TransWikia.com

Time evolution of a state on adding a constant to the Hamiltonian of the Schrodinger equation

Physics Asked by a_point_particle on February 21, 2021

Adding a constant to a potential energy function does not change the dynamics (time evolution) of an object in Newtonian physics. I expect the same to be true in quantum mechanics but for some reason, I’m getting a result that contradicts this.

$$ ihbar ,partial_t Psi = hat{H}Psi $$
$$ hat{H}phi_j=E_jphi_j$$
$$implies ihbar,partial_t phi = E_jphi_j$$
$$impliesphi_j(t)=phi_j;expleft(-frac{i,E_j,t}{hbar}right)$$
$$Psi(0) = sum_j c_j,phi_j implies Psi(t)=sum_j c_j,phi_j,expleft(-frac{i,E_j,t}
{hbar}right)$$

Now consider adding a constant $V_0$ to the hamiltonian, the eigenfunctions will be the same $phi_j$ whereas the energy eigenvalues will become $E_j+V_0$.

$$hat{H}^prime=hat{H}+V_0$$
$$ihbar ,,partial_txi = (hat{H}+V_0),xi$$
$$ (hat{H}+V_0)phi_j=(E_j+V_0)phi_j $$
$$xi(0) = Psi(0) =sum_j c_j,phi_j implies xi(t)=sum_j c_j,phi_j,expleft(-frac{i,(E_j+V_0),t}{hbar}right)$$

Now despite $xi(0) = Psi(0)$ and the only difference between the two hamiltonians being a constant $V_0$ the time evolution of $Psi$ and $xi$ are different (the argument in the exponential term is different).

I have a strong feeling that I have made a mistake in this but I can’t see where and would be grateful for any insights.

2 Answers

You can simply factor out $expleft(-frac{i,V_0,t}{hbar}right)$ from $$xi(t)=sum_j c_j,phi_j,expleft(-frac{i,(E_j+V_0),t}{hbar}right) $$ to get $$xi(t)=expleft(-frac{i,V_0,t}{hbar}right)sum_j c_j,phi_j,expleft(-frac{i,E_j,t}{hbar}right) $$ which is a global phase shift, and so has no observable effect since states are represented by rays in Hilbert space. Naturally, though, the energy eigenvalues are shifted evenly as $E_j rightarrow E_j + V_0$.

Correct answer by Nihar Karve on February 21, 2021

What you've essentially done is added a universal phase factor to the total wavefunction: $$xi(t) = sum_j c_j phi_j , e^{frac{-i (E_j + V_0) t}{hbar}} = sum_j c_j phi_j , e^{frac{-i E_j t}{hbar}}left(e^{frac{-i V_0 t}{hbar}}right).$$

This phase factor $e^{frac{-i V_0 t}{hbar}},$ which affects all the eigenfunctions, does not affect the dynamics of the system, since it disappears when you calculate the expectation value of an observable, such as momentum. That is, it cancels out with its own complex conjugate, giving just unity. The coefficients $c_j$ are still the same.

The expected energy values are still shifted by $V_0$, however, which is unsurprising.

Answered by Yejus on February 21, 2021

Add your own answers!

Ask a Question

Get help from others!

© 2024 TransWikia.com. All rights reserved. Sites we Love: PCI Database, UKBizDB, Menu Kuliner, Sharing RPP