# What can we expect about the new state of |𝜓⟩ after its unitary transformation? [closed]

I am new to quantum computation. Please bear with my question if it is a strange one and let me know if it needs any clarifications.

Suppose there is a closed single qubit system as follows:

$$|\psi\rangle = \alpha|0\rangle + \beta|1\rangle$$

Then we apply some unitary transformation to $$|\psi\rangle$$. Generally speaking, what can we predict or expect about the new state of $$|\psi\rangle$$?

• The unitary operator $| \psi \rangle \langle \chi | + | \chi \rangle \langle \psi |$ maps $| \psi \rangle$ to $| \chi \rangle$ meaning that by an appropriate choice of unitary, $| \psi \rangle$ can be mapped to any other state in the Hilbert space. Therefore, without additional information, you can say absolutely nothing about the "new state of $| \psi \rangle$". Commented Jan 18, 2020 at 21:15
• @Bsh - That question is far too open ended. It totally depends on what type of prediction you are interested in and what class of operators you are considering. Commented Jan 18, 2020 at 21:22
• @Prahar How is $|\psi\rangle\langle \chi| + |\chi\rangle\langle \psi|$ unitary? Commented Jan 18, 2020 at 21:52
• @Prahar I think by zero state OP means $|0\rangle$ not the null vector.
– nox
Commented Jan 18, 2020 at 22:06
• People who are new will naturally have questions that aren't perfectly formed. I think it does a disservice to the community by closing these kinds of questions. This is a positively received question and has positively received answers. Commented Jan 20, 2020 at 1:28

Since it is not clear in the question I treat two possible interpretations of unitary transformations. First a transformation on the whole Hilbert space.

Quantum mechanics is invariant under unitary transformations in the following sense: Given a state $$|\Psi\rangle$$ we can carry out a unitary transformation $$|\Psi\rangle \rightarrow \hat{U}|\Psi \rangle := |\Phi\rangle$$ which maps to another state $$|\Phi\rangle$$. Under this transformation, since $$\hat{U}^{\dagger} \hat{U} = \hat{I}$$ is the identity operator, we have a few results:

1. The normalisation (magnitude) of the new state is equal to that of the old: $$\langle \Phi | \Phi \rangle = \langle \Psi | \hat{U}^{\dagger} \hat{U} | \Psi \rangle = \langle \Psi | \Psi \rangle$$ 1.b The inner product (overlap) between any two states is unchanged if both state are transformed in the same way by $$\hat{U}$$.

2. Expectation values of operators remain unchanged if they transform as $$\hat{A} \rightarrow \hat{U}\hat{A}\hat{U}^{\dagger} := \hat{B}$$: $$\langle \Phi | \hat{B} | \Phi \rangle = \langle \Psi | \hat{U}^{\dagger} \hat{U} A \hat{U}^{\dagger} \hat{U} | \Psi \rangle = \langle \Psi | \hat{A} | \Psi \rangle$$

This is what allows for the freedom of description of QM in terms of Schrödinger, Heisenberg, Interaction pictures etc. We know we can make a unitary transformation from one picture to another without changing the physics (it just corresponds to different choices of bases within the Hilbert space).

Secondly, a unitary transformation may apply only to a single state. E.g. time evolution of the state is produced by $$|\Psi(t)\rangle = \exp(-i / \hbar \hat{H}t) |\Psi(0)\rangle.$$ Other example, an apparatus, A, (maybe a quantum circuit) acts upon a state to produce a new state that is related by a unitary transformation to the old: $$|\Psi\rangle \rightarrow \hat{U}_{A} |\Psi\rangle$$ leaving the rest of the Hilbert space alone. Now we will still have the result $$\mathbf{1}$$ above but we lose $$\mathbf{1b}$$ and $$\mathbf{2}$$ if other states an operators have been unchanged.

Now let's look at your question. Let's say the overlap of $$|\Psi\rangle$$ with a state $$|\phi\rangle$$ before the transformation is $$p$$: $$\langle \phi | \Psi \rangle = p$$. After the transformation we can find out the overlap of $$|\Psi'\rangle = \hat{U}_{A} |\Psi\rangle$$ with the same state: $$p' := \langle \phi | \Psi' \rangle = \langle \phi | \hat{U}_{A} | \Psi \rangle$$ Now without knowing the explicit for of $$\hat{U}_{A}$$ I'm not sure what you expect to say. In general this overlap will be different from $$p$$ in a way that depends strictly on the form of the operation. What we could say is that if $$\hat{U}_{A}^{\dagger} |\phi\rangle$$ happens to be orthogonal to $$|\Psi\rangle$$ then $$p' = 0$$. If $$\hat{U}_{A}^{\dagger} |\phi\rangle$$ happens to equal $$|\Psi\rangle$$ then $$p' = 1$$.

