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| classes:2009:fall:phys4101.001:lec_notes_1021 [2009/10/23 22:45] – yk | classes:2009:fall:phys4101.001:lec_notes_1021 [2009/10/23 23:09] (current) – yk |
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| Only if BA=AB can we get that <math>(AB)^+=BA=AB</math>. AB is Hermitian. QED | Only if BA=AB can we get that <math>(AB)^+=BA=AB</math>. AB is Hermitian. QED |
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| ==== Part 3. Orthonormality for eigenstates ==== | ==== Part 3. Orthonormality for eigenstates of Hermitian operators ==== |
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| * We can verify that for infinite square well and simple harmonic oscillator, the eigenstates are orthonormal. (This is not proved in class, but if you are interested, you can see it in the end of this lecture notes) | * We can verify that for infinite square well and simple harmonic oscillator, the eigenstates are orthonormal. (This is not proved in class, but if you are interested, you can see it in the end of this lecture notes) |
| <math><\psi_n|\psi_m>=<n|m>=\int \psi_n^* \psi_m\, dx =\delta_n_m</math> | <math><\psi_n|\psi_m>=<n|m>=\int \psi_n^* \psi_m\, dx =\delta_n_m</math> |
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| * For degenerate states, we can use the Gram-Schmidt orthogonalization procedure to construct orthogonal eigenfunctions within each degenerate subspace. | Again, as it's proven in the textbook, this can be shown using the nature of the dot product, complex numbers, etc. This method works only when the eigenvalue of the two states are different. So what will happen is the eigenvalues happen to be the same for the two state? |
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| | * For degenerate states (sharing the same eigenvalues), we can use the Gram-Schmidt orthogonalization procedure to construct orthogonal eigenfunctions within each degenerate subspace. Since no students raised this as an interesting problem, we did not discussed in class. |
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| | ====Part 4: expanded statistical interpretation ==== |
| * Any state can be expressed by linear combination of eigenstates. | * Any state can be expressed by linear combination of eigenstates. |
| <math>|\psi>=\sum C_n |\psi_n> </math> | <math>|\psi>=\sum C_n |\psi_n> </math> |
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| <math>|C_n|^2</math> is the probability that the particle is found to have energy <math>E=E_n</math> | <math>|C_n|^2</math> was INTERPRETED AS the probability that the particle is found to have energy <math>E=E_n</math> |
| * <math>\sum |C_n|^2 = 1 </math> | |
| | If this interpretation to make sense (self consistent), at least a few relation must be satisfied. For example, |
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| | * <math>\sum |C_n|^2 = 1 </math> must be true. Otherwise, calling <math>|C_n|^2</math> probability is laughable. |
| proof: because of orthonormality, <math><\psi|\psi>=1</math>, <math><\psi_n|\psi_m>=\delta_n_m</math> | proof: because of orthonormality, <math><\psi|\psi>=1</math>, <math><\psi_n|\psi_m>=\delta_n_m</math> |
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| so <math>1=<\psi|\psi>=\sum_{n} \sum_{m} C_n^* C_m <\psi_n|\psi_m>=\sum_{n} \sum_{m} C_n^* C_m \delta_n_m=\sum_{n} |C_n|^2 </math>. QED | so <math>1=<\psi|\psi>=\sum_{n} \sum_{m} C_n^* C_m <\psi_n|\psi_m>=\sum_{n} \sum_{m} C_n^* C_m \delta_n_m=\sum_{n} |C_n|^2 </math>. QED |
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| * The expectation value of energy can be expressed as <math><\hat{H}>=\sum_{n} |C_n|^2 E_n </math> | * If <math>|C_n|^2</math> represents a probability, the expectation value should agree with the following: |
| | <math><\hat{H}>=\sum_{n} |C_n|^2 E_n </math> |
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| proof: in eigenstates, the expectation values are eigenvalues, which means <math><\hat{H}|\psi_n>=E_n|\psi_n></math> | proof: in eigenstates, the expectation values are eigenvalues, which means <math><\hat{H}|\psi_n>=E_n|\psi_n></math> |
