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I Two quantum states and qubits...

  1. Jan 10, 2017 #1
    Hello everyone,

    My understanding is that a two-quantum state system is simply a system that can only be in two states. That is equivalent to say that the observable of interest that is being considered can only have possible values. Is that the case?

    If so, a classical bit can have two values (either 1 or 0) while a quantum bit can have value 1, 0 or a superposition of both. The state with eigenvalue 1 is indicated by ##|1>## while the state with eigenvalue 0 by ##|0>##. The superposition state is give by $$|\Psi=c_1 |1> +c_2 |0>$$

    This two state system could be achieved using the spin (which has values 1/2 and -1/2) or even the energy observable if there are only two energy states available, correct? The system could be in a superposition of the two energy eigenstates...

    What is a system of two qubits? Is it a physical system with two particles? In the case of two qubits, what would the notation ##|11>## or ##|10>## indicate? Does ##|10>## indicate the one particle is in state 1 while the other is in state 0?
    The superposition state for 2 qubits appears to be

    $$|\Psi> = c_{11} |11> + c_{10} |10> + c_{01} |01> +c_{00} |00> $$
     
  2. jcsd
  3. Jan 10, 2017 #2

    PeterDonis

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    This is not correct. A correct statement would be that a "two-quantum state system", by which I assume you mean something like a qubit, has a Hilbert space that is two-dimensional. That means we can construct a basis for the Hilbert space with two states. But there are an infinite number of states in the Hilbert space, since all possible complex linear combinations of the two basis states are also valid states.

    Note also that whatever basis we construct will not be unique; there are an infinite number of possible pairs of basis states on this Hilbert space. Basically, if you take any given basis, you can construct another basis by taking any chosen complex linear combination of the two states in the given basis as state "0" of the new basis, and then constructing state "1" of the new basis by finding the complex linear combination of the original basis states that is orthogonal to your state "0" (there will be only one such state once state "0" is chosen). This is all assuming we are working with normalized states (so all complex linear combinations must have squared moduli that sum to 1).

    I assume you mean "two possible values". This is not quite correct either, for two reasons: first, there is not just one observable of interest, there are an infinite number (one corresponding to each of the infinite number of possible pairs of basis states); and two, the "values" in question are measurement results, so it would be more accurate (and less likely to cause confusion) to say that for each observable of interest, there are two possible measurement results.

    This is the sort of confusion that is avoided by stating things the way I did above, rather than the way you did. There is no difference in the measurement results you can get on a classical bit vs. a qubit. The difference is in the space of possible states.

    Assuming we are talking about a spin-1/2 system (a qubit), yes. There are other possible spins, but we probably shouldn't get into that.

    Yes, with the clarification that by "energy states" is really meant "eigenstates of the energy observable", or, in the language I used above, the "basis states" in the energy basis (which is just one of the infinite number of possible pairs of basis states).

    A system whose Hilbert space is the tensor product of two qubit Hilbert spaces. This Hilbert space will have four basis states, and any complex linear combination of the four basis states will also be a valid state.

    Two qubits. Qubits are not "particles" except as an approximation, which doesn't always apply.

    The usual usage of this notation is that the first number is the state of the first qubit, and the second number is the state of the second qubit (where the qubits are labeled by something such as spatial location to distinguish them). A less confusing notation would be ##|1>|1>## and ##|1>|0>## for the two states you give.

    If you adopt the basis you have adopted, yes. But that's not the only possible basis. See above.

    Note also that for two-qubit states, not all of the states you can write down in this notation are allowed. Qubits are fermions, so a two-qubit state must change sign under exchange of the two qubits. So, for example, the state ##|10>## is not an allowed state, but the state ##|10> - |01>## is allowed, since it changes sign under exchange (just switch the order of the numbers in each term). So the basis you adopted is not very useful, since none of your basis states are actually allowed.
     
  4. Jan 11, 2017 #3
    Thanks PeterDonis!

    Since you mention the tensor product, I started looking into it and how it works and why it is used. Before I get to that, let me ask you this question since you were so clear in your explanation:

    I know that the state of a physical system is ##\Psi>## lives in a single and unique linear Hilbert vector space. Each operator (observable) has its own basis of eigenvectors and the state ##\Psi>## can be written as a linear superposition of those eigenvectors. That makes it look like the state ##\Psi>## is in the same vector space as the basis vector of the chosen operator. We could do the same using the basis of any other hermitian operators. It seems that ##\Psi>## is now living in many different vector spaces at the same time since each operator, with its basis, defines a different vector space. So if we consider 3 operator, we would have three difference vector spaces constructed using the bases of each operator....Is that true? In which vector space does the vector state ##\Psi>## live? Does ##\Psi>## live in the tensor product of those multiple vector spaces? I am still confused.