Choosing $$|\phi \rangle = |0\rangle$$ with $$|0\rangle$$ the state zero (rather than the null state) then $$|p'|^{2}$$ is the probability of finding the state $$|\Psi'\rangle$$ in the state $$|0\rangle$$: $$\mathbb{P}(|\Psi'\rangle = |0\rangle) = \left| \langle 0 | \Psi'\rangle \right|^{2} = \left| \langle 0 | \hat{U}_{A} |\Psi\rangle \right|^{2}$$ but without knowing the form of $$\hat{U}_{A}$$ we cannot determine its numerical value.

Final point:

In this problem the basis is two dimensional so we can explicitly write $$|\Psi \rangle = \alpha |0\rangle + \beta |1\rangle \longrightarrow \hat{U}_{A} |\Psi\rangle = \alpha' |0\rangle + \beta' |1\rangle = |\Psi'\rangle$$ where $$|\alpha|^{2} + |\beta|^{2} = 1 = |\alpha'|^{2} + |\beta'|^{2}$$ for normalisation (the latter following from point $$\mathbf{1}$$). In fact the unitary transformation can be represented by the linear transformation on the coefficients: $$\begin{pmatrix}\alpha \\ \beta \end{pmatrix} \longrightarrow \begin{pmatrix}\alpha'\\ \beta' \end{pmatrix} = U_{A} \begin{pmatrix}\alpha \\ \beta\end{pmatrix}$$ where now $$U_{A}$$ is a $$2\times2$$ unitary matrix. If you know $$\alpha$$ and $$\beta$$ and the explicit form of the matrix you can determine $$\alpha'$$ and $$\beta'$$. If you can find $$\alpha'$$ then the probability you want is $$\mathbb{P}(|\Psi'\rangle = |0\rangle) = |\alpha'|^{2}$$.

A unitary "transformation" basically means you're changing the state without adding or losing anything from it. This means that if your probability amplitudes sum up to 1 to begin with, the transformation will not change that.

The comments are suggesting that there is nothing else to it, that nothing can be learned or understood without providing the specific transformation, but I will show you probably the most important thing to know about unitary transformations in the general case, which is how the Schrodinger's Equation will change after transformation.

When we start out with Schrodinger's Equation: $$-ih \frac{d\psi}{dt} = H \psi \\$$

Then we provide a unitary transformation, we need to change the coordinates that we are working in for all things involving matrices or vectors. (Because everything at this point was written in terms of the previous coordinates, which we have just "transformed").

This means that: $$\psi_{new} \rightarrow U\psi_{old}$$ AND $$H_{new} \rightarrow U H_{old} U^\dagger$$

Therefore when we plug it all into the schrodinger's equation we get:

$$-ih \frac{d}{dt} (U^\dagger\psi_{new}) = U^\dagger H_{new} U U^\dagger\psi_{new} \\$$

Unitary operators mean that $$U^\dagger U = 1$$ (an identity matrix)

Next we expand out $$\frac{d}{dt} (U^\dagger\psi)$$ using product rule:

$$\frac{d}{dt} (U^\dagger\psi) = \psi \frac{dU^\dagger}{dt} + U^\dagger \frac{d \psi}{dt}$$

And now we have:

$$\psi \frac{dU^\dagger}{dt} + U^\dagger \frac{d \psi}{dt} = U^\dagger H$$

Bringing $$\frac{d \psi}{dt}$$ to one side:

$$U^\dagger \frac{d \psi}{dt} = (U^\dagger H - \frac{dU^\dagger}{dt}) \psi$$

Reduces to (multiply by U), finally we get: $$\frac{d \psi_{new}}{dt} = ( H_{new} - U \frac{dU^\dagger}{dt}) \psi_{new}$$

This is an interesting result, because it says that if we transform our state, we cannot expect that we can just transform all our linear operators in the same way to get what we want. Instead we see that this works only when this rotation is either slow or constant $${dU^\dagger}{dt}$$. You can think of it like a "centrifugal force" that adds an extra component to the mix when things are transformed too fast.