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if $\hat{A}$ is act on $\psi$ with an eigenvalue of $a$

$$\hat{A}\psi=a\psi$$

then how we can calculate

$$\left(\tfrac{1}{\sqrt{2}}\right)^\hat{A}\psi=??$$

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closed as off-topic by stafusa, sammy gerbil, ACuriousMind Dec 1 '17 at 12:18

This question appears to be off-topic. The users who voted to close gave this specific reason:

  • "Homework-like questions should ask about a specific physics concept and show some effort to work through the problem. We want our questions to be useful to the broader community, and to future users. See our meta site for more guidance on how to edit your question to make it better" – stafusa, sammy gerbil, ACuriousMind
If this question can be reworded to fit the rules in the help center, please edit the question.

  • $\begingroup$ Hi and welcome to physics.SE! Please note that homework-like questions and check-my-work questions are generally considered off-topic here. We intend our questions to be potentially useful to a broader set of users than just the one asking, and prefer conceptual questions over those just asking for a specific computation. $\endgroup$ – ACuriousMind Dec 1 '17 at 12:18
  • $\begingroup$ This question could be on-topic if it were instead about how to define the odd object $(1/\sqrt{2})^A$ to begin with, but as written you give no context for why one should be interested in that operator, or that it is its definition that is the problem here. Please include sufficient context so that answerers know what exactly you're struggling with here, and why one would want to compute this in the first place. $\endgroup$ – ACuriousMind Dec 1 '17 at 12:20
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Well, the question, after clarification, could be given a certain mathematical meaning by writing:

$$\left(\frac{1}{\sqrt 2}\right)^{A} \psi = e^{\left(\ln\frac{1}{\sqrt 2}\right)A} \psi $$

then invoke the operational calculus for the (presumably self-adjoint) operator A to obtain the result.

Indeed, it could be a typo in what the OP is reading and no exponential there, case in which one could also give a physical meaning to the mathematical expression.

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  • $\begingroup$ replace $\hat{A}$ with number operator $\hat{n}$ what will happen with the state? $\endgroup$ – Faheem Akhtar Dec 1 '17 at 8:11
  • $\begingroup$ The number operator doesn't have an effect on its eigenstate, since, in general, $f(\hat{A}) \psi = f(a) \psi$. $\endgroup$ – DanielC Dec 1 '17 at 8:27
  • $\begingroup$ means answer will be $$$$ $\endgroup$ – Faheem Akhtar Dec 1 '17 at 10:26
  • $\begingroup$ means that answer will be $$\left(\frac{1}{\sqrt{2}}\right)^\hat{n}\psi=\left(\frac{1}{\sqrt{2}}\right)^n\psi$$ $\endgroup$ – Faheem Akhtar Dec 1 '17 at 10:30
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The other solutions are correct in general. For the specific case where $\psi$ is an eigenvalue of $\hat{A}$, it's even easier.

For any diagonal operator $\hat{O}$ with entries $a_{ii}$, the entries of $e^\hat{O}$ are $e^{a_{ii}}$ (this is easy to prove using the Taylor expansion shown above). Since the entries of a diagonal operator are the eigenvalues of that operator, then exponentiation turns an eigenvalue $a$ into $e^a$.

To transform between bases of exponentiation, note that

$$b^x=e^{\ln(b)x}$$

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Found this "exponential operators"

A function of an operator is defined through its expansion in a Taylor series, for instance

$$\hat{T} = e^{-i\hat{A}}=\sum_{n=0}^{\infty}\frac{(-i\hat{A})^n}{n!}=1-i\hat{A}-\frac {\hat{A}\hat{A}} 2 - \cdots$$

In analogy you would have to expand this square root of 1/2 exponential operator in a Taylor series in order to see what value it gives when acting on psi.

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  • $\begingroup$ remember e is a number, it is not the number that is expanded but the whole function. you should ask at math stack exchange,math.stackexchange.com $\endgroup$ – anna v Dec 1 '17 at 13:51
  • $\begingroup$ Yes I suppose you can make it into a function and formally expand. Good point. $\endgroup$ – ZeroTheHero Dec 1 '17 at 13:54
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I'm not quite sure what you are asking. Is the $\dfrac{1}{\sqrt(2)}$ raised to the $\hat A$ power? If it inst then I mean nothing would happen, then I think the answer would just be $$\dfrac{1}{\sqrt(2)} *a * \psi $$ because you aren't changing anything, just multiplying by a constant

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  • $\begingroup$ Sorry for language problem and weak Communication. Yes! It is $\frac{1}{\sqrt{2}}$ raised to the $\hat{A}$ power. $\endgroup$ – Faheem Akhtar Dec 1 '17 at 6:57
  • $\begingroup$ Yes! It is $\frac{1}{\sqrt{2}}$ raised to the $\hat{A}$ power $\endgroup$ – Faheem Akhtar Dec 1 '17 at 7:06

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