New User Special Price Expires in

Let's log you in.

Sign in with Facebook


Don't have a StudySoup account? Create one here!


Create a StudySoup account

Be part of our community, it's free to join!

Sign up with Facebook


Create your account
By creating an account you agree to StudySoup's terms and conditions and privacy policy

Already have a StudySoup account? Login here

Introduction to Quantum Mechanics I

by: Ms. Bryce Wisoky

Introduction to Quantum Mechanics I PHY 567

Marketplace > Syracuse University > Physics 2 > PHY 567 > Introduction to Quantum Mechanics I
Ms. Bryce Wisoky
GPA 3.93


Almost Ready


These notes were just uploaded, and will be ready to view shortly.

Purchase these notes here, or revisit this page.

Either way, we'll remind you when they're ready :)

Preview These Notes for FREE

Get a free preview of these Notes, just enter your email below.

Unlock Preview
Unlock Preview

Preview these materials now for free

Why put in your email? Get access to more of this material and other relevant free materials for your school

View Preview

About this Document

Class Notes
25 ?




Popular in Course

Popular in Physics 2

This 4 page Class Notes was uploaded by Ms. Bryce Wisoky on Wednesday October 21, 2015. The Class Notes belongs to PHY 567 at Syracuse University taught by Staff in Fall. Since its upload, it has received 12 views. For similar materials see /class/225636/phy-567-syracuse-university in Physics 2 at Syracuse University.

Similar to PHY 567 at Syracuse


Reviews for Introduction to Quantum Mechanics I


Report this Material


What is Karma?


Karma is the currency of StudySoup.

You can buy or earn more Karma at anytime and redeem it for class notes, study guides, flashcards, and more!

Date Created: 10/21/15
Lecture 4 Ehrenfest s theorem The connection to classical physics can be made more explicit in a celebrated theorem due to Paul Ehrenfest which shows that quantum expectation values evolve according to Newton s 2nd law Consider the time rate of change of the expectation value of the momentum lt p gt dltPgt 2 2 dt dl8tqji8xqj Using Leibnitz we nd dltpgt dx8j i It at 81 8x87 Substituting in from the Schroedinger equation and using integration by parts we can SOC d ltp gt dt The right hand side is nothing but the expectation value of the force on the particle and we have the result Similarly it is to see that dltxgtlt gt dt p Summary so far QM tells us that the most information we can obtain about a microscopic particle is contained in its wavefunction x Once we know the wave function at one time Schroedinger s equation allows you to calculate it at any later time Physical particles are described by normalized wavefunctions flIll 1 In fact QM tells us to interpret llll a probability density This then allows us to write expressions for the expectation value of some observable quantity Q l p lt Q gt Vow where the QM operator Q is just obtained from its classical expression by replacing x with just 1 and p by Time independent Schroedinger equation OK we have the Schroedinger equation but how do we go about solving it It turns out that if the potential V is independent of time this may be accomplished by a method termed the separation of variables What this means is that we seek solutions of the form We 75 Motif The justi cation for this is three fold o More general solutions can be built up from these separable solutions 0 They turn out to be states of de nite energy 0 Expectation values in these states are independent of time They are also termed stationary states If we do this we nd that the Schroedinger equation reduces to two ordinary differential equations ft exp iEth 1 MW E l 2 E is a constant which we will identify shortly the energy of the state and H is the Hamiltonian 72 d2 I a H 2m l This second equation involving the Hamiltonian 2 is called the timeindependent Schroedinger equation Notice that advertised the probability density V is time independent Furthermore the operator H is a QM version of the classical energy function for the system and from eqn2 has an expecta tion value equal to E Furthermore any power of H has expectation value just equal to E raised to that power Thus the variance of the probability distribution for the energy is zero the distribution is trivial Thus any mea surement of the energy will return exactly E stationary states are states of xed energy But what is the energy E so far we have not speci ed it In general we will see that the energy E can take on an in nite number of discrete values dependent on the nature of the potential V We will call these values E1 E2 E3 and to each allowed value of E there will be an associated solution to the timeindependent Schroedinger equation p1xp2x It is a theorem we will not attempt to prove it that the most general solution to Schroedinger s equation is a linear combination of these stationary state solutions the cn s are constants x t Z cn nl E iEquot h These coefficients on can be usually found from a knowledge of the wave function at t 0 and the solution of the timeindependent problem The moral of the story is that once we have solved the timeindependent equa tion we have very little left to do to find the most general solution to the time dependent Schroedinger equation Furthermore the timeindependent equation does not contain i and so we can just look for real solutions of this equation Examples The in nite square well Suppose V 0 for 0 g x g a and is in nite elsewhere A particle is permanently confined inside this potential well It could be thought of a very crude model for a single electron atom Classically a particle confined to such a system would just bounce back and forth at constant speed Its energy could take on any value We will see that in QM the allowed possible energies are discrete First notice that p 0 for 1 lt 0 and 1 gt a since there is no probability of finding the particle outside the well Inside the well where V 0 the timeindependent equation reduces to 4 d2 2m 112 E Assuming that E gt 0 we may introduce the variable k xQmEh2 and write this equation 2 L k2ltp dxz The general solution to this is Asinkx8coskx 3 The constants A and B are xed by applying the boundary conditions M0 0 and Ma 0 This yields 8 0 and the quantizatian canditian sin ka 0 The latter means that ka mr Thus not all wavelengths are allowed only those which correspond to standing waves in the well We have seen that the Bohr quantization condition for the hydrogen atom could be understood on a similar basis here we see for the rst time that the formal theory of QM is able to explain many of the quantum phenomena which had been observed and which had proven so difficult for classical physics to account for What remains well we still have to normalize the solution that is the origin of the remaining freedom in the constant A Thus we nd that the stationary states of this potential are of the form sinnlx an a a The energy of this state is En We speak of the ground state39s the state of lowest energy which here corresponds to n 1 with E1 Classically the state of lowest energy corresponds to the particle at rest with E 0 We see in QM that such a state is impossible The minimum energy the particle can have is El which increases we con ne the particle to smaller and smaller regions a gt 0 This is a corollary of a very general theorem in QM called Heisenberg s uncertainty principle This roughly states that the more accurately ones knows the position of a quantum particle the less certain we are about its momentum and hence its energy In fact the product of the uncertainty in its position times the uncertainty in its momentum is always greater than some minimum which is equal to hQ We will prove this theorem later in the course but one immediate corollary is that no particle can ever be at rest at a point since then it would have a wellde ned position and momentum zerol Thus even at zero temperature particles always su 39er fluctuations in their positions and momenta they are somewhat smeared out This of course is required if they sometimes behave like waves Other points to notice 91 increases the number of zero crossings of the wavefunction increases By symmetry the expectation value for x is at x aQ for all states


