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Discrepancy between eigenvalues and eigenvectors derived from analytical solution and matlab code.

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Hello,
I have this matrix [ep+V/2 t*phi; t*conj(phi) eb-V/2].
The analytical solution for eigenvalues of this matrix is E=(eb+ep)/2+v*sqrt((eb-ep+V)/2+t^2*|phi|^2).
But matlab solution is different from this.
Can someone help me for solve this chalenge?
  2 件のコメント
KSSV
KSSV 2021 年 7 月 23 日
Show us the code which you tried.
mohammad mortezaie
mohammad mortezaie 2021 年 7 月 23 日
syms eb ep t V phi
H=[ep+V/2 t*phi; t*conj(phi) eb+V/2];
[E,v]=eig(H);

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採用された回答

Chunru
Chunru 2021 年 7 月 23 日
編集済み: Chunru 2021 年 7 月 23 日
First, the sign in the last element of H should be '-' rather than '+' as in your question. Second, "doc eig" command for the order of output variables. Third, make sure your analytical result is correct. Try manual simplification then. You may want to verify the symbolic expressions with some numerical values to see if they agree.
syms eb ep t V phi
H=[ep+V/2 t*phi; t*conj(phi) eb-V/2]
H = 
[v,d]=eig(H) % not [E, v]
v = 
d = 

その他の回答 (1 件)

Steven Lord
Steven Lord 2021 年 7 月 23 日
syms eb ep t V phi
H=[ep+V/2 t*phi; t*conj(phi) eb+V/2]
H = 
[E,v]=eig(H)
E = 
v = 
Let's check if the elements in E and v satisfy the definition of the eigenvectors and eigenvalues for H.
simplify(H*E-E*v)
ans = 
The elements in E and v satisfy the definition of the eigenvectors and eigenvalues for H, so they are eigenvectors and eigenvalues of H. What did you say you expected the eigenvalues to be?
Are you sure you're not missing a ^2 somewhere in the first term under the radical symbol?

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