# Principle of superposition of states

• I
Upon reading Landau QM, the Principle of superposition of states, I got confused. It states (and i quote):

"Suppose that, in a state with wave function Ψ1(q), some measurement leads with certainty to a definite result 1, while in a state with Ψ2(q) it leads to a different result 2. Then it is assumed that every linear combination of Ψ1(q) and Ψ2(q) gives a state in which that measurement leads to either result 1 or result 2."

Cool I said, this implies that Ψ1,2(q) = Ψ1(q) +Ψ2(q) no?
Well I carried on reading of a system composed of two well known parts and found:

"Ψ1,2(q) =Ψ1(q)Ψ2(q)"

Be gentle with the explanation please!!

dextercioby
Homework Helper
There are essentially two systems. The first is "unbreakable", while the second is decomposable in two distinct subsystems. Hence the mathematical description must be different.

I dont see how they are different

vanhees71
Gold Member
LL talk about one system and two states ##|\Psi_1 \rangle## and ##|\Psi_2 \rangle## that are eigenstates to different eigenvalues of some self-adjoint operator ##\hat{A}## that represents an observable,
$$\hat{A} |\Psi_1 \rangle=a_1 |\Psi_1 \rangle, \quad \hat{A} |\Psi_2 \rangle=a_2 |\Psi_2 \rangle.$$
Then, according to the fundamental postulates of QT, if the system is prepared in a state
$$|\psi \rangle=c_1 |\Psi_1 \rangle + c_2 |\Psi_2 \rangle$$
with
$$\langle \psi|\psi \rangle=1$$
the probability to get any eigenvalue ##a## of ##\hat{A}## when measuring the observable is
$$P(a|\psi)=\sum_{\beta} |\langle a,\beta|\psi \rangle|^2,$$
where ##|a,\beta \rangle## denotes the orthonormalized basis of the eigenspace of ##\hat{A}## to the eigenvalue ##a##. Since now ##|\Psi_1 \rangle## and ##|\Psi_2 \rangle## are eigenvectors to different eigenvalues they are orthogonal to each other, and we can make them part of the complete orthonormal basis of eigenvectors of ##\hat{A}##. Using this basis, it's immidiately clear that if the system is prepared in the state represented by ##|\psi \rangle##, then you find always either the value ##a_1## or ##a_2## when measuring the observable represented by ##\hat{A}##, and the probabilities are ##|c_1|^2## and ##|c_2|^2##, respectively.

In other words, ##|\psi \rangle## represents a state, for which the value of this observable is indetermined, and it tells you the probabilities for finding any of the possible values.

Alex Cros
martinbn
I dont see how they are different
The notation confuses you. The same symbol is used in two different contexts. In the first case you have one system, which is in a state that is a superposition of two other states. In the second case you have a system that consists of two subsystems and the state is the the product of the state of each system. For example one system "a cat" can be in a state |alive>, it can also be in a state |dead>, it can also be in a state |alive>+|dead>. Second case two things "a cat" and "an observer". If the cat is in the state |alive> and the observer in the state |happy>, together the composite system can be in the state |alive>|happy>. Of course it can be in many other states as well.

Alex Cros
The notation confuses you. The same symbol is used in two different contexts. In the first case you have one system, which is in a state that is a superposition of two other states. In the second case you have a system that consists of two subsystems and the state is the the product of the state of each system. For example one system "a cat" can be in a state |alive>, it can also be in a state |dead>, it can also be in a state |alive>+|dead>. Second case two things "a cat" and "an observer". If the cat is in the state |alive> and the observer in the state |happy>, together the composite system can be in the state |alive>|happy>. Of course it can be in many other states as well.
So let me get this straight, a system composed of two other states is simply the linear superposition of them both whereas two different systems with states |A> & |B> form the together system in the state |A>|B>?

vanhees71
Gold Member
Somehow you should get more precise in your language. States cannot be "composed". Hilbert-space vectors can be linearly combined, as is implied by the mathematical structure of a vector space.