| The expectation value of H in state <math>\psi</math> is: <math><\hat{H}>=<\psi|\hat{H}|\psi>=\sum_{n} \sum_{m} C_n^* C_m <\psi_n|\hat{H}|\psi_m> = \sum_{n} \sum_{m} C_n^* C_m E_m \delta_n_m = \sum_{n} |C_n|^2 E_n </math> QED. | The expectation value of H in state <math>\psi</math> is: <math><\hat{H}>=<\psi|\hat{H}|\psi>=\sum_{n} \sum_{m} C_n^* C_m <\psi_n|\hat{H}|\psi_m> = \sum_{n} \sum_{m} C_n^* C_m E_m \delta_n_m = \sum_{n} |C_n|^2 E_n </math> QED. |
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| **Proof of orthonormality of infinite square well and simple harmonic oscillator(this part is not mentioned in class)** | Based on these self consistency checks which work out, we keep(better word?) this generalized probability interpretation of <math>|C_n|^2</math>. |
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| | **Proof of orthonormality of infinite square well and simple harmonic oscillator(this part was not mentioned in class)** These are very complicated, so it's amazing that the proof for general case is much easier than these individual cases. |
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| * infinite square well | * infinite square well |
| The wave function is <math>\psi_n(x)=\sqrt{\frac{2}{L}} \sin\frac{n \pi x}{L} </math> | The wave function is <math>\psi_n(x)=\sqrt{\frac{2}{L}} \sin\frac{n \pi x}{L} </math> |
| The left hand side <math>e^{2ts} \int_{-\infty}^{+\infty} e^{-(t+s-\xi)^2} d\xi =e^{2ts} \int_{-\infty}^{+\infty} e^{-z^2} dz =e^{2ts}sqrt{\pi}=sqrt{\pi} \sum_{n=0}^{\infty} \frac{(2ts)^n}{n!} </math> | The left hand side <math>e^{2ts} \int_{-\infty}^{+\infty} e^{-(t+s-\xi)^2} d\xi =e^{2ts} \int_{-\infty}^{+\infty} e^{-z^2} dz =e^{2ts}sqrt{\pi}=sqrt{\pi} \sum_{n=0}^{\infty} \frac{(2ts)^n}{n!} </math> |
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| So we get <math>sqrt{\pi} \sum_{n=0}^{\infty} \frac{(2ts)^n}{n!}=\sum_{m=0}^{\infty}\sum_{n=0}^{\infty} \frac{t^m s^n}{m!n!} \int_{-\infty}^{+\infty} H_m(\xi) H_n(\xi)e^{-\xi^2} d\xi</math> | So we get <math>sqrt{\pi} \sum_{n=0}^{\infty} \frac{(2ts)^n}{n!}=\sum_{m=0}^{\infty}\sum_{n=0}^{\infty} \frac{t^m s^n}{m!n!} \int_{-\infty}^{+\infty} H_m(\xi) H_n(\xi)e^{-\xi^2} d\xi</math>. |
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| | This equality should hold for any values of //t// and //s//. Since both sides of the equation is polynomials of //t// and //s//, the coefficients for <math>t^m s^n</math> must equal to each other. |
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| If <math>n \neq m </math>, the left hand side is independent of m, while the right hand side is dependent of m. So in order to satisfy the equation the integral must be zero. <math>\int_{-\infty}^{+\infty} H_m(\xi) H_n(\xi)e^{-\xi^2} d\xi =0</math> | If <math>n \neq m </math>, the left hand side has no corresponding term (coefficient = zero), while the right hand side has those terms. So in order to satisfy the equation the integral must be zero. <math>\int_{-\infty}^{+\infty} H_m(\xi) H_n(\xi)e^{-\xi^2} d\xi =0</math> |
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| If n=m then <math>sqrt{\pi} \sum_{n=0}^{\infty} \frac{(2ts)^n}{n!}= \sum_{n=0}^{\infty} \frac{(ts)^n}{2(n!)} \int_{-\infty}^{+\infty} H_m(\xi) H_n(\xi)e^{-\xi^2} d\xi </math>, so <math>\int_{-\infty}^{+\infty} H_m(\xi) H_n(\xi)e^{-\xi^2} d\xi = sqrt{\pi}2^n n! </math> | If n=m then <math>sqrt{\pi} \frac{(2ts)^n}{n!}= \frac{(ts)^n}{(n!)^2} \int_{-\infty}^{+\infty} H_m(\xi) H_n(\xi)e^{-\xi^2} d\xi </math>, so <math>\int_{-\infty}^{+\infty} H_m(\xi) H_n(\xi)e^{-\xi^2} d\xi = sqrt{\pi}2^n n! </math> |
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| So <math>\int_{-\infty}^{+\infty} H_m(\xi) H_n(\xi)e^{-\xi^2} d\xi = sqrt{\pi}2^n n! \delta_m_n </math> | So <math>\int_{-\infty}^{+\infty} H_m(\xi) H_n(\xi)e^{-\xi^2} d\xi = sqrt{\pi}2^n n! \delta_m_n </math> |