ps6 - E x ( z ) is ˆ E x ( z ) = E sin( kz ) ± a + a †...

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Harvard University Physics 143b: Quantum Mechanics II Problem Set 6 due Friday November 5 Problem numbers refer to Griffiths, second edition 1. Griffiths 9.11 2. Griffiths 9.14 (read Section 9.3.2 and 9.3.3, including material not covered in class). 3. Griffiths 9.21 Do this by the quantized electric field method developed in class. We wrote the electric field as ~ E ( ~ r ) i X ~ k,s ~e ~ k,s ± a ~ k,s e i ~ k · ~ r - a ~ k,s e - i ~ k · ~ r ² where ~e ~ k,s is the polarization vector, and a ~ k,s is the photon annihilation operator. We approximated e i ~ k · ~ r 1 in class. Here keep the next term with e i ~ k · ~ r 1 + i ~ k · ~ r. Hence obtain the answer in (9.95) with ˆ n the polarization vector. Proceed using (9.95) to work out (b) and (c). 4. Consider an optical cavity with a single mode of the electromagnetic field, with an electric field polarized in the x direction, and frequency ω . So the electric field operator ˆ
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Unformatted text preview: E x ( z ) is ˆ E x ( z ) = E sin( kz ) ± a + a † ² where E is a normalization constant, and a is the photon annihilation operator. We also define the photon number operator ˆ n by ˆ n = a † a. (a) Let the state of cavity field at time t = 0 be | Ψ(0) i = | m i , where m is an integer. Determine | Ψ( t ) i . Then compute the average values of ˆ E x and ˆ n at time t , along with their variances σ E x ( t ) and σ n ( t ) as defined in Section 3.5.1. (b) As above, but for | Ψ(0) i = ( | m i + e iϕ | m + 1 i ) / √ 2, where m is a positive integer, and ϕ is a phase. 1 (c) Use (3.62) to establish an uncertainty relation between the electric field and the photon number. Verify that your answers above are consistent with this uncer-tainty relation. 2...
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This note was uploaded on 02/15/2011 for the course PHYS 143B taught by Professor Unknow during the Spring '10 term at Harvard.

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ps6 - E x ( z ) is ˆ E x ( z ) = E sin( kz ) ± a + a †...

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