Take the 2-minute tour ×
Physics Stack Exchange is a question and answer site for active researchers, academics and students of physics. It's 100% free, no registration required.

Consider a coupled harmonic oscillator with their position given by $x_1$ and $x_2$. Say the normal coordinates $x_{\pm}={1\over\sqrt{2}} (x_1\pm x_2)$, in which the harmonic oscillators decouple, exist.

When you quantize this theory, the corresponding operators are $\hat x_1,~ \hat x_2,~ \hat x_{\pm}$. The operators $\hat x_1,~\hat x_2$ commutes and $$\hat x_{\pm}\equiv{1\over\sqrt{2}} (\hat x_1 \otimes \hat 1_2 \pm \hat 1_1\otimes \hat x_2),$$ where $\hat 1_1,~\hat 1_2$ are identity operators. So this operator $\hat x_{\pm}$ acts in a direct product basis given by $$|x_1,x_2\rangle=|x_1\rangle \otimes |x_2\rangle$$ where $|x_1\rangle$ and $|x_2\rangle$ are eigen states of $\hat x_1$ and $\hat x_2$ respectively.

But the state $|x_1,x_2\rangle$ is eigenstate of both $\hat x_{\pm}$.

How can we construct unique eigenstates of the operator $\hat x_{\pm}$ in the basis $|x_1,x_2\rangle$? Or is it that it is not possible?

share|improve this question
3  
Nitpicking: "direct product" ($\times$) $\neq$ "tensor product" ($\otimes$) for vector spaces or more generally modules. In fact, for finite number of spaces being "(co)producted," the direct product is equivalent to the "direct sum" ($\oplus$). –  Chris White Jul 24 '13 at 18:50

2 Answers 2

up vote 1 down vote accepted

Both $\hat{x}_{\pm}$ commute with both $\hat{x}$ (product with identity). So they do (as you verified) share a common set of eigenvectors.

In this case the set contains any product of two position eigenstates. That's why any state $|x,y\rangle$ is an eigenvector of all four of your operators.

So I guess the final answer is: No, that's not possible.

share|improve this answer

As you note, the states $|x_1, x_2\rangle$ are eigenstates of $x_\pm$; $$ \hat x_\pm |x_1, x_2\rangle = \frac{1}{\sqrt{2}}(x_1 + x_2)|x_1,x_2\rangle $$ Moreover, these states form an orthonormal basis (in the Dirac sense) for the tensor product Hilbert space. It follows that every eigenvector of either of these operators must be a scalar multiple of one of the states $|x_1, x_2\rangle$.

To see this, let's suppose that there is a state $|\psi\rangle$ that is not a scalar multiple of one of the vectors $|x_1, x_2\rangle$, then there exists some other $|x_1',x_2'\rangle$ and nonzero complex numbers $a$ and $b$ for which $$ |\psi\rangle = a|x_1,x_2\rangle + b|x_1',x_2'\rangle $$ and therefore $$ \hat x_\pm|\psi\rangle = \frac{1}{\sqrt{2}}\left(a(x_1 + x_2)|x_1,x_2\rangle + b(x_1'+x_2')|x_1',x_2'\rangle\right) $$ this state is not an eigenvector of $x_{\pm}$ unless $x_1'+x_2' = x_1 + x_2$.

Edit. In response to the comment below, the following statement is false:

in this basis both eigen states of x± are same. So it implies x^+=x^−

Two operators having a simultaneous eigenbasis does not mean that they are equal! What matters is the action of these operators on each of these basis states. In the case at hand, the eigenvalues of $\hat x_+$ and $\hat x_-$ are distinct, so we can clearly see that they are distinct operators.

share|improve this answer
    
I agree with you and the answer before by Bruce Connor. –  Nilanjan Jul 24 '13 at 18:27
    
@user27470 Cool. Does this response not answer the question? Perhaps I misinterpreted it? –  joshphysics Jul 24 '13 at 18:29
    
I agree with you and the answer before by Bruce Connor. To be more clear about my question let us consider the product hilbert space is called $H=H_1 \otimes H_2$. Clearly $\hat x_{\pm}$ are operators in the hilbert space $H$. One of the possible basis of $H$ is $|x_1,x_2>$. What is confusing me is that I thought that every opertor eigenstate in $H$ can be expressed as linear combination of its basis states, here one choice is the simple states $|x_1,x_2>$. But in this basis both eigen states of $x_{\pm}$ are same. So it implies $\hat x_+=\hat x_-$. But we know this cannot be true! –  Nilanjan Jul 24 '13 at 18:36
    
The first response went by mistake. You are correct. I agree with you but it does not answer my question –  Nilanjan Jul 24 '13 at 18:38
    
@user27470 See my edit. –  joshphysics Jul 24 '13 at 18:42

Your Answer

 
discard

By posting your answer, you agree to the privacy policy and terms of service.

Not the answer you're looking for? Browse other questions tagged or ask your own question.