# Is it possible for the measurement to yield a superposition of states in the original space?

Suppose we have a wave function $$|\Psi\rangle=\sum_i c_i|\psi_i\rangle$$, where the original probability probability amplitude was in a distribution of, i.e. $$c_1^*c_1=\frac{1}{2},...,c_i^*c_i=\frac{1}{2^n}$$.

Suppose we perform a measurement $$M$$ on $$|\Psi\rangle$$ that measures all the states except for $$|\psi_1\rangle$$ and $$|\psi_2\rangle$$ (possibly simultaneously by a plates). Furthermore, suppose that the measurement $$M$$ does not change the probability distribution of the states, i.e., $$M$$ only collapses the measured states, but does not affect the rest of the wave function.

Then will the resulting quantum state be a superposition of states, i.e. $$|\Psi_M\rangle= \sqrt{2/3}|\psi_1\rangle+\sqrt{1/3}|\psi_2\rangle$$?

• You can't choose to measure certain states – Aaron Stevens Sep 11 at 19:46
• What do you mean by "measured all the states except 1 and 2?" What motivates the question? Why do you want $c_n^*c_n=1/2^n$? – Ben Crowell Sep 11 at 23:11
• @AaronStevens An easier case: suppose a quantity of three discrete states, ($E_1,E_2,E_3$), such as the energy of big separation; then we will have three plates specifically to be responsive to each of the energy states. Then we only insert the one of the plates to perform the measurement. Mathematically, $M_{new}=(1-M*\delta_i^3)$ which make perfectly sense as well. If the measurement read at $M_{new}=0$, then the states is in superposition of states $E_1$ and $E_2$. – user9976437 Sep 12 at 2:45
• @BenCrowell It's just an example so that the states is normalized. – user9976437 Sep 12 at 2:45

In QM, a measurement always amounts to a choice of a basis (or more generally, a set of projectors summing to the identity) with respect to which the wavefunction collapses. In other words, any measurement of a state $$|\Psi\rangle$$ can be described via a set of orthogonal projectors $$P_k$$ such that $$\sum_k P_k=I$$, by writing the state as $$|\Psi\rangle=\sum_k P_k|\Psi\rangle$$ and destroying all the coherence between the subspaces corresponding to each projector. Mathematically, this amounts to the following mapping $$|\Psi\rangle\simeq\mathbb P(|\Psi\rangle)\mapsto\sum_k \mathbb P(P_k|\Psi\rangle),$$ where I used the shortcut notation $$\mathbb P(|\phi\rangle)\equiv|\phi\rangle\!\langle\phi|$$.

When the projectors $$P_k$$ have unit trace, and thus can be written as $$P_k=\mathbb P(|\phi_k\rangle)$$, you recover the standard notion of measuring $$|\Psi\rangle$$ in an orthonormal basis $$\{\lvert\phi\rangle\}_k$$.

This is the most general way in QM in which you can "ask a question" to a state, which is what measurements fundamentally amount to. For this reason, you cannot "measure all $$|\psi_k\rangle$$ except for some of them". Quite simply, such a statement means nothing. You don't "measure some of the $$|\psi_k\rangle$$", you measure $$|\Psi\rangle$$ in a given basis, and observe one of the elements of the basis.

In QM the act of measurement $$M$$ collapses the wavefunction into one of the eigenstates of the measured observable $$\hat{O}$$. One does not 'measure states', rather one measures an observable.

Assume the initial wavefunction is of the form $$\left|\Psi \right> = \sum_i c_i\left| \psi_i\right>$$, where $$\left|\psi_i \right>$$ form an orthonormal basis and are not the eigenstates $$\left| e_i \right>$$ of $$\hat{O}$$. Then some superposition of the basis eigenstates $$\left|\psi_i \right>$$ will be an eigenstate of the observable $$\hat{O}$$, i.e. $$\left| e_i \right> = \sum_j c_{ij}\left|\psi_j \right>$$. Therefore the collapsed wave function does not have to be a pure eigenstate in the original basis $$\left|\psi_i \right>$$.

• yes, but if the particular states wasnt there, there's no collapse and end up a 0, trivally. So although M may be thought as a projection, if the results was 0, it essentially did not perform any projecton on the original states. – user9976437 Sep 11 at 20:13
• That's simply not the case. I think it may be worth for you to go through the mathematics. If the eigenstate $\left|e_n\right>$ is not present in the spectral decomposition of the wavefunction, then the probability of collapsing into $\left|e_n\right>$ is zero and the measurement will simply yield a different eigenstate. – Akerai Sep 11 at 20:29
• It is important to remember that the eigenstates of an observable form a complete orthogonal set. Therefore if you have a properly normalised wavefunction - a measurement of an observable will always yield a non-zero state. – Akerai Sep 11 at 20:30
• @user9976437 The above comments are intended for you. Akerai, if you want to guarantee a user will be notified of your comments, make sure to tag that user. – Aaron Stevens Sep 11 at 23:29
• @Akerai See my comment after the OP, the new measurement $M_{new}=(1-M*\delta_i^1-M*\delta_i^2)$ which make perfect sense as well. – user9976437 Sep 12 at 2:49

As the other answers point out, your question is very confusing and it's not entire clear what you're asking. I think what you're asking is what happens if you measure whether the system is in any of the states $$|\psi_i\rangle$$ with $$i \geq 3$$, i.e. you measure the value of the observable that is the projection operator

$$\hat{P} = \sum_{i \geq 3} |\psi_i\rangle \langle \psi_i|,$$

and you happen to find that the answer is "no", i.e. you get the eigenvalue $$0$$ (which will occur with probability 3/4). If this is what you are asking, then the answer to your question is yes: the state is projected down to the subspace spanned by $$\{|\psi_1\rangle, |\psi_2\rangle$$ }. The new state after the measurement will be $$|\psi'\rangle = \sqrt{\frac{4}{3}} \left( \frac{1}{\sqrt{2}} e^{i \theta_1} |\psi_1\rangle + \frac{1}{\sqrt{4}} e^{i \theta_2} |\psi_2\rangle \right) = \sqrt{\frac{2}{3}} e^{i \theta_1} |\psi_1\rangle + \sqrt{\frac{1}{3}} e^{i \theta_2} |\psi_2\rangle,$$

since you didn't provide enough information in your question to specify the phase factors.

This counterintuitive behavior is known as the Renninger negative-result experiment.