    I read that even the simple wavefunction ##Psi(x,y,z)##, where x,y,z are the three independent variables, can be seen as the result of a tensor product. But the function ##\Psi(x,y,z)## will not necessarily be a product ##\Psi(x) \Psi(y) \Psi(z)##.

    thanks.
     
  5. Jan 11, 2017 #4

    PeterDonis

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    That's correct (with the one minor correction that it should be "basis vectors" plural).

    Yes.

    No. Remember that I said the Hilbert space has an infinite number of possible sets of basis vectors. All of those sets are vectors in the same space (the Hilbert space). Each set of basis vectors is, in principle, the set of eigenstates of some Hermitian operator; all of those operators are operators on the same space (again, the Hilbert space).

    No; the Hilbert space that the wave function "lives" in can be viewed as the tensor product of three "smaller" Hilbert spaces ("smaller" is in quotes because in this case, where the wave function is a continuous function of position, the Hilbert spaces are infinite dimensional, so "smaller" does not have its usual meaning of "fewer dimensions").

    That's correct. But the product here is not a tensor product; it's just ordinary multiplication of three functions, each of one variable, to obtain a function of three variables. All this is saying is that there will be many possible wave functions on the ##x, y, z## Hilbert space that are not simple products of a function of ##x## only, a function of ##y## only, and a function of ##z## only.
     
  6. Jan 11, 2017 #5
    Hi again and thanks.

    You say that "Each set of basis vectors is, in principle, the set of eigenstates of some Hermitian operator; all of those operators are operators on the same space (again, the Hilbert space)".

    From linear algebra, as you mention above, there are many possible bases in the same vector space to describe the same particular vector. But there the bases have (must have) have the same number of basis vectors. For example, vectors in a vector space with dimensionality=3 are decomposed using bases that ALL have 3 linearly independent vectors. In QM, if the vector space is the same, the bases associated to each operator should contain the same number of eigenvectors but they don't (spin has a two-dimensional basis, energy and position may have bases with multiple eigenstates...)
     
  7. Jan 11, 2017 #6

    PeterDonis

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    No, there are many possible bases in the same vector space, period. Once you have a basis, you can describe any vector in the vector space using that basis.

    Spin of a spin-1/2 particle has a two-dimensional basis. Spin of other types of particles can have more (or less--there are spin-0 particles as well, whose "spin state" is trivial, it's just the number 1). Energy and position of a free particle have bases with an infinite number of eigenstates. But energy of a particle in a bound state might have as few as two eigenstates--if it happens to be bound in an atom with only two energy levels, for example. (Yes, this is an idealized case; but there are practical situations where it is a good approximation.) You can't just look at the name of the observable and deduce from that the number of basis states; you have to look at the actual specifics of the system you are interested in.
     
  8. Jan 11, 2017 #7
    Great, I am on board with what you are saying: for a certain vector space there is an infinity of possible bases (each containing as many basis vectors as the dimension of the hosting vector space).

    At the same time, I have some QM notes that talk about a vector space which is the tensor product of the vector spaces associated to each individual (commuting) observable under consideration. Commuting observables form independent degrees of freedom. That seems to say that different operators have each a vector space associated to them and the total vector space is the tensor product of those spaces. That sounds slightly different from what your are saying, i.e. the operator bases are just different and possible bases inside the same vector space.

    From linear algebra, I know we can have intersection, sum and products between different vector spaces. The tensor product of two different vector spaces S and W gives a new vector space whose elements are matrices...
     
  9. Jan 11, 2017 #8

    PeroK

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    This is why you need maths, not words!
     
  10. Jan 11, 2017 #9

    PeterDonis

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    That's because you're now talking about something different. Let me illustrate with a concrete example.

    Suppose I have a qubit. This qubit has a Hilbert space which is two-dimensional. That means I need two orthogonal vectors to specify a basis.

