Responsible party: Schrödinger's Dog, Devlin

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Main points

• Simple Harmonic Potential and how to recover the the wave-function from the Schrodinger Equation algebraically. It was also mentioned that it can be solved analytically(through the use of power series)
• While we were solving the Schrodinger equation, we learned the properties of operators and how to test equality using a “test function” (i.e. <math> \Psi(x), f(x) </math>, etc…)
• We learned about commutators, which is instrumental in finding the wave-function algebraically.

In lecture, we started by discussing how to solve Schrodinger's equation for a simple harmonic potential algebraically. We started with the basic Schrodinger's equation:

<math> \hat{H}\Psi(x)=E\Psi(x),</math>

where <math> \hat{H}</math>, is our Hamiltonian, E is the Energy, and <math> \Psi(x)</math> is the wave-function. We try and solve for <math> \Psi(x)</math>. Substituting our simple harmonic potential and the momentum operator into our Hamiltonian, we get:

<math> \hat{H}=\frac{p^2}{2m}+V(x)=-\frac{\hbar^2}{2m}\frac{\partial^2}{\partial x^2}+\frac{1}{2} k \omega x^2. </math>

Putting the above expression in the Schrodinger equation, gives us:

<math> [-\frac{\hbar^2}{2m}\frac{\partial^2}{\partial x^2}+\frac{1}{2}k \omega^2 x^2]\Psi(x)=E\Psi(x). *</math>

Now that we have our Schrodinger equation, we will “play” around with it, as Yuichi suggested. By “playing” with the last expression, we hopefully will be successful in arriving to a first order differential equation. What we do first is look algebraic statement:

<math> x^2-y^2=(x+y)(x-y).</math>

We similarly wanna factor the left hand side of *. For simplicity, we let the constants be 1, so we arrive at the expression of:

<math> -\frac{\partial^2}{\partial x^2}+x^2,</math>

which we want to factor. Problem is, we are dealing with operators, which acts significantly different from variables we are used to dealing with.

Now that we have obtained our lowering and raising operators, we can “play” around with them again. Let us use the fact that

<math> [a_-,a_+]=1.</math>

Now, suppose we found a solution <math>\Psi</math>, which satisfies <math> \hat{H}\Psi=E\Psi,</math>.

Now suppose <math> a_-\Psi </math> is also solution (i.e. <math> \hat{H}(a_-\Psi)=E(a_-\Psi),</math>). Then we should find that the eigenvalue of this solution is <math> E-\hbar\omega </math>:

Proof:

<math> \hat{H}=(a_-a_-\frac{1}{2})\hbar\omega,</math>

Then

<math> \hat{H}(a_-\Psi)=(a_-a_-\frac{1}{2})\hbar\omega a_\Psi-= \hbar\omega[a_-a_+a_-\frac{a_-}{2}]\Psi=[a_-(a_+a_-\frac{1}{2}]\Psi=a_-(a_-a_+ -1-frac{1}{2})\Psi=\hbar\omega a_-(\frac{\hat{H}}{\hbar\omega}-1]\Psi= a_-[\hat{H}-\hbar\omega]\Psi </math>

The defintion of a commutators is:

<math> [x,y]=xy-yx.</math>

We use this definition to find relations for the operators we have worked with to find commutative relation between them(i.e. what needs to happen in order to switch the order of the operators which you are commuting). The way we use commutative relations to solve the Schrodinger's equation is illustrated above in part3 and part 4.

The rest of this section is extracted from wikipedia, if you are interested in learning more about commutators and there propeties.

The 'commutator' of two elements a and b of a ring or an associative algebra is defined by :[a, b] = ab -ba.

It is zero if and only if a and b commute. In linear algebra, if two endomorphisms of a space are represented by commuting matrices with respect to one basis, then they are so represented with respect to every basis. By using the commutator as a Lie bracket, every associative algebra can be turned into a Lie algebra. The commutator of two operators defined on a Hilbert space is an important concept in quantum mechanics since it measures how well the two observables described by the operators can be measured simultaneously. The uncertainty principle is ultimately a theorem about these commutators via the Robertson-Schrödinger relation.

The commutator has the following properties:

Lie-algebra relations: * <math>[A,A] = 0</math>

* <math>[A,B] = -[B,A]</math>

* <math>[A,[B,C]] + [B,[C,A]] + [C,[A,B]] = 0</math>

The second relation is called anticommutativity, while the third is the Jacobi identity.

Additional relations:

* <math> [A,BC] = [A,B]C + B[A,C]</math>

* <math> [AB,C] = A[B,C] + [A,C]B</math>

* <math> [A,BC] = [AB,C] + [CA,B]</math>

* <math> [ABC,D] = AB[C,D] + A[B,D]C + [A,D]BC</math>

* <math> [A,B], C], D] + [[[B,C], D], A] + [[[C, D], A], B] + [[[D, A], B], C] = [[A, C], [B, D</math>

* <math> [AB,C]=\{A,B\}C-C\{A,B\}+\{C,B\}A-B\{A,C\}</math>, where {A,B}=AB+BA is the anticommutator.

-Schrodinger's Dog is gonna take a quick nap, but will finish this before you nap!

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