Skip to main content
Physics

Return to Question

deleted 5 characters in body; edited tags
Source Link
Qmechanic ♦
Qmechanic ♦
  • 213.1k
  • 48
  • 590
  • 2.3k

Does anyone know why in quantum mechanics the second statement is always true?

"When the spectrum of an operator $A$ has a continuous part, we associate a bra $\langle a|$ and a ket $|a \rangle$ to each element $a$ of the continuous spectrum of $A$. Obviously, the bras $\langle a|$ and kets $|a \rangle$ are not in the Hilbert space."

Thanks.

"When the spectrum of an operator $A$ has a continuous part, we associate a bra $\langle a|$ and a ket $|a \rangle$ to each element $a$ of the continuous spectrum of $A$. Obviously, the bras $\langle a|$ and kets $|a \rangle$ are not in the Hilbert space."

Does anyone know why in quantum mechanics the second statement is always true?

"When the spectrum of an operator $A$ has a continuous part, we associate a bra $\langle a|$ and a ket $|a \rangle$ to each element $a$ of the continuous spectrum of $A$. Obviously, the bras $\langle a|$ and kets $|a \rangle$ are not in the Hilbert space."

Thanks.

Does anyone know why in quantum mechanics the second statement is always true?

"When the spectrum of an operator $A$ has a continuous part, we associate a bra $\langle a|$ and a ket $|a \rangle$ to each element $a$ of the continuous spectrum of $A$. Obviously, the bras $\langle a|$ and kets $|a \rangle$ are not in the Hilbert space."

Source Link
Alex
Alex
  • 1.1k
  • 2
  • 15
  • 41

Bras and kets of continuous spectrum

Does anyone know why in quantum mechanics the second statement is always true?

"When the spectrum of an operator $A$ has a continuous part, we associate a bra $\langle a|$ and a ket $|a \rangle$ to each element $a$ of the continuous spectrum of $A$. Obviously, the bras $\langle a|$ and kets $|a \rangle$ are not in the Hilbert space."

Thanks.