States in QT are represented by a positive semidefinite self-adjoint operator of trace 1, the Statistical Operator. Pure states are those states which are represented by projection operators of the form ##|\psi \rangle \langle \psi|## with ##|\psi \rangle## a normalized vector in Hilbert space. It is determined up to a phase factor. In this sense pure states are equivalently determined by rays in Hilbert space.

Modulo superselection rules any normalized superposition of two or more Hilbert-space vectors can be used to represent unit rays in Hilbert space and thus pure states.

If you have composite systems, e.g., two disinguishable particles the Hilbert space of the composite system is given by the tensor-product of the Hilbert spaces of the systems they consist of. For two dinstinguishable particles the Hilbert space is given by ##\mathcal{H}_1 \otimes \mathcal{H}_2##. If ##|u_n \rangle## and ##|v_n \rangle## are complete orthonormal systems of ##\mathcal{H}_1## and ##\mathcal{H}_2##, then ##|u_1,v_1 \rangle \equiv |u_1 \rangle \otimes |v_1 \rangle## is a complete orthonormalized basis of ##\mathcal{H}_1 \otimes \mathcal{H}_2##.

The tensor product of Hilbert spaces creates another Hilbert space, while superpositions (linear combinations) of vectors refer to just one Hilbert space.

Somehow you should get more precise in your language. States cannot be "composed". Hilbert-space vectors can be linearly combined, as is implied by the mathematical structure of a vector space.

States in QT are represented by a positive semidefinite self-adjoint operator of trace 1, the Statistical Operator. Pure states are those states which are represented by projection operators of the form ##|\psi \rangle \langle \psi|## with ##|\psi \rangle## a normalized vector in Hilbert space. It is determined up to a phase factor. In this sense pure states are equivalently determined by rays in Hilbert space.

Modulo superselection rules any normalized superposition of two or more Hilbert-space vectors can be used to represent unit rays in Hilbert space and thus pure states.

If you have composite systems, e.g., two disinguishable particles the Hilbert space of the composite system is given by the tensor-product of the Hilbert spaces of the systems they consist of. For two dinstinguishable particles the Hilbert space is given by ##\mathcal{H}_1 \otimes \mathcal{H}_2##. If ##|u_n \rangle## and ##|v_n \rangle## are complete orthonormal systems of ##\mathcal{H}_1## and ##\mathcal{H}_2##, then ##|u_1,v_1 \rangle \equiv |u_1 \rangle \otimes |v_1 \rangle## is a complete orthonormalized basis of ##\mathcal{H}_1 \otimes \mathcal{H}_2##.

The tensor product of Hilbert spaces creates another Hilbert space, while superpositions (linear combinations) of vectors refer to just one Hilbert space.

Yes, but you guys must understand that up until the point where I was quoting Landau´s (when I wrote the question) he had not said anything about all that mathematical formalim you guys are stating, so I guess my question is more basic than all that?

stevendaryl
Staff Emeritus
Yes, but you guys must understand that up until the point where I was quoting Landau´s (when I wrote the question) he had not said anything about all that mathematical formalim you guys are stating, so I guess my question is more basic than all that?

Maybe it helps to consider concrete examples. Suppose you have a single electron: It can be in two possible "spin states": $|u\rangle$ ("spin up") or $|d\rangle$ ("spin down"). Up and down are relative to the z-axis. If you measure its spin along the z-axis, you will get result "spin up" if the electron is in state $|u\rangle$, and you'll get result "spin down" if the electron is in state $|d\rangle$.

Now, what Landau is saying in the first section is that for any two numbers $\alpha$ and $\beta$ we can form another state $|\psi\rangle = \alpha |u\rangle + \beta |d\rangle$. That's a linear combination of $|u\rangle$ and $|d\rangle$. If an electron is in state $|\psi\rangle$ and you measure the spin in the z-direction, you'll either get "spin up" or "spin down". You won't get some third result that is half-way between "up" and "down".

