Does Subtracting a Constant from a Hamiltonian's Diagonal Change Its System?

[fargoth]

correct me if I am wrong..
as i see it, i can subtract any constant from the diagonal of the hamiltonian without really changing the system it describes... am i right?

if i got a two state system, the hamiltonian can look like this:

$$\left( <br /> \begin{array}{cc}<br /> 0 & P_{2->1} \\<br /> P_{1->2} & \Delta E \\<br /> \end{array} <br /> \right)$$

where $$\Delta E=H_{22}-H_{11}$$

if I am right so far... then here's my problem - when you diagonize the matrix it doesn't have the same eigenvalues (i mean not even the same delta...) if you add or subtract from the diagonal... and that can't be right... so what's wrong in my view of things?

 

[vanesch]

correct me if I am wrong..
as i see it, i can subtract any constant from the diagonal of the hamiltonian without really changing the system it describes... am i right?

if i got a two state system, the hamiltonian can look like this:

$$\left( <br /> \begin{array}{cc}<br /> 0 & P_{1->2} \\<br /> P_{2->1} & \Delta E \\<br /> \end{array} <br /> \right)$$

where $$\Delta E=H_{22}-H_{11}$$

if I am right so far... then here's my problem - when you diagonize the matrix it doesn't have the same eigenvalues (i mean not even the same delta...) if you add or subtract from the diagonal... and that can't be right... so what's wrong in my view of things?

You must be making a mistake.
Imagine H0 the original hamiltonian (the one you quote, for instance), and |e1> an eigenvector.
Then we have H0 |e1> = E1 |e1>

Now consider that we add C 1 to H0 (unit matrix times constant C).

Then of course C 1|e1> = C |e1>

Define H1 = H0 + C 1

H1 |e1> = H0 |e1> + C 1 |e1> = E1 |e1> + C |e1> = (E1 + C) |e1>

So we see that |e1> is again an eigenvector of H1, but this time with eigenvalue E1 + C. All eigenvalues of H1 will be the eigenvalues of H0 plus C, and the eigenvectors will be the same. The differences between the eigenvalues will of course remain the same (C will be eliminated).

cheers,
Patrick.

 

[fargoth]

yeah, i found my mistake... the delta remains the same...