Buy Material

Are you sure you want to buy this material for

25 Karma

Buy Material

BOOM! Enjoy Your Free Notes!

We've added these Notes to your profile, click here to view them now.


You're already Subscribed!

Looks like you've already subscribed to StudySoup, you won't need to purchase another subscription to get this material. To access this material simply click 'View Full Document'

Why people love StudySoup

Bentley McCaw University of Florida

"I was shooting for a perfect 4.0 GPA this semester. Having StudySoup as a study aid was critical to helping me achieve my goal...and I nailed it!"

Anthony Lee UC Santa Barbara

"I bought an awesome study guide, which helped me get an A in my Math 34B class this quarter!"

Jim McGreen Ohio University

"Knowing I can count on the Elite Notetaker in my class allows me to focus on what the professor is saying instead of just scribbling notes the whole time and falling behind."

Parker Thompson 500 Startups

"It's a great way for students to improve their educational experience and it seemed like a product that everybody wants, so all the people participating are winning."

Become an Elite Notetaker and start selling your notes online!

Refund Policy


All subscriptions to StudySoup are paid in full at the time of subscribing. To change your credit card information or to cancel your subscription, go to "Edit Settings". All credit card information will be available there. If you should decide to cancel your subscription, it will continue to be valid until the next payment period, as all payments for the current period were made in advance. For special circumstances, please email


StudySoup has more than 1 million course-specific study resources to help students study smarter. If you’re having trouble finding what you’re looking for, our customer support team can help you find what you need! Feel free to contact them here:

Recurring Subscriptions: If you have canceled your recurring subscription on the day of renewal and have not downloaded any documents, you may request a refund by submitting an email to

Satisfaction Guarantee: If you’re not satisfied with your subscription, you can contact us for further help. Contact must be made within 3 business days of your subscription purchase and your refund request will be subject for review.

Please Note: Refunds can never be provided more than 30 days after the initial purchase date regardless of your activity on the site.