When a Hamiltonian operator apply to a wavefunction, how could we write the hamiltonian as, $$H \psi = (E_n-\hbar \omega_0) \psi \ \ ? $$
Is this because $E_n= H+ \hbar \omega_0$?
where $\omega_0$ is the angular frequency.
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2$\begingroup$ Do you know what are eigenvectors and eigenvalues of an operator/matrix ? These are the most basic definitions you need to learn before trying to understand any equations of quantum mechanics. $\endgroup$– AdamCommented Oct 20, 2013 at 16:56
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$\begingroup$ yup I know it. But the confusions arises, because as far as I know the Eigen value of Hamiltonian Is energy. If this is true then why zeropoint energy appeared in the eigen values with negative sign? $\endgroup$– user2378Commented Oct 20, 2013 at 17:55
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1$\begingroup$ You'll need to provide more information than you already have, since it's hard to understand what you're talking about. What exactly is the system, and what form does its Hamiltonian take? What degrees of freedom does it have? What is $\omega_0$ and $E_n$? $\endgroup$– DumpsterDoofusCommented Oct 20, 2013 at 18:58
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$\begingroup$ Please see this question. grephysics.net/ans/9677/100 $\endgroup$– user2378Commented Oct 20, 2013 at 19:27
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$\begingroup$ You're referring to the harmonic oscillator, one of the most basic applications of quantum mechanics. Go read any elementary textbook on the topic, such as the widely used book Introduction to Quantum Mechanics by D. Griffiths. There, you'll find sufficient information & discussion of the results, which really are not that difficult once you've had a proper introduction. $\endgroup$– DanuCommented Oct 20, 2013 at 19:43
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It is not true that \begin{align} E_n = H+\hbar\omega_0. \end{align} Why? Well, $H$ is a linear operator while $E_n$ and $\hbar\omega_0$ are real numbers; an operator cannot equal a real number. You might try to fix this by multiplying each of the real numbers by the identity operator and then claim that \begin{align} E_n I = H+\hbar\omega_0 I \end{align} This is still not correct. You can immediately tell that it can't be correct because the left hand side depends on $n$, a non-negative integer, by the right hand side does not. So what's going on? As essentially mentioned in the comments, the following statement is true:
If $\psi$ is an eigenvector of the harmonic oscillator, then there exists a non-negative integer $n$ for which \begin{align} H\psi = \left(n+\frac{1}{2}\right)\hbar\omega\psi \end{align}
where here we are using notation in which the harmonic oscillator hamiltonian is given by \begin{align} H = \frac{1}{2m} P^2 + \frac{1}{2}m\omega^2X^2 \end{align} where $P$ and $X$ are the position and momentum operators respectively. In other words, if you act the hamiltonian on an eigenvector, then it acts simply by multiplying that eigenvector by a real number, the corresponding eigenvalue. But when the Hamiltonian acts on a vector that is not an eigenvector, this doesn't happen.
answered Oct 20, 2013 at 20:42
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$\begingroup$ I understand what you wrote but how can we write $H \psi = (E_n-\hbar \omega_0) \psi \ \ ?$ $\endgroup$– user2378Commented Oct 21, 2013 at 6:47
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$\begingroup$ @user2378 I don't understand what you mean. If we define $E_n = (n+\frac{1}{2})\hbar\omega$, then $H\psi = E_n\psi$ is just the energy eigenvalue equation. If the equation you are trying to interpret is the energy eigenvalue equation, then it's simply being written using unconventional notation. $\endgroup$ Commented Oct 21, 2013 at 7:08
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$\begingroup$ grephysics.net/ans/9677/100 In this link, a operator is represented by $a^*$,and when this opertor acts on $\psi$ then it gives eigen value k. Why cant we say there that H and $a^*$ commutes by stateing like this, $$a^* H \psi_n= k E_n \psi_n$$ And $$H a^* \psi_n= Hk \psi_n = k E_n \psi_n$$. And this is why I asked you that how can we write like this $H \psi = (E_n-\hbar \omega_0) \psi \ \ ?$ $\endgroup$– user2378Commented Oct 21, 2013 at 7:18
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$\begingroup$ @user2378 If $\psi$ is an eigenstate of $H$, then it will not be an eigenstate of $a^\dagger$; your second string of equations does not hold. In fact, then $a^\dagger$ acts on an eigenstate of $H$, it returns a different eigenstate corresponding to a larger eigenvalue. $\endgroup$ Commented Oct 21, 2013 at 7:35
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