# How to assign a value to an observable when the statevector is not an eigenvector of the operator?

We get the value of an observable $$A$$ for a given state $$|\lambda\rangle$$ of a system from the eigenequation $$\hat{A} |\lambda\rangle = \lambda |\lambda\rangle$$ where $$\hat{A}$$ is the operator corresponding to observable $$A$$ and $$\lambda$$ is the required value of the observable. But this is true only when the statevector $$|\lambda\rangle$$ is an eigenvector of the operator $$\hat{A}$$. How do we assign a value to observable $$A$$ when the statevector $$|\lambda\rangle$$ is not an eigenvector of the operator $$\hat{A}$$?

I am a beginner in quantum physics with elementary understanding of linear algebra so please try to answer accordingly.

PS: This question is related and its accepted answer is helpful. But from it I could not understand how to assign a value to the observable.

• Thank you @Ghoster for the edit!
– t2m
Sep 20, 2022 at 17:07

How do we assign a value to observable $$A$$ when the statevector $$|\lambda\rangle$$ is not an eigenvector of the operator $$\hat{A}$$?

We don't.

One of the toughest parts of learning quantum mechanics is learning to let go — in particular, learning to let go of the requirement that everything must have a value. In QM, if you have a system in a given state $$|\psi⟩$$ and a given observable $$\hat A$$ representing a physical quantity $$A$$, the generic setting is that the system cannot be assigned a value for $$A$$.

In some lucky occasions, of course, it turns out that $$A$$ does have a well-defined value (and these are, as you'll have guessed, when $$|\psi⟩$$ is an eigenstate of $$\hat A$$).

But, in general, observables in quantum mechanics do not have value until and unless a projective measurement of that observable is performed on the system.

• According to answer of @Arturodonjuan, we can find average value of observable when the state vector is not eigenvector of operator. I assume that both answers spell truth with different perspectives. But then I can't reconcile both answers. Can you please explain on that line?
– t2m
Sep 21, 2022 at 8:05
• The 'expectation value' of $A$ in the state $|\psi⟩$ is what you get if you take the system in the state $|\psi⟩$ and then (i) perform a projective measurement of $A$, which will then (ii) change the state of the system to one of the eigenstates of $\hat A$ and produce the corresponding eigenvalue as the value of $A$, and then (iii) you repeat this procedure multiple times (typically on a large ensemble), which will then typically (iv) give a wide spread of different values of $A$ with different frequency; and then finally you (v) take the average of the value of $A$ over this distribution. Sep 21, 2022 at 16:57
• However, this does not mean that the expectation value of $A$ is in any way assignable as "the value of $A$ prior to the measurement". That value does not exist, and cannot be assigned. Sep 21, 2022 at 16:57
• I get your point that there is an 'expectation value' but don't get that it is not assignable as a value of $A$ prior to the measurement. I think, even the definite outcome given by the eigenvalues too is what we get after measurement. Isn't it?
– t2m
Sep 21, 2022 at 17:13
• Thank you for helping with insightful answers.
– t2m
Sep 22, 2022 at 12:25

A pure physical state is comprised of a linear sum of eigenstates:

$$|\psi\rangle = \sum_{i} c_i |\lambda_i\rangle$$

where $$|\lambda_i\rangle$$ are complete eigenstates of a self-adjoint operator $$A$$, $$\langle \lambda_i |\lambda_j\rangle = \delta_{ij}$$ and $$\sum_i |c_i|^2=1$$.

The probability that we will find $$|\psi \rangle$$ in the $$|\lambda_i\rangle$$ state (i.e. having value $$\lambda_i$$ when acted upon by operator $$A$$) is given by $$|\langle \lambda_i | \psi\rangle|^2=|c_i|^2$$.

The average value of the operator $$A$$ in the state $$|\psi\rangle$$ is given by $$\langle \psi | A |\psi \rangle = \sum_i |c_i|^2\lambda_i$$. This is the average value of $$A$$ you would measure if you prepared an infinite number of identical systems described by the state-vector $$|\psi\rangle$$.

• Is there a reason for saying a state as 'pure' physical state?
– t2m
Sep 21, 2022 at 8:54
• Yes because a physical state is in general described by as a 'mixed' state, represented by a density matrix/operator. A special case is when the physical state/system can be represented by a single state vector, i.e. it is a 'pure' state. See the wikipedia article here. Sep 21, 2022 at 17:02
• Can you explain in the post, or point me to an easy to understand online resource, why the average value (expectation value) is as you have written?
– t2m
Sep 22, 2022 at 9:39
• @Aarone Do you mean why the expectation value of an observable $\hat A$ in a physical state $|\psi\rangle$ is given by $\langle \psi | \hat A |\psi\rangle$? This is sort of by definition, an axiom of the framework of quantum mechanics. With this, the relation $\langle \psi | \hat A |\psi\rangle = \sum_i |c_i|^2 \lambda_i$ follows by direct substitution of $|\psi\rangle = \sum_i c_i |\lambda_i\rangle$, and using that the set of eigenstates $|\lambda_i\rangle$ is orthonormal, i.e. $\langle \lambda_i |\lambda _j\rangle = \delta_{ij}$. Sep 22, 2022 at 18:06
• I was asking about how $\langle \psi | \hat A |\psi\rangle$ is equal to $\sum_i |c_i|^2 \lambda_i$. Your hint is sufficient. Thank you.
– t2m
Sep 23, 2022 at 2:08