Stationary State: Energy & Time-Independent Wave Function

[KFC]

For stationary state, the energy of the system is constant and time-independent, so for some specific form of potential, the Schrodinger can be variable separated such that the general solution will be


<br /> \psi(x, t) = \phi(x) \exp(-iE t /\hbar)<br />

The probability is given by the square of the wave function so the arbitrary phase factor \exp(iE t/\hbar) in above solution doesn't matter. However, if we consider the wave function itself, what's the meaning of the time-dependent factor \exp(iE t/\hbar). If it is stationary, why wavefunction is depending on time?


By the way, if the solution is non-stationary, is it no more any definite energy level? So in this case, there is no corresponding eigenvalue problem?

 

[meopemuk]

In quantum mechanics, wave function is actually not a good description of a quantum state. It is a redundant description. Two wave functions that differ only by a multiplicative (phase) factor correspond to the same physical state. So, the exponential time-dependent factor present in the wave function of any stationary state has no physical significance. It is just a mathematical redundancy.

 

[olgranpappy]

For stationary state, the energy of the system is constant and time-independent, so for some specific form of potential, the Schrodinger can be variable separated such that the general solution will be


<br /> \psi(x, t) = \phi(x) \exp(-iE t /\hbar)<br />

The probability is given by the square of the wave function so the arbitrary phase factor \exp(iE t/\hbar) in above solution doesn't matter. However, if we consider the wave function itself, what's the meaning of the time-dependent factor \exp(iE t/\hbar).

The meaning is just what you have indicated by using the symbol 'E'; the time-dependence of the state tells us the energy 'E'.

If it is stationary, why wavefunction is depending on time?

Even the "stationary" state depends on time but the dependence is "trivial". I.e., a simple exponential e^(-iEt).

By the way, if the solution is non-stationary, is it no more any definite energy level? So in this case, there is no corresponding eigenvalue problem?

You can still solve the eigenvalue problem
<br /> \hat H \psi = E \psi\;,<br />
and obtain the energy eigenvalues and they are still useful. For example, suppose you know that the state of the systems at t=0 is
<br /> \Phi=a\phi_1+b\phi_2\;,<br />
where \phi_1 and \phi_2 are eigenfunctions of the eigenvalue problem with eigenvalues E_1 and E_2, respectively. Knowing this you can determine the time dependence of the state \Phi, which is not "trivial" since \Phi is not an eigenstate of the hamiltonian (unless E_1=E_2).

 

[DeShark]

The probability is given by the square of the wave function

No it's not, it's given by the square of the modulus of the wave function. The modulus of exp(ikt) will always be 1, so it has no direct affect on the probabilities. However, when you combine two systems, then there is a possibility of interference, which is explained by the combination of the phases.

 

[dx]

So, the exponential time-dependent factor present in the wave function of any stationary state has no physical significance. It is just a mathematical redundancy.

Time dependent phase factors are not redundant. A constant phase factor is redundant, but multiplying by a time dependent factor changes many things of physical relevance, such as the energy.

 

[meopemuk]

Time dependent phase factors are not redundant. A constant phase factor is redundant, but multiplying by a time dependent factor changes many things of physical relevance, such as the energy.

In the Hilbert space the true representation of a state is not a vector |\Psi \rangle but a ray (1-dimensional subspace containing |\Psi \rangle). All vectors of the form \alpha(t)|\Psi \rangle (where \alpha(t) is any numerical function of t) belong to the same ray. Expectation values of all operators (e.g., energy) do not depend on the form of \alpha(t), so this phase factor is completely irrelevant for physics.