Unitarity of Time-evolution Operator

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  • #1
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I am reading a quantum mechanics book. I did not clearly understand one particular idea.

When the book talks about the time-evolution operator [itex]U(t,t_0)[/itex], it says that one very important property is the unitary requirement for [itex]U(t,t_0)[/itex] that follows from probability conservation.

My question is, provided that the time-evoluation operator [itex]U(t,t_0)[/itex] satisfies the unitary requirement, that is, [itex]U(t,t_0)^{\dagger}U(t,t_0)=\mathbb{1}[/itex], how can I see and then proof explicitly that it indeed follows from probability conservation, that is, [itex]\sum_{a'}\left|c_{a'}(t_0)\right|^2=\sum_{a'}\left|c_{a'}(t)\right|^2[/itex]?

Thanks in advance for giving me a helping hand.

Reference: P.67, Modern Quantum Mechanics by Sakurai.
 

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  • #2
dextercioby
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This is typically contained in the discussion of the Wigner's theorem in textbooks treating symmetries in Quantum Theory. Look up this subject in the book by Fonda and Ghirardi (1970).
 
  • #3
tom.stoer
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My question is, provided that the time-evoluation operator [itex]U(t,t_0)[/itex] satisfies the unitary requirement, that is, [itex]U(t,t_0)^{\dagger}U(t,t_0)=\mathbb{1}[/itex], how can I see and then proof explicitly that it indeed follows from probability conservation, that is, [itex]\sum_{a'}\left|c_{a'}(t_0)\right|^2=\sum_{a'}\left|c_{a'}(t)\right|^2[/itex]?
I don't know what you're ca's are, but the general idea is the following

[tex]\langle \psi|\psi\rangle = 1\;\Rightarrow\;\langle \psi,t|\psi,t\rangle =
\langle U\psi|U\psi\rangle = ( \langle \psi|U^\dagger)\,(U|\psi \rangle ) = \langle \psi|(U^\dagger\,U)|\psi \rangle = \langle \psi|\psi\rangle = 1[/tex]
 
  • #4
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I read few pages of the book by Fonda and simply ensure that I am really understand.

Quote:
--- --- --- --- ---
Wigner's Theorem
Suppose that we have chosen a particular one-to-one vector mapping [itex]T[/itex] compatible with [itex]\mathbf{T}[/itex], of the coherent subspace [itex]\mathcal{H}_c[/itex] onto the coherent subspace [itex]\mathcal{H}_c[/itex], satisfying then [itex]\left|\left(T\psi,T\phi\right)\right|=\left|(\psi,\phi)\right|[/itex]
--- --- --- --- ---

From the conservation of probability, that is [itex]{\left|\left(U\psi,U\phi\right)\right|}^{2}=\left|(\psi,\phi)\right|^2[/itex], according to the Wigner's theorem we obtain that ray mapping [itex]\mathbf{T}[/itex] can be realised by a linear unitary vector mapping [itex]U[/itex].
 
  • #5
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I don't know what you're ca's are, but the general idea is the following

[tex]\langle \psi|\psi\rangle = 1\;\Rightarrow\;\langle \psi,t|\psi,t\rangle =
\langle U\psi|U\psi\rangle = ( \langle \psi|U^\dagger)\,(U|\psi \rangle ) = \langle \psi|(U^\dagger\,U)|\psi \rangle = \langle \psi|\psi\rangle = 1[/tex]
:surprised Nice. Thanks.
 
  • #6
dextercioby
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The c_a's are the typical notation for the Fourier coefficients.

From the Wigner theorem you get that the U's can be either linear and unitary or antilinear and antiunitary. Since time evolution is a continuous process/symmetry, it follows that the operators must be unitary.
 
  • #7
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dextercioby and tom.stoer, thanks very much.
 

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