So that's about states of a single electron. Now, suppose we have two electrons? Then it's perfectly possible for the first to be in state $|u\rangle$ while the second is in state $|d\rangle$. Then the two-electron system would be described by a "composite" state $|\Psi\rangle = |u\rangle |d\rangle$. (Actually, since electrons are indistinguishable, the two-particle state will be in a linear combination of the state $|u\rangle |d\rangle$ and $|d\rangle |u\rangle$, so that you can't say the first electron is in state $|u\rangle$ and the second is in state $|d\rangle$.)

So the first way of combining states---linear combinations---produces a new one-particle state from two different one-particle states. The second way of combining states---products--produces a two-particle from two one-particle states.

vanhees71 and Alex Cros
vanhees71
Gold Member
Yes, but you guys must understand that up until the point where I was quoting Landau´s (when I wrote the question) he had not said anything about all that mathematical formalim you guys are stating, so I guess my question is more basic than all that?
Maybe, you should read another introduction to quantum mechanics first. Landau and Lifhitz are usually pretty brief and deep, i.e., there's much more information per page than in many other textbooks. A good introductory book is, e.g., J. J. Sakurai, Modern Quantum Mechanics, 2nd edition.

Maybe it helps to consider concrete examples. Suppose you have a single electron: It can be in two possible "spin states": $|u\rangle$ ("spin up") or $|d\rangle$ ("spin down"). Up and down are relative to the z-axis. If you measure its spin along the z-axis, you will get result "spin up" if the electron is in state $|u\rangle$, and you'll get result "spin down" if the electron is in state $|d\rangle$.

Now, what Landau is saying in the first section is that for any two numbers $\alpha$ and $\beta$ we can form another state $|\psi\rangle = \alpha |u\rangle + \beta |d\rangle$. That's a linear combination of $|u\rangle$ and $|d\rangle$. If an electron is in state $|\psi\rangle$ and you measure the spin in the z-direction, you'll either get "spin up" or "spin down". You won't get some third result that is half-way between "up" and "down".

So that's about states of a single electron. Now, suppose we have two electrons? Then it's perfectly possible for the first to be in state $|u\rangle$ while the second is in state $|d\rangle$. Then the two-electron system would be described by a "composite" state $|\Psi\rangle = |u\rangle |d\rangle$. (Actually, since electrons are indistinguishable, the two-particle state will be in a linear combination of the state $|u\rangle |d\rangle$ and $|d\rangle |u\rangle$, so that you can't say the first electron is in state $|u\rangle$ and the second is in state $|d\rangle$.)

So the first way of combining states---linear combinations---produces a new one-particle state from two different one-particle states. The second way of combining states---products--produces a two-particle from two one-particle states.
You absolutely hit the right spot, thank you so much!!
Sorry for being a pest <3

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Maybe it helps to consider concrete examples. Suppose you have a single electron: It can be in two possible "spin states": $|u\rangle$ ("spin up") or $|d\rangle$ ("spin down"). Up and down are relative to the z-axis. If you measure its spin along the z-axis, you will get result "spin up" if the electron is in state $|u\rangle$, and you'll get result "spin down" if the electron is in state $|d\rangle$.

Now, what Landau is saying in the first section is that for any two numbers $\alpha$ and $\beta$ we can form another state $|\psi\rangle = \alpha |u\rangle + \beta |d\rangle$. That's a linear combination of $|u\rangle$ and $|d\rangle$. If an electron is in state $|\psi\rangle$ and you measure the spin in the z-direction, you'll either get "spin up" or "spin down". You won't get some third result that is half-way between "up" and "down".

So that's about states of a single electron. Now, suppose we have two electrons? Then it's perfectly possible for the first to be in state $|u\rangle$ while the second is in state $|d\rangle$. Then the two-electron system would be described by a "composite" state $|\Psi\rangle = |u\rangle |d\rangle$. (Actually, since electrons are indistinguishable, the two-particle state will be in a linear combination of the state $|u\rangle |d\rangle$ and $|d\rangle |u\rangle$, so that you can't say the first electron is in state $|u\rangle$ and the second is in state $|d\rangle$.)

So the first way of combining states---linear combinations---produces a new one-particle state from two different one-particle states. The second way of combining states---products--produces a two-particle from two one-particle states.

