many abe fact, we'll use it in Chapter 4 to treat the hydrogen atom). The second is adi algebraic technique, using so-called ladder operators. I'll show you te first, because it is quicker and simpler (and a lot more fun);- if you war series method for now, that's fine, but you should certainly plan to study lra We need slippery to them a "tes you'll be le [x. F .23 Dropping th 2.3.1 Algebraic Method This lovely + (max)²| = Ev, 2m trode 24 Put a hat C 25 In a deep 22 do not com not even approximately simple harmonic. it to derive (Chapter
Expert Solution
trending now

Trending now

This is a popular solution!

steps

Step by step

Solved in 3 steps

SEE SOLUTIONCheck out a sample Q&A here
Blurred answer