    Here is one possible choice of basis for this Hilbert space: ##|+z>## and ##|-z>##, meaning "spin up in the z direction" and "spin down in the z direction". I can express any state in this Hilbert space as a complex linear combination of these two basis vectors. And these two basis vectors are eigenvectors of a Hermitian operator, ##S_z##, which means "measure the spin in the z direction".

    But I can also choose a different basis for this same Hilbert space; for example: ##|+x>## and ##|-x>##, meaning "spin up in the x direction" and "spin down in the x direction". I can also express any state in the Hilbert space as a complex linear combination of these two basis vectors. (Exercise: express the previous set of basis vectors, ##|+z>## and ##|-z>##, in terms of the new basis vectors ##|+x>## and ##|-x>##.) And these two basis vectors are eigenvectors of a different Hermitian operator, ##S_x##, which means "measure the spin in the x direction".

    Both of these sets of basis vectors, and their corresponding operators, are vectors/operators in the same Hilbert space. Everything I said above relates to that Hilbert space. No tensor products anywhere; no different Hilbert spaces; but an infinite number of possible sets of basis vectors and corresponding Hermitian operators (since I can pick any direction in space at all, not just the z or x directions, as the orientation to use to specify a basis and the associated Hermitian operator describing a spin measurement).

    Now suppose I have two qubits. This quantum system has a Hilbert space which is four-dimensional. That means I need four orthogonal vectors to specify a basis.

    One simple way to construct such a basis is to consider this four-dimensional Hilbert space as a tensor product of two one-qubit Hilbert spaces, i.e., as a tensor product of two copies of the two-dimensional Hilbert space I described above. Then we can just pick a basis for the first copy, and a basis for the second copy, and construct a basis for the four-dimensional Hilbert space by taking all possible pairs of basis vectors from the two copies. For example, suppose we pick the z direction for both copies; then our basis for the four-dimensional Hilbert space will be ##|+z, +z>##, ##|+z, -z>##, ##|-z, +z>##, ##|-z, -z>##. (Note that nothing requires us to pick the same direction for both copies. Exercise: what if we pick the z direction for the first copy and the x direction for the second? What will the four basis vectors be?)

    One nice thing about this basis is that each basis state is a simple product of basis states of the two copies of the two-dimensional Hilbert space. However, as noted before, it also has the flaw that the basis states themselves cannot be physically realized, since they are not antisymmetric under particle exchange. (Note that there are cases where this is actually not an issue; these are cases where this four-dimensional Hilbert space is not the complete Hilbert space of the system, but only a part of it. For example, in the H2 molecule, the spins of the two electrons form a quantum system described by this four-dimensional Hilbert space; but the spins alone don't have to be antisymmetric under particle exchange, because the total Hilbert space of the H2 molecule also includes a part for the positions of the electrons. But I'm leaving cases like that out of consideration here.) We can construct other sets of basis states that are antisymmetric under particle exchange, but they will not be simple products of basis states of the two copies of the two-dimensional Hilbert space. (Exercise: show that this is true and construct a specific example. Google "Bell states" if you want hints.)

    It is also true that each set of basis vectors for this four-dimensional Hilbert space has a corresponding Hermitian operator for which the basis vectors are eigenvectors. For example, the basis I described above, where we pick the z direction for both copies, obviously corresponds to the Hermitian operator ##S_z S_z##, which means "measure the spin of both qubits in the z direction". Note that, as with the basis vectors themselves, this Hermitian operator can be viewed as simply the product of operators on each of the two copies of the two-dimensional Hilbert space. This is the sense in which some (but not all! see below) Hermitian operators on a tensor product space can be viewed as products of operators on the spaces we took the tensor product of.

    For sets of basis vectors which cannot be expressed as simple products, the above will not be true; the Hermitian operators will be more complicated as regards their physical interpretation, and will not have an interpretation as a simple product (meaning they won't be viewable simply as "make measurement A on one qubit, and make measurement B on the other"), but they will still exist.
     
  11. Jan 11, 2017 #10
    Thanks for the patience, PeterDonis.
    I am starting to get this.

    What if we had just a single particle? That system (single particle) has several degrees of freedom. I have seen the wavefunction for that particle expressed as the product of a spatial wavefunction and a spin wavefunction: ##\Psi(x,y,z) \Phi(s)## and I think this is rooted in a tensor product and in the idea of independent degrees of freedom.