So for a complete description of a two-electron system one can say that the most general form is:

$|\Psi\rangle = \alpha|u\rangle |d\rangle + \beta|d\rangle |u\rangle + \gamma|u\rangle |u\rangle + \delta|d\rangle |d\rangle$

stevendaryl
Staff Emeritus
So for a complete description of a two-electron system one can say that the most general form is:

$|\Psi\rangle = \alpha|u\rangle |d\rangle + \beta|d\rangle |u\rangle + \gamma|u\rangle |u\rangle + \delta|d\rangle |d\rangle$

Yes, that's correct. Except that for Fermions (such as electrons), the state has to be antisymmetric (change sign) under interchange of two identical particles. We've only mentioned the spin state; the total state also includes the spatial part, but the total state has to be antisymmetric for Fermions.

vanhees71
So that's about states of a single electron. Now, suppose we have two electrons? Then it's perfectly possible for the first to be in state |u⟩|u⟩|u\rangle while the second is in state |d⟩|d⟩|d\rangle. Then the two-electron system would be described by a "composite" state |Ψ⟩=|u⟩|d⟩|Ψ⟩=|u⟩|d⟩|\Psi\rangle = |u\rangle |d\rangle. (Actually, since electrons are indistinguishable, the two-particle state will be in a linear combination of the state |u⟩|d⟩|u⟩|d⟩|u\rangle |d\rangle and |d⟩|u⟩|d⟩|u⟩|d\rangle |u\rangle, so that you can't say the first electron is in state |u⟩|u⟩|u\rangle and the second is in state |d⟩|d⟩|d\rangle.)
I found the @vanhees71 post #4 satisfactory though more general than need be to obtain simplicity. However, now I'm confused.
If you have two electrons (a left one and a right one) and one is in state |u⟩ and the other in state |d⟩, but you don't know which, then you have a mixture of the two states |u⟩|d⟩ and |d⟩|u⟩. On the other hand a linear combination (superposition) of |u⟩|d⟩ and |d⟩|u⟩ is a single entangled state, not a (tensor) product of the two simple states.

Why isn't possible to know the one on the left is in state |u⟩ (it was prepared that way), while the one on the right is in state |d⟩, so we have |u⟩|d⟩?

vanhees71
Gold Member
The complete state of a single electron is determined by preparing (with some given accuracy) a complete set of compatible observables. For a non-relativistic electron you can choose position and the spin-##z## component (or any spin component, but usually one uses the ##z## component).

For two electrons, you have to take into account that (a) their positions and spin-##z## components form a set of compatible observables, which implies that the Hilbert space of the two-electron system can be built as the span of the (generalized) product states ##|\vec{x}_1,\sigma_1 \rangle \otimes |\vec{x}_2,\sigma_2 \rangle##, but (b) you have to take into account that electrons are indistinguishable particles and (since they have spin 1/2 according to the spin-statistics theorem of local relativistic QFT) they are fermions. Thus only the antisymmetrized basis vectors are a complete orthonormalized set of general eigenvectors, i.e., the two-electron Hilbert space is spanned by
$$\frac{1}{\sqrt{2}} (|\vec{x}_1,\sigma_1 \rangle \otimes |\vec{x}_2,\sigma_2 \rangle-|\vec{x}_2,\sigma_2 \rangle \otimes |\vec{x}_1,\sigma_1 \rangle.$$
If you know only the probabilities for finding the electrons in the four possible spin-##z## states ##(1/2,1/2)##, ##(1/2,-1/2)##, ##(-1/2,1/2)##, and ##(-1/2,-1/2)##, you should choose the maximum-entropy state given this information, in order not to impose any prejudice. This state is given by the statistical operator in two-electron spin space
$$\hat{\rho}=\sum_{\sigma_{i},\sigma_j=-1/2}^{1/2} p_{ij} |\sigma_i \rangle \otimes |\sigma_j \rangle \langle \sigma_i \rangle \otimes \langle \sigma_j|,$$
i.e., a mixed state.

Note that due to the spatial part of the above choosen basis, you can have all four two-electron spin states, although the total basis states must be antisymmetrized due to the fermionic nature of the electrons.