/*! This file is auto-generated */ .wp-block-button__link{color:#fff;background-color:#32373c;border-radius:9999px;box-shadow:none;text-decoration:none;padding:calc(.667em + 2px) calc(1.333em + 2px);font-size:1.125em}.wp-block-file__button{background:#32373c;color:#fff;text-decoration:none} Q34E Because we have found no way to ... [FREE SOLUTION] | 91影视

91影视

Because we have found no way to formulate quantum mechanics based on a single real wave function, we have a choice to make. In Section 4.3,it is said that our choice of using complex numbers is a conventional one. Show that the free-particle Schrodinger equation (4.8) is equivalent to two real equations involving two real functions, as follows:

-221(x,t)m=2(x,t)tand

-222(x,t)m=1(x,t)t

where (x,t)is by definition 1(x,t)+i2(x,t). How is the complex approach chosen in Section4.3more convenient than the alternative posed here?

Short Answer

Expert verified

We should be able to recover the original Schrodinger equation using the two supplied equations, using the information that=1+i2.

Step by step solution

01

Combining two partial differential equations.

So, as suggested in the section, combining the two partial differential equations into a single equation should greatly simplify our analysis. We should be able to reconstruct the original Schrodinger equation using the two supplied equations and the information that=1+i2.

22m21(x,t)x2=2(x,t)t

-22m22(x,t)x2=1(x,t)t

Subtracting equation(1) from the equation (2), after multiplying it by i with equation(2).

22m-i22(x,t)x2-21(x,t)x2=i1(x,t)t-2(x,t)t

-22m21(x,t)+i2(x,t)x2=i1(x,t)+i2(x,t)t

=1+i2


-22m2x2=it

This is the same equation as in the section 4.3, we only have to solve one differential equation (for a complex function) rather than two differential equations (for real functions), which is a significant simplification given the cost of complexification.

02

Conclusion

Using =1+i2we should be able to recover the original Schrodinger equation.

Unlock Step-by-Step Solutions & Ace Your Exams!

  • Full Textbook Solutions

    Get detailed explanations and key concepts

  • Unlimited Al creation

    Al flashcards, explanations, exams and more...

  • Ads-free access

    To over 500 millions flashcards

  • Money-back guarantee

    We refund you if you fail your exam.

Over 30 million students worldwide already upgrade their learning with 91影视!

One App. One Place for Learning.

All the tools & learning materials you need for study success - in one app.

Get started for free

Most popular questions from this chapter

An electron in an atom can "jump down" from a higher energy level to a lower one, then to a lower one still. The energy the atom thus loses at each jump goes to a photon. Typically, an electron might occupy a level for a nanosecond. What uncertainty in the electron's energy does this imply?

In the hydrogen atom, the electron鈥檚 orbit, not necessarily circular, extends to a distance of a about an angstrom (1脜=0.1鈥塶尘)from the proton. If it is to move about as a compact classical particle in the region where it is confined, the electron鈥檚 wavelength had better always be much smaller than an angstrom. Here we investigate how large might be the electron鈥檚 wavelength. If orbiting as a particle, its speed at 1脜could be no faster than that for circular orbit at that radius. Why? Find the corresponding wavelength and compare it to1脜 . Can the atom be treated classically?

One of the cornerstones of quantum mechanics is that bound particles cannot be stationary-even at zero absolute temperature! A "bound" particle is one that is confined in some finite region of space. as is an atom in a solid. There is a nonzero lower limit on the kinetic energy of such a particle. Suppose minimum kinetic energy of width L. Obtain an approximate formula for its minimum kinetic energy.

A beam of particles, each of mass m and (nonrelativistic) speed v, strikes a barrier in which there are two narrow slits and beyond which is a bunk of detectors. With slit 1 alone open, 100 particles are detected per second at all detectors. Now slit 2 is also opened. An interference pattern is noted in which the first minimum. 36 particles per second. Occurs at an angle of 30ofrom the initial direction of motion of the beam.

(a) How far apart are the slits?

(b) How many particles would be detected ( at all detectors) per second with slit 2 alone open?

(c) There are multiple answers to part (b). For each, how many particles would be detected at the center detector with both slits open?

10A beam of electrons strikes a barrier with two narrow but equal-width slits. A screen is located beyond the barrier. And electrons are detected as they strike the screen. The "center" of the screen is the point equidistant from the slits. When either slit alone is open,electrons arrive per second in a very small region at the center of the screen. When both slits are open, how many electrons will arrive per second in the same region at the center of the screen?

See all solutions

Recommended explanations on Physics Textbooks

View all explanations

What do you think about this solution?

We value your feedback to improve our textbook solutions.

Study anywhere. Anytime. Across all devices.