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| classes:2009:fall:phys4101.001:lec_notes_0918 [2009/09/20 17:42] – gebrehiwet | classes:2009:fall:phys4101.001:lec_notes_0918 [2009/09/20 20:11] (current) – yk |
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| Main points | Main points |
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| Schrodinger's Dog: | Schrodinger's Dog - Yuichi edited some which are often indicated by //italics//: |
| ==The main points of today's lecture:== | ==The main points of today's lecture:== |
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| *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) | *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...) | *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. | *We learned about commutators, which is instrumental in finding the wave-function algebraically. |
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| 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: | 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: |
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| <math> \hat{H}\Psi(x)=E\Psi(x),</math> | <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.// |
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| 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: | 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: |
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| <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> | <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> |
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| Putting the above expression in the Schrodinger equation, gives us: | Putting the above expression in the Schrodinger equation, gives us: |
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| <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> | <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>. |
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| ==Solving the Schrodinger's equation for a Simple Harmonic Potential Algebraically (Part 2):== | ==Solving the Schrodinger's equation for a Simple Harmonic Potential Algebraically (Part 2):== |
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| 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: | 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. <del>By "playing" with the last expression, we hopefully will be successful in arriving to a first order differential equation.</del> //This idea may be inspired by the observation that the LHS has a structure similar to // <del>What we do first is look algebraic statement</del>: |
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| <math> x^2-y^2=(x+y)(x-y).</math> | <math> x^2-y^2=(x+y)(x-y).</math> |
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| | 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: |
| We similarly wanna factor the left hand side of *. For simplicity, we let the constants be 1, so we arrive at the expression of: | |
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| <math> -\frac{\partial^2}{\partial x^2}+x^2,</math> | <math> -\frac{\partial^2}{\partial x^2}+x^2,</math> |
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| which we want to factor. Problem is, we are dealing with operators, which acts significantly different from variables we are used to dealing with. Fortunely, this isn't to bad of an expression to play around with. We first see if | which we want to factor. |
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| | //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 |
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| <math> \partial_x x=x\partial_x</math>. | <math> \partial_x x=x\partial_x</math>. |
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| To check this, we use the test function <math> \Psi </math>, giving us: | To check this, we use the test function <math> \psi </math>, giving us: |
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| <math> (\partial_x x)\Psi=(x\partial_x)\Psi</math> | <math> (\partial_x x)\psi=(x\partial_x)\psi</math> |
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| The left hand side gives us: | The left hand side gives us: |
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| <math> \Psi+x(\partial_x \Psi)</math> | <math> \psi+x(\partial_x \psi)</math> |
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| and the right hand side gives us: | and the right hand side gives us: |
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| <math> x(\partial_x \Psi) </math>. | <math> x(\partial_x \psi) </math>. |
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| Clearly, LHS does not equal the right hand side. But the expression, | Clearly, LHS does not equal the right hand side. But the expression, |
| is true (you can check this yourself using a test function such as before). | is true (you can check this yourself using a test function such as before). |
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| We find using commutaor notation that (look in the last section to learn about commutators) that | We find using commutator notation that (look in the last section to learn about commutators) that |
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| <math> [x,\partial_x]=-1 </math> | <math> [x,\partial_x]=-1 </math> |
| <math> \hat{H}=(x+\partial_x)(x-\partial_x)-1 </math> | <math> \hat{H}=(x+\partial_x)(x-\partial_x)-1 </math> |
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| Using the fact that <math> p \approx i \partial_x</math>, we define | Using the fact that <math> p \approx -i \partial_x</math>, we define |
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| <math> a_+=x+\partial_x=-ip+x</math> | <math> a_+=x+\partial_x=-ip+x</math> |
| <math> a_-=x-\partial_x=ip+x, </math> | <math> a_-=x-\partial_x=ip+x, </math> |
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| where <math> a_+/a_- </math> are the raising and lowering operators respectively. | where <math> a_+/a_- </math> turn out to be something called the "raising" and "lowering" operators, respectively for a reason we will find out. |
