Have you ever seen the preparation of the state $a^{\star}|0\rangle+b^{\star}|1\rangle$ and $a|0\rangle+b|1\rangle$ from one initial state?

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    $\begingroup$ It's not a good idea to completely modify the question when you already have answers to it. Now the existing answers, which answered the original question, are unrelated to the apparent question. Could you please restore the original question and then ask a new one? $\endgroup$
    – Rammus
    Dec 14 '20 at 11:03

$\langle \psi|$ is not a quantum state, but a linear functional on the set of quantum states. $|\psi\rangle$ is a quantum state, any gates that you can apply to it can only take it to quantum states, so in particular not to $\langle \psi|$.

Of course, mathematically it is trivial to transform $|\psi\rangle$ into $\langle \psi|$ and vice-versa, but this is not a physical transformation.

  • $\begingroup$ Ok, Thanks, Ans what about construction of the state $a^{\star}|0\rangle+b^{\star}|1\rangle$ if I can create $a|0\rangle+b|1\rangle$? $\endgroup$
    – Kim
    Dec 12 '20 at 7:56
  • $\begingroup$ That is a valid quantum state, but it cannot be done as this transformation is nonlinear. $\endgroup$ Dec 13 '20 at 13:00
  • $\begingroup$ I know, but maybe some state preparation method? I did not find in literature something like this unfortunatly. Maybe it is not usefull $\endgroup$
    – Kim
    Dec 14 '20 at 10:49
  • $\begingroup$ It's hard to find that in the literature because people usually talk about the transposition of the density matrix, which is equivalent to the complex conjugation of the state vector. A nice reference about the problem is this one. Roughly speaking, they show that exact transposition is impossible, but if you're happy with a noisy version than it becomes possible with multiple copies of the input state. $\endgroup$ Dec 14 '20 at 13:10

$\left<\psi\right|$ is not a state of a quantum system, it is a linear functional that takes a quantum state and returns a scalar. In terms of basic Linear Algebra, it is a row vector rather than a column vector, and the conjugate transpose of $\left|\psi\right>$. So $\left<\phi|\psi\right>$ is a scalar (inner product) while $\left|\psi\right>\left<\phi\right|$ is a matrix (outer product).

You may have wanted instead to convert $\left|\psi\right>$ to another ket that has all of its probability amplitudes on a basis complex conjugated, without transposing the vector, e.g. turning $\left|\psi\right> = \sum_{x = 0}^Na_x\left|x\right>$ to $\sum_{x = 0}^N \bar a_x\left|x\right>$. There is no single unitary transformation that can take any arbitrary state and complex conjugate this way since unitary transformations by their definition preserve complex inner product and if two different states go through this transformation their new inner product together would be conjugated compared to prior, so any inner products that aren't real numbers wouldn't be preserved.

If, however, you know in advance what the probability amplitudes of $\left|\psi \right>$ are in the computational basis, then a unitary transform that will complex conjugate at least that quantum state is $C_{jk} = \delta_{jk}e^{-2i \phi_j}$ where $\phi_j$ is the argument of $a_j$ in $\left|\psi\right>$. If you want to convert the probability amplitudes of $U\left|p_0\right>$ to their conjugates given knowledge of $U$ and $\left|p_0\right>$, then such a transformation is $\bar U C\left|p_0\right>$, where $C$ conjugates the amplitudes of $\left|p_0\right>$.

  • $\begingroup$ And If I know the initial state $\rho_0$ and the unitay evolution $U$ I cant do both states $a|0\rangle+b|1\rangle$ and $a^{\star}|0\rangle+b^{\star}|1\rangle$? $\endgroup$
    – Kim
    Dec 12 '20 at 9:01
  • $\begingroup$ Edited to answer that as well. $\endgroup$ Dec 12 '20 at 17:18

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