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Note: $\psi,\psi^{\dagger} :\Bbb{R} \to \Bbb{C}$ and $x, \lambda_i , \hbar, m \in \Bbb{R}$

Say we know that $\lambda_1$ is a solution to the eigenvalue equation: $$\hat{\Pi}\psi(x)= \lambda_1 \psi(x) .$$ Now for the complex conjugate of $\psi$, to be denoted as $\psi^{\dagger}$, is the eigenfunction for the eigenvalue problem $$\hat{\Pi}\psi^{\dagger}(x)=\lambda_2\psi^{\dagger}(x)$$ the same as the first, or its negative, or totally different? In other words, is $$\lambda_1=\lambda_2 \text{ ?}$$

Thank you!

This is in reference to a quantum mechanics problem, in which I try to prove the following: Given $$j(x,t)=\frac{\hbar}{2 i m} \left[ \psi(x)^{\dagger}\frac{\partial \psi(x)}{\partial x}-\psi(x)\frac{\partial \psi(x)^{\dagger}}{\partial x } \right]$$ Show it can be represented in the form $$j(x,t) =\operatorname{Re} \left[ \psi(x)^{\dagger}\frac{\hbar}{i m}\frac{\partial \psi(x)}{\partial x} \right]$$

This is related because the momentum operator is defined as $\hat{P}= -i \hbar \hat{\nabla}=-i \hbar \dfrac{\partial}{\partial x}$ in one dimension, and this can be substituted into the Eigenvalues problem and simplified.

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    $\begingroup$ Does $x \in \Bbb{C}?$ And $\psi (x)$ is a function $\Bbb{C}\rightarrow \Bbb{C}$? $\endgroup$ – MathematicianByMistake Mar 27 '16 at 21:17
  • $\begingroup$ Added answer to your comment to question. $\endgroup$ – Shinaolord Mar 27 '16 at 21:22
  • $\begingroup$ I am not familiar with Quantum Mechanics I fear, but it is known that eigenvalues of real matrices do come in conjugate pairs. That is, if $A$ is a square matrix with entries real numbers and $x$ is an eigenvector of $A$ for the eigenvalue $λ$, then $\bar x$ is an eigenvector of A for the eigenvalue $\bar λ $. The problem lies in the value $\hat{\Pi}\psi(x)= \lambda_1 \psi(x)$ as far as I can tell. If $Π$ corresponds to a matrix with real values I think you have your answer. $\endgroup$ – MathematicianByMistake Mar 27 '16 at 21:28
  • $\begingroup$ Since $\lambda_i \in \Bbb{R}$, $\text{Im}[\lambda_i]=0 \implies \lambda_1=\lambda_1^{\dagger}=\lambda_2$? $\endgroup$ – Shinaolord Mar 27 '16 at 21:30
  • $\begingroup$ Does $\hat{\Pi}\psi(x)= \lambda_1 \psi(x)$ represent a linear transformation from $\Bbb{R}$ to $\Bbb{R}$? If so, I believe your assertion is correct and $λ_1= \bar λ_1= λ_2$. $\endgroup$ – MathematicianByMistake Mar 27 '16 at 21:38
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What you really need for your problem is the following: if $z$ is a complex number, then $$ \frac{z + z^\dagger}{2} = \text{Re}(z), \qquad \frac{z - z^\dagger}{2i} = \text{Im}(z), \qquad \text{Im}(z) = \text{Re}(\tfrac{z}{i}). $$ See if you can prove these by writing $z = a + bi$, where $a,b \in \mathbb R$, and performing the indicated operators.

Turning to your equation for probability current, we can set $z = \psi^\dagger(x)\tfrac{\partial \psi(x)}{\partial x}$, so $z^\dagger = \psi(x) \tfrac{\partial \psi^\dagger(x)}{\partial x}$. Then $$ j(x,t) = \frac{\hbar}{2 i m} (z - z^\dagger) = \frac{\hbar}{m} \cdot \frac{z - z^\dagger}{2i} = \frac{\hbar}{m} \text{Im}(z) = \frac{\hbar}{m} \text{Re}\left(\frac{z}{i}\right) = \text{Re}\left( \frac{\hbar}{mi} z \right) = \text{Re}\left( \frac{\hbar}{mi} \psi^\dagger(x) \frac{\partial \psi(x)}{\partial x} \right) $$ as desired.

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  • $\begingroup$ I was hoping to solve it myself, but that answers my question as well. Thank you. $\endgroup$ – Shinaolord Mar 27 '16 at 22:29
  • $\begingroup$ Sorry for the spoiler :( $\endgroup$ – Jon Warneke Mar 27 '16 at 22:30
  • $\begingroup$ It's alright. Saved me 20 minutes of work, though now I'll feel like I cheated. And I can't remove it from my head. $\endgroup$ – Shinaolord Mar 27 '16 at 22:31
  • $\begingroup$ But still. Thank you so much! $\endgroup$ – Shinaolord Mar 27 '16 at 22:31
  • $\begingroup$ Could u see why I was asking about operators though? It seemed to me the easiest attack plan was to sub in the momentum operator P and then lambda it's Eigenvalue, rearrange to get the sum of the two, than expand back out to P and back out to the original definition. But this requires the conjugate Eigenvalue is -1*eigenvalue. Would that have worked? $\endgroup$ – Shinaolord Mar 27 '16 at 22:33

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