[問題] 基底向量與本徵值

看板Physics作者 (約束された勝利の剣)時間17年前 (2009/04/18 14:24), 編輯推噓3(3推 0噓 8→)
留言11則, 4人參與, 最新討論串1/1
as title 如果我一個Hamiltonian : H 使用某一基底向量展開 可得一組本徵值 若是使用么正矩陣將其對角化 (U+)H(U) (意即將H的基底向量換為其本徵向量) 那H的本徵值會改變嗎? 我如果使用不同的矩陣: <ri| H |r'j> 可得到另一矩陣(表其用另一組基底向量展開) 則這個Hamiltonian的本徵值會改變嗎? -- Our destiny have been entwined,Jenny, but never joined... -- ※ 發信站: 批踢踢實業坊(ptt.cc) ◆ From: 140.115.218.227 ※ 編輯: nightkid 來自: 140.115.218.227 (04/18 14:27)
推
04/18 16:21, , 1F
不會
04/18 16:21, 1F
→
04/18 16:22, , 2F
no.
04/18 16:22, 2F
推
04/18 16:58, , 3F
From some points of view, when you measure a quantity
04/18 16:58, 3F
→
04/18 16:59, , 4F
"A" what you will get is nothing but the expectation
04/18 16:59, 4F
→
04/18 17:00, , 5F
of quantity "A", and we know that the expectation of
04/18 17:00, 5F
→
04/18 17:00, , 6F
"A" is just some linear combination of its eigenvalues
04/18 17:00, 6F
→
04/18 17:02, , 7F
so the eigenvalues of "A" are invariant under basis
04/18 17:02, 7F
→
04/18 17:02, , 8F
transformation.
04/18 17:02, 8F
推
04/18 17:06, , 9F
Sorry,I should pinpoint that expectation values of "A"
04/18 17:06, 9F
→
04/18 17:07, , 10F
are independent of representation of "A".
04/18 17:07, 10F
→
04/18 23:22, , 11F
感謝您 thank a lot!
04/18 23:22, 11F
‣ 更多 nightkid 的文章
文章代碼(AID): #19wN8Y09 (Physics)