# Observable Operator on a Superposition?

I'm probably missing something obvious and basic here but I can't make sense of certain usages of Observables as present in basic treatments of Quantum Mechanics that i've come across.

$$\hat{A}|\Psi\rangle = a|\Psi\rangle$$

The above equation implies to me that a single eigenket gives a single eigenvalue of $\hat{A}$.

However Ket Vectors that are composed of superpositions have multiple possible eigenvalues. Which leads me to believe that that equation is only valid for eigenkets which are Basis States.

However in the Schrödinger equation we have an Observable (Hamiltonian) acting on Wave Functions in Position Space which are composed of an infinite number of Basis States.

In this usage is it somehow assumed that every Basis State in the Position Basis corresponds to a single Energy Eigenstate? (I wouldn't think this would be the case. But what is the point/result of applying the Hamiltonian to any given Wave Function then?)

Further confusion arises from this because if the Energy is exactly known then shouldn't there be some sort of maximal uncertainty in time?

As a final question is there any kind of useful interpretation of multiplying the eigenket by it's eigenvalue as appears in the above Observable Equation? In all treatments I've seen this multiplication is simply ignored and the eigenvalue itself is the only focus.

However Ket Vectors that are composed of superpositions have multiple possible eigenvalues. Which leads me to believe that that equation is only valid for eigenkets which are Basis States.

The equation \begin{align} \hat A|\Psi\rangle = a|\Psi\rangle \end{align} holds only for eigenvectors of the operator $\hat A$. In general, there is a mathematical theorem, the spectral theorem, that says that for any hermitian (self-adjoint) operator $\hat A$ acting on a Hilbert space $\mathcal H$, there exists a basis of the Hilbert space composed of eigenvectors of $\hat A$. This tells us that any vector $|\psi\rangle$ in the Hilbert space can be written as a linear combination of eigenvectors of any given observable. Let's say, for example, that the basis of eigenvectors corresponding to observable $\hat A$ is denoted by $\{|a_1\rangle, |a_2\rangle, \dots\}$ where the vector $|a_i\rangle$ has eigenvalue $a_i$. Then for any state $|\psi\rangle$ in the Hilbert space, we can write \begin{align} |\psi\rangle = \sum_i c_i|a_i\rangle \end{align}

is it somehow assumed that every Basis State in the Position Basis corresponds to a single Energy Eigenstate?

No. An eigenstate of one operator is not necessarily an eigenstate of another operator. If, however, two operators commute, then it is possible to find a basis for the Hilbert space comprised of vectors that are eigenstates of both operators (we usually call these "simultaneous eigenstates" of the two operators).

is there any kind of useful interpretation of multiplying the eigenket by it's eigenvalue as appears in the above Observable Equation?

I'm not sure what you're exactly looking for here, but one fact is that if $|\Psi\rangle$ satisfies the eigenvalue equation, and if the system is prepared in that state, then a measurement of the observable $\hat A$ will return the corresponding eigenvalue with probability $1$.

• Perhaps the third point can be solved by noting that multiplication by a complex scalar is an equivalence relation, so $\lvert \Psi \rangle$ and $a\lvert \Psi \rangle$ are physically identical states? Though I'm also sort of guessing as to what is being sought.
– user10851
Commented Nov 1, 2013 at 21:45
• @ChrisWhite Yeah that's a good point. I'm guessing that's probably more in line with what the OP wants. Maybe we'll get some clarification. Commented Nov 1, 2013 at 21:52
• Yeah I was stupidly forgetting the fact that multiplying a State Vector by a Real Number won't have any meaningful effect. Also I suspect the Spectral Theorem is the key piece I was missing. I'll learn more about it, have a think, then come back to this. Commented Nov 1, 2013 at 22:24
• @jcelios You may find the following posts useful: physics.stackexchange.com/q/54154, physics.stackexchange.com/q/74943, physics.stackexchange.com/q/9551 Commented Nov 2, 2013 at 2:47

As a final question is there any kind of useful interpretation of multiplying the eigenket by it's eigenvalue as appears in the above Observable Equation?

This is the math that describes the idea that $\langle \Psi \vert A \vert \Psi \rangle = \langle A \rangle$ is the expectation value of the observable quantity assocated with $A$ when the system is in the state $\Psi$.

Using @joshphysic's expansion in term of the eigenvectors $$\langle \Psi \vert A \vert \Psi \rangle = \sum_{ij} c_i c_j^* a_i a_j \langle a_j \vert a_i \rangle = \sum_i \vert c_i \vert^2 a_i$$ where the final expression arises because the eigen-basis is orthonormal.

Final expression is just the weighted-average of the operator's value.