    Also, last night I run into this other thread, available at http://physics.stackexchange.com/qu...nsor-product-originate-in-the-history-quantum, where I read about the function ##\Psi(x,y,z)## involving a tensor product in some sense since the spatial variables are independent degrees of freedom...
     
  12. Jan 11, 2017 #11

    PeterDonis

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    If the physical situation is appropriate, yes. For example, a free particle in 3-dimensional space.

    Yes, this description takes the position Hilbert space and the spin Hilbert space and forms their tensor product to get the Hilbert space of the system. Here the issue with antisymmetry doesn't arise (because there's only one particle), so there won't be any vector in this Hilbert space that doesn't represent a physically realizable state.

    However, it is worth noting that even in this case, the position and spin degrees of freedom won't necessarily be "independent". They could be entangled, meaning that the system could be in a state that is not expressible as a product of a single position state and a single spin state. For example, a common device for measuring the spin of charged particles is called a Stern-Gerlach device; it puts the particle through a magnetic field oriented in a particular direction (the direction in which we want to measure the spin), and the particle will be deflected either "up" or "down" depending on its spin in that direction. So after passing through this device, the particle's position is entangled with its spin, and its state is not expressible as a product. (This is easier to see if we use the momentum basis rather than the position basis for the "position" part of the state; the direction of the momentum is what is entangled with the spin.)
     
  13. Jan 12, 2017 #12
    Hi,

    Thanks. I am familiar with the SG experiment and how two sub-beams are formed (one for each spin value). I never think of this as being an entangled situation. Usually, for entanglement I thing of two or more particles entangled together and the entanglement involves the same observable for the two particles. Based on what you say, two different observables for the same particle can be entangled together and we cannot express the wavefunction for that particle as a product....is that correct?

    If two observables don't commute, it means they are not independent from each other. Does dependence automatically imply entanglement?
     
  14. Jan 12, 2017 #13

    PeterDonis

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    That description is not often used, but it's correct.

    That's one kind of entanglement, but not the only one. The more general view is that different degrees of freedom can be entangled. See below.

    More precisely, two different degrees of freedom for the same particle can be entangled (in the case of the SG apparatus, it's the momentum and spin degrees of freedom). And in that case, yes, you cannot express the wave function for the particle as a product.
     
  15. Jan 12, 2017 #14

    PeterDonis

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    Why do you think that? What do you mean by "not independent"?

    Entanglement is not a concept that applies to observables. It applies to states. The state of a quantum system can be either entangled or not. If it can be expressed as a product, it's not entangled; if it can't, it is.
     
  16. Jan 12, 2017 #15

    Zafa Pi

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    That's not necessary. The tensor product S⊗W is a vector space of dimS•dimW, e.g. [a,b]⊗[c,d] = [ac,ad,bc,bd}. Notice that [1,0,0,1] is in the 4D tensor product space, but is not itself a tensor product of two 2D vectors, and in that case [1,0,0,1] is said to be an entangled state.
     
  17. Jan 21, 2017 #16
    Hello PeterDonis,

    If I may, let me ask you something I have been struggling with in the past months. I am trying to see the parallelism between concept in linear algebra and concepts in QM.
    In the linear algebra course I took, a vector is an entity that lives inside a vector space U together with an infinity of other vectors. With the vector space U we can find and infinity of possible basis sets. Each basis set is as good as any other one. Each basis set always contains as many vectors as the dimensionality of vector space U itself.

    That said, back to QM. The state A of a quantum system is represented also by a vector living in a Hilbert vector space U. Each operator has its own basis set. But the basis set for different operators can have different dimensionality from each other, i.e. contain a different number of eigenstates, correct? Do all the eigenvectors for each different operator live in the same Hilbert vector space U?

    Do the eigenvectors associated to each operator form subsets of vectors that are part of the entire Hilbert vector space U?

    thanks for any clarification.
     
  18. Jan 21, 2017 #17

    vanhees71

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    That's not entirely correct. The confusion comes from the sloppy use of definitions by physicists. In quantum theory you have a true Hilbert space, i.e., an infinite-dimensional Hilbert space. In usual quantum mechanics the Hilbert space is a separable Hilbert space, i.e., there exists a complete countable set of normalized orhogonal vectors, which form a basis (in the sense of convergence with respect to the norm induced by the scalar product of the Hilbert space). This is called a separable Hilbert space.