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| Substituting <math> a_+/a_- </math> this into our Hamiltonian expression, we get: | Substituting <math> a_+/a_- </math> this into our Hamiltonian expression, we get: |
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| <math> \hat{H}=p^2+x^2=(x+ip)(x-ip)-1=a_-a_{+}-\frac{1}{2}=a_{-}a_{+}+\frac{1}{2}. </math> | <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> |
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| This gives us the commutative relation: | <del>This gives us</del>Using the commutative relation between //x// and <math>\partial_x</math>, one can find the commutative relation: |
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| <math> [a_-,a_+]=1.</math> | <math> [a_-,a_+]=1.</math> |
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| ==Solving the Schrodinger's equation for a Simple Harmonic Potential Algebraically (Part 3):== | ==Solving the Schrodinger's equation for a Simple Harmonic Potential Algebraically (Part 3):== |
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| Now that we have obtained our lowering and raising operators, we can "play" around with them again. Let us use the fact that | Now that we have <del>obtained our lowering and raising operators</del> 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. |
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| <math> [a_-,a_+]=1.</math> | <del>Let us use the fact that </del> |
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| Now, suppose we found a solution <math>\Psi</math>, which satisfies <math> \hat{H}\Psi=E\Psi,</math>. | <del><math> [a_-,a_+]=1.</math> </del> |
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| 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>: | 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') <del>Then we should find that the eigenvalue of this solution is <math> E-\hbar\omega </math>:</del> |
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| Proof: | Proof: |
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| <math> \hat{H}=(a_-a_-\frac{1}{2})\hbar\omega,</math> | <math> \hat{H}=(a_-a_-\frac{1}{2})\hbar\omega,</math> (equation A) is a given. |
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| Then | |
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| <math> \hat{H}(a_-\Psi)=(a_-a_- -\frac{1}{2})\hbar\omega a_\Psi-= \hbar\omega[a_-a_+a_- -\frac{a_-}{2}]\Psi= | <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 = a_-[E-\hbar\omega]\Psi=[E-\hbar\omega]a_-\Psi </math> | [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> |
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| Hence we have shown that the eigenvalues of <math> a_-\Psi </math> is <math>E-\hbar\omega</math>. | Hence we have shown that the eigenvalues of <math> a_-\Psi </math> is <math>E-\hbar\omega</math>. |
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| 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 | //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.// |
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| | <del>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</del> |
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| <math> a_- \Psi=0 </math>. | <math> a_- \Psi=0 </math>. |
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| Now that we have a first order differential equation, we can find the solution of the Schrodinger equation and use the raising operator on it. This concludes the end of lecture. | 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. |
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| Devlin: | ==== Devlin's version! ==== |
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| <math> \hat{H}\Psi(x)=E\Psi(x),</math> | <math> \hat{H}\psi(x)=E\psi(x),</math> |
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| ==Simple Harmonic Oscillator:== | ==Simple Harmonic Oscillator:== |
| <math> (x^2-y^2) = (x+y)(x-y) (1*)</math> | <math> (x^2-y^2) = (x+y)(x-y) (1*)</math> |
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| except in the Hamiltonian case, we are looking at something like this | except in the Hamiltonian case, we are looking at something like this (ignoring various factors involving mass, Planck constant, angular frequency, ... |
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| <math> (x^2-\partial^2 x)=(x+\partial x)(x-\partial x). (2*) </math> | <math> (x^2-\partial^2 x)=(x+\partial x)(x-\partial x). (2*) </math> |
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| Only there is one key difference. The Hamiltonian is an operator and does not 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>. | 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>. |
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| To deal with this difference, we introduce the idea of raising and lowering operators <math> (a_-,a_+) </math>. They are called this because 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 | 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> | <math> [a_-,a_+]=1. </math> |
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| With these, the Hamiltonian can be written as | With these, the Hamiltonian can be written as |
| <math> H=\hbar\omega(a_+,a_- + 1/2) </math>. | <math> H=\hbar\omega(a_+a_- + 1/2) </math>. |
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| Which makes the Schrodinger equation | Which makes the Schrodinger equation |
| <math> \hbar\omega(a_-,a_+ (\+/-) (1/2))\psi=E\psi </math> | <math> \hbar\omega(a_\mp a_\pm \pm (1/2))\psi=E\psi </math> |
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| With this, we claim that if <math> \psi </math> satisfies the Schrodinger equation with energy E, then <math> a_+\psi </math> satisfies the Schrodinger equation with energy <math> (E+\hbar\omega). </math> We can use this result/assumption to find new solutions for higher and lower energies. | 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. |
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| 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 | 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 |