Responsible party: Schrödinger's Dog, Devlin

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Main points

Schrodinger's Dog - Yuichi edited some which are often indicated by italics:

• 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> since this wave function is stripped of the time dependence part, the small letter <math>\psi</math> is better.

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 \strike{\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 \strike{\omega^2} x^2]\psi(x)=E\psi(x). *</math> or <math>k=m\omega^2</math>.

Now that we have our Schrodinger equation, we will “play” around with it, hopefully to make it simpler. One way to make a differential equation simpler is to reduce the order of the differentials. Factoring the LHS may do this trick as Yuichi suggested. By “playing” with the last expression, we hopefully will be successful in arriving to a first order differential equation. This idea may be inspired by the observation that the LHS has a structure similar to What we do first is look algebraic statement:

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

We may similarly wanna factor the left hand side of *. For simplicity so that a “big” picture is easier to see, we let the constants be 1. Then we arrive at the expression:

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

which we want to factor.

A small problem is, we are dealing with operators, which may act significantly differently from variables we are used to dealing with. Fortunately, this isn't too bad of an expression to play around with. We first see if

<math> \partial_x x=x\partial_x</math>.

To check this, we use the test function <math> \psi </math>, giving us:

<math> (\partial_x x)\psi=(x\partial_x)\psi</math>

The left hand side gives us:

<math> \psi+x(\partial_x \psi)</math>

and the right hand side gives us:

<math> x(\partial_x \psi) </math>.

Clearly, LHS does not equal the right hand side. But the expression,

<math> \partial_x x = x\partial_x +1 </math>

is true (you can check this yourself using a test function such as before).

We find using commutator notation that (look in the last section to learn about commutators) that

<math> [x,\partial_x]=-1 </math>

Hence we find that the Hamiltonian is:

<math> \hat{H}=(x+\partial_x)(x-\partial_x)-1 </math>

Using the fact that <math> p \approx -i \partial_x</math>, we define

<math> a_+=x+\partial_x=-ip+x</math>

and

<math> a_-=x-\partial_x=ip+x, </math>

where <math> a_+/a_- </math> turn out to be something called the “raising” and “lowering” operators, respectively for a reason we will find out.

Substituting <math> a_+/a_- </math> this into our Hamiltonian expression, we get:

<math> \hat{H}=p^2+x^2=(x+ip)(x-ip)-1</math> When proper constants (which have been ignored up to this point) are properly accounted for, this equals to <math>\hbar\omega(a_- a_{+}-\frac{1}{2})=\hbar\omega(a_{+}a_{-}+\frac{1}{2}). </math>

This gives usUsing the commutative relation between x and <math>\partial_x</math>, one can find the commutative relation:

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

Now that we have obtained our lowering and raising operators factored the Hamiltonian and named each of the factored terms <math>a_+</math> and <math>a_-</math>, can we obtain a 1st-order differential equation instead of 2nd order? No great idea to do this. So, we should “play” around with them again.

Let us use the fact that

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

Now, suppose we have found a solution <math>\psi</math>, which satisfies the Schrodinger equation. i.e.<math> \hat{H}\psi=E\psi,</math>. Sometimes, in math, we are lucky so that once we have one solution, we can find more by manipulating it. Perhaps, <math>a_+</math> and <math>a_-</math> can help in this respect! Would it be possible that <math> a_-\psi </math> is also a solution? (i.e. <math> \hat{H}(a_-\Psi)=E'(a_-\Psi),</math>)? (Since <math> \psi </math> and <math> a_-\psi </math> are different solutions, their energies don't have to be the same, thus E') 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> (equation A) is a given.

<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</math> Since <math>\psi</math> is the solution, it would be good to bring back <math>\hat{H}</math> to use the equation (A) above: <math> =a_-(a_-a_+ -1-\frac{1}{2})\psi=\hbar\omega a_-[\frac{\hat{H}}{\hbar\omega}-1]\Psi= a_-[\hat{H}-\hbar\omega]\Psi = a_-[E-\hbar\omega]\Psi=[E-\hbar\omega]a_-\Psi </math>

Hence we have shown that the eigenvalues of <math> a_-\Psi </math> is <math>E-\hbar\omega</math>.

By applying <math>a_-</math> many times, we can create many solutions, while their energies keep decreasing. Sooner or later, the energy will reach a negative value, but we have learned in Chap 1 that there cannot be a solution if the energy is smaller than the minimum potential energy, which in our case is zero. So when the energy reaches a negative value, i.e. suppose <math>\psi_0</math> is the last solution with a positive energy, <math>E_0</math> and <math>a_-\psi</math> would have a negative energy, this wave function cannot be a sensible wave function. One possibility is the it is a non-normalizable function, but zero is another way to get out of this nonsense. If the latter gives us a sensible solution, we will be happy. The former will be much harder to deal with so we want to avoid. (If you feel I am a bit wishy washy, I am.

Since E cannot be negative, because we are dealing with a potential V>0, then E can never be negative. This means that if we apply the lowering operator a number of times, we should get that

<math> a_- \Psi=0 </math>.

Now that we have a first order differential equation that we were hoping to find instead of 2nd-order one, we can find the solution of the Schrodinger equation more easily and once that's done use the raising operator on it to find many more solutions. This concludes the end of lecture.

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 (algebra) 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 Uncertainty relation|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.

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

• We discussed how to algebraically solve Schrodinger's equation for a SHO using ladder operators.

To do this we began with the idea of simplifying the Hamiltonian. One way is to factor it. When factoring we can change the expression into a first order differential. We do this with a price though, it adds an extra variable. It is easy to see how the Hamiltonian could be factored when it is arranged like so

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

There, the two different square factors are easily identified. This makes it look like it could factor the same way <math> (x^2-y^2) </math> would.

<math> (x^2-y^2) = (x+y)(x-y) (1*)</math>

except in the Hamiltonian case, we are looking at something like this (ignoring various factors involving mass, Planck constant, angular frequency, …

<math> (x^2-\partial^2 x)=(x+\partial x)(x-\partial x). (2*) </math>

Only there is one key difference. The Hamiltonian is an operator and does not quite behave like a normal variable. In equation (1*), <math> yx=xy.</math> But in equation (2*), <math> (x)(\partial x)\not= (\partial x)(x). </math>.

To deal with this difference, we will correct for the difference in the two sides of the inequality above. We also introduce the idea of raising and lowering operators <math> (a_-,a_+) </math>. They are called this because it turns out that you can use them to “climb up” (raise) the “ladder of solutions” and vice versa. We also introduce the canonical commutation relation <math> [x,p]=i\hbar. </math> Using these, we come up with the equation <math> [a_-,a_+]=1. </math>

With these, the Hamiltonian can be written as <math> H=\hbar\omega(a_+a_- + 1/2) </math>.

Which makes the Schrodinger equation <math> \hbar\omega(a_\mp a_\pm \pm (1/2))\psi=E\psi </math>

With this, we claim that if <math> \psi </math> satisfies the Schrodinger equation with energy E, then <math> a_\pm\psi </math> satisfies the Schrodinger equation with energy <math> (E\pm\hbar\omega). </math> We can use this result/assumption to find new solutions for higher and lower energies.

This works well, but a few things must be noted. Ladder operators can not be used universally. Also, there is no guarantee that each solution will be normalizable so we define a “lowest rung” to be <math> a_-\psi_0=0. </math> We can use this to find the energy state at the lowest rung and then continuously use the raising operator (<math> a_+ </math>) to find the excited states.

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