Quantum Mechanics:Evolution of Superposition state

Classical Mechanics Level 2
  • Let ψ 1 ( x ) \psi_1(x) and ψ 2 ( x ) \psi_2(x) be two normalized stationary states with energies E 1 E_1 and E 2 E_2 ,respectively with E 2 > E 1 E_2> E_1 .At t=0 the state of the system Ψ \Psi Ψ ( x , 0 ) = 1 2 ( ψ 1 ( x ) + ψ 2 ( x ) ) . \begin{aligned} \Psi (x,0)=\frac{1}{\sqrt {2}}\left(\psi _1(x)+\psi _2(x)\right). \end{aligned}
  • Find Ψ ( x , t ) \Psi(x,t)
  • 1. Ψ ( x , t ) = 1 2 ( e i E 1 t / ℏ ψ 1 ( x ) + e i E 2 t / ℏ ψ 2 ( x ) ) \Psi(x,t)=\frac{1}{\sqrt{2}}(e^{iE_1t/\hbar}\psi_1(x)+e^{iE_2t/\hbar}\psi_2(x))
  • 2. Ψ ( x , t ) = 1 2 ( e − i E 1 t / ℏ ψ 1 ( x ) + e − i E 2 t / ℏ ψ 2 ( x ) ) \Psi(x,t)=\frac{1}{\sqrt{2}}(e^{-iE_1t/\hbar}\psi_1(x)+e^{-iE_2t/\hbar}\psi_2(x))
  • 3. Ψ ( x , t ) = 1 2 ( e − i E 1 t / ℏ ψ 1 ( x ) + i e − i E 2 t / ℏ ψ 2 ( x ) ) \Psi(x,t)=\frac{1}{\sqrt{2}}(e^{-iE_1t/\hbar}\psi_1(x)+ie^{-iE_2t/\hbar}\psi_2(x))
  • 4. Ψ ( x , t ) = 1 2 ( ψ 1 ( x ) + ψ 2 ( x ) ) e i ( E 1 + E 2 ) t / ℏ \Psi(x,t)=\frac{1}{\sqrt{2}}(\psi_1(x)+\psi_2(x))e^{i(E_1+E_2)t/\hbar}
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∆Lex Palacin by ∆lex Palacin , 32, Spain

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1 solution

∆Lex Palacin
Dec 30, 2018

Since ψ 1 ( x ) \psi_1(x) and ψ 2 ( x ) \psi_2(x) are energy eigenstates, their evolution in time is given simply by ψ 1 ( x ) → e − i E 1 t / ℏ ψ 1 ( x ) \psi_1(x)\rightarrow e^{-iE_1t/\hbar}\psi_1(x) and ψ 2 ( x ) → e − i E 2 t / ℏ ψ 2 ( x ) \psi_2(x)\rightarrow e^{-iE_2t/\hbar}\psi_2(x) Using superposition, we then find Ψ ( x , t ) = 1 2 ( e − i E 1 t / ℏ ψ 1 ( x ) + e − i E 2 t / ℏ ψ 2 ( x ) ) \Psi(x,t)=\frac{1}{\sqrt{2}}(e^{-iE_1t/\hbar}\psi_1(x)+e^{-iE_2t/\hbar}\psi_2(x))

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