    Now physicists use also the word "basis" in a wider sense. It would be better to name it "generalized basis". Take as an example a spin-0 particle moving in 1 direction. Then usually one uses as a generalized "basis" the "eigenvectors" of position or momentum. Use the realization of the vectors as wave functions ##\psi(x)=\langle x | \psi \rangle##. Then the momentum operator is given by
    $$\hat{p} \psi(x)=-\mathrm{i} \hbar \partial_x \psi(x).$$
    First of all you should note that this operator is only defined on part of the Hilbert space, which is here realized by the square-integrable complex functions, i.e., ##\mathrm{L}_2(\mathbb{R},\mathbb{C})##.

    Now let's look for the eigenstates of the momentum operator. It's defined by
    $$\hat{p} u_p(x)=p u_p (x) \; \Rightarrow \; -\mathrm{i} \hbar \partial_x u_p(x)=p u_p(x),$$
    which has the solution
    $$u_p(x)=N \exp \left (\frac{\mathrm{i} p x}{\hbar} \right ), \quad N =\text{const}.$$
    Obviously, that's not a square integrable function and thus doesn't belong to the Hilbert space!

    It's rather a "generalized function" in the sense of a distribution. It's meaning is that you can express any wave square integrable function by the Fourier transformation
    $$\psi(x) = \int_{\mathbb{R}} \mathrm{d} p \tilde{\psi}(p) u_p(x).$$
    The most convenient say to choose the normalization factor of the generalized momentum eigenstates is such that it's "normalized to a Dirac ##\delta## distribution", i.e., such that
    $$\langle u_{p'}|u_p \rangle=\int_{\mathbb{R}} \mathrm{d} p u_{p'}^*(x) u_{p}(x)= |N|_2 \int_{\mathbb{R}} \mathrm{d}x \exp \left [\frac{\mathrm{i}}{\hbar} x (p-p') \right] = 2 \pi |N|^2 \delta \left (\frac{p-p'}{\hbar} \right ) = 2 \pi \hbar |N|^2 \delta(p-p').$$
    The most simple choice thus is ##N=1/\sqrt{2 \pi \hbar}##. Then you can Fourier transform back and force as if ##|u_p \rangle## is a usual Hilbert space basis but with a continuous label ##p## for these basis vectors. However, as the (mathematically of course not rigorous!) analysis above shows, it's in fact a generalized kind of a basis, it's rather a distribution than a true Hilbert space basis. So normalizing the momentum eigenvectors as written above, you have
    $$u_p(x)=\langle x|u_p \rangle = \frac{1}{\sqrt{2 \pi \hbar}} \exp \left (\frac{\mathrm{i} p x}{\hbar} \right), $$
    and the Fourier transform from the momentum to the position representation and back is given by
    $$\psi(x)=\langle x|\psi \rangle = \int_{\mathbb{R}} \mathrm{d} p \langle x|u_p \rangle \langle p|\psi \rangle = \int_{\mathbb{R}} \mathrm{d} p u_p(x) \tilde{\psi}(p)$$
    and
    $$\tilde{\psi}(p) = \label p|\psi \rangle = \int_{\mathbb{R}} \mathrm{d} x \langle p|x \rangle \langle x|\psi \rangle = \int_{\mathbb{R}} \mathrm{d} x u_p^*(x) \psi(x).$$
     
  19. Jan 21, 2017 #18
    Thanks vanhees71,

    So you are saying, just to paraphrase, that the Hilbert vector space the state lives in is a separable space. When I think of a separable function, I think of something that can be expressed as the product like ##f(x,y)= p(x) g(y)##.

    Several vector spaces can be added to create a new vector space or we can do a tensor product. What you are saying is that the complete Hilbert space is the tensor product of the vector spaces of each operator which are spanned by the basis or each operator.

    From a slightly intuitive point of view where we see vectors as these entities of a vector space, what does a vector that lives in the full Hilbert space have in common with the vectors that inhabit in the various vector spaces (of possibly different dimensionality) associated to each operator (observable)?
     
  20. Jan 21, 2017 #19

    vanhees71

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    A Hilbert space is called separable if there exists a countable orthonormal basis. There's only one separable Hilbert space (up to equivalence). In non-relativistic QT of a single particle moving in one dimension, e.g., you can take ##\mathrm{L}^2(\mathbb{R},\mathbb{C})## as this Hilbert space (position representation aka. wave mechanics). An example for a countable orthonormal basis are the energy-eigen states of the harmonic oscillator.
     
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