Heisenberg Equation

  • Thread starter Calabi
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  • #1
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Hello when I try to go form the Schrödinger picture to the Heisenberg picture I get this equality : [tex]
\begin{align}
{d \over dt} A(t)
= {i \over \hbar} H e^{iHt / \hbar} A e^{-iHt / \hbar} + e^{iHt / \hbar} \left(\frac{d A}{d t}\right) e^{-iHt / \hbar} + {i \over \hbar} e^{iHt / \hbar} A \cdot (-H) e^{-iHt / \hbar} \\
= {i \over \hbar} e^{iHt / \hbar} \left( H A - A H \right) e^{-iHt / \hbar} + e^{iHt / \hbar} \left(\frac{d A}{d t}\right) e^{-iHt / \hbar} \\
= {i \over \hbar } \left( H A(t) - A(t) H \right) + e^{iHt / \hbar} \left(\frac{d A}{d t}\right)e^{-iHt / \hbar} .
\end{align}[/tex]

I right [tex]A(t)[/tex] the Heisenberg picture of the operator(which depend of the time.). and [tex]A[/tex] the Schrödinger picture of the operator.

What permit me to say that : [tex]\frac{\partial A(t)}{\partial t} = e^{iHt / \hbar} \left(\frac{d A}{d t}\right)e^{-iHt / \hbar}[/tex] please?

I want to know if it's a Mathemtical definition or a physical definition.

Then I can say that with the classical Poisson brackets [tex]\{H, A\}[/tex] between the Hamiltonian and the Physical Quantity [tex]A[/tex] correspond to the [tex]\frac{i}{\hbar}[H, A][/tex] commutator between those 2 operator.

By the correspondance principle.

Thank you in advance and have a nice afternoon:biggrin:.
 

Answers and Replies

  • #2
140
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So what do you think please?

Thank you in advance and have a nice afternoon:biggrin:.
 
  • #3
140
2
So could someone help me please? I really need to have an answer please please please.

Thank you in advance and have a nice afternoon:biggrin:.
 
  • #4
bapowell
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dA/dt is zero because you are in the Schroedinger picture -- the time dependence is carried by the exp(iHt) and exp(-iHt) terms.
 
  • #5
vanhees71
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Let's look at it more carefully. For simplicity we assume that we have a non-explicitly time-dependent Hamiltonian. Now we consider two pictures of time evolution, namely the Schrödinger and the Heisenberg picture. It is of utmost importance to clearly distinguish all mathematical objects in the two pictures. One must never mix two pictures when evaluating physically observable quantities.

The Schrödinger picture is defined such that observables which are not explicitly time-dependent are also time-independent in the mathematical description, and the entire time evolution is on the states, which we can without loss of generality assume as pure states described by normalized Hilbert-space vectors. The time dependence of any self-adjoint operator ##\hat{O}_S(t)##, representing a not explicitly time-dependent observable, and the state ket ##|\psi_S(t) \rangle## in the Schrödinger picture thus is
$$\hat{O}_S(t)=\hat{O}_s(0)=\text{const}, \quad |\psi_S(t) \rangle=\exp(-\mathrm{i} \hat{H} t) |\psi_S(0) \rangle.$$
I've chosen ##t=0## as the initial time, and set ##\hbar=1##.

In the Heisenberg picture the operators representing observables carry the full time dependence, i.e.,
$$\hat{O}_H(t)=\exp(\mathrm{i} \hat{H} t) \hat{O}_H(0) \exp(-\mathrm{i} \hat{H} t), \quad |\psi_H(t) \rangle=|\psi_H(0) \rangle=\text{const}.$$
The transformation between the two pictures is unitary. We can easily find this transformation with the above definitions of the Schrödinger and Heisenberg pictures. Without loss of generality we can assume that both pictures coincide at ##t=0##. If this is not the case, there must be a time-independent unitary transformation of operators and kets to achieve this. So we can assume
$$\hat{O}_H(0)=\hat{O}_S(0), \quad |\psi_S(0) \rangle=|\psi_H(0) \rangle.$$
Then we have
$$|\psi_S(t) \rangle=\exp(-\mathrm{i} \hat{H} t) |\psi_S(0) \rangle = \exp(-\mathrm{i} \hat{H} t) |\psi_H(0) \rangle = \exp(-\mathrm{i} \hat{H} t) |\psi_H(t) \rangle.$$
Thus the unitary transformation is
$$\hat{U}_{SH}(t)=\exp(-\mathrm{i} \hat{H} t).$$
Now we have to check that this is consistent with the corresponding transformation for the operators:
$$\hat{O}_S(t)=\hat{O}_S(0)=\hat{O}_H(0)=\exp(-\mathrm{i} \hat{H} t) \hat{O}_H(t) \exp(\mathrm{i} \hat{H} t) = \hat{U}_{SH}(t) \hat{O}_H(t) \hat{U}_{SH}^{\dagger}(t),$$
which indeed shows the consistency between the transformation of the operators that represent observables with the transformation of the state